QI III: Entanglement applications

6.Entanglement applications

We focus on bipartite systems which are in pure states. From a QI point of view separable states are not interesting. If we have a separable pure state, we simply have two separate quantum systems, each with their own pure state. There is no quantum correlation between such systems so at most there are classical correlations. However, there are interesting effects when we have an entangled state. Then it turns out to be possible to do things which could not be done otherwise. Some examples we explore are:

Teleportation – The transmission of a quantum state using classical communication, but no quantum communication.
Quantum Key Distribution – The ability to share a secret random key, using classical and quantum communication, with no possibility that the key could be know by Eve.
Superdense coding – The ability to transmit 2 classical bits of information by sending just one qubit.

Note that entanglement implies a quantum correlation which is different from classical correlations. In particular even though we have a bipartite system, measurement in one subsystem can instantaneously affect the other. It is this “spooky action at a distance” which Einstein objected to.

However, to reiterate our discussion from above: note that this instantaneous effect does not imply faster than light communication. The main point is that say if Alice makes a measurement, although this can alter Bob's state the measurement process is random, so she cannot choose how Bob's state is altered. Furthermore, Bob cannot determine which measurement Alice made, or even if she made a measurement at all. This is because Bob's reduced density matrix is not changed by any unitary transformations of measurements made by Alice, it describes the same mixed state. If Alice tells Bob the result of a measurement, Bob can then learn something about his state which would change his mixed state (perhaps to a pure state). However, this process requires classical communication so there cannot be any transmission of information faster than the speed of light. Another way to say this is that if Alice and Bob both make a measurement, the probability distribution of their combined results does not depend on whether Alice measures first, Bob measure first or if they measure simultaneously.

A simple example of classical correlation is the following. Charlie gives Alice and Bob each a box with a coin in it. One coin is heads up, the other tails, and Alice and Bob both know this. Alice can open her box to find out whether her coin is heads or tails, and she would then know immediately that Bob's coin was the other, whether or not Bob had opened his box. Obviously the result is independent of whether or not Bob already opened his box, and also obviously Alice's measurement has no effect on Bob's coin.

A similar example of quantum correlation would be that Charlie gives Alice and Bob each a box containing a ‘quantum coin’, again with the guarantee that if they both measure to heads/tails, one of them will finds heads, the other tails. Now the difference to the classical case is that before measuring, neither coin is heads or tails, measuring changes the state to either heads or tails at random. Then if Alice measures heads, Bob's coin will instantaneously become tails (assuming Bob has not yet made a measurement.) Similarly if Bob measures first, his measurement will affect Alice's coin and her coin will instantly be heads if he found tails. However, in QM we do not have any mechanism to describe a signal travelling from one system to the other. Indeed, the measurements could be made simultaneously and they will still yield opposite results, yet we cannot claim that one affected the other ( In Special Relativity the concept here is spacelike separation – there is no possibility for any signal at or below the speed of light to travel from Alice to Bob or Bob to Alice to allow the result of one measurement to influence the other. In fact in relativity there is no concept of which measurement was made before the other if they were made at a spacelike separation.).

Now, a natural question is, can we really distinguish this quantum correlation from classical correlation? More generally, how can we be sure that the state is not either heads or tails before we measure? For any given process it is easy to come up with possible ways that the measurement result is pre-determined, i.e. where our uncertainty about the result of a measurement is simply due to our lack of knowledge about some “hidden variables”.

It sounds very difficult to argue against this possibility, but in fact Bell derived an inequality that must hold in any theory with the property of local realism. This means any theory where the result of any measurement is pre-determined (i.e. is determined by the state and does not involve any randomness) and local in the sense that no event can affect any other event unless some sort of signal travels (no faster than the speed of light) to communicate the first event to the second. It is easy to show that QM can violate Bell inequalities, and experiment has confirmed that they are violated in nature. This proves that no theory obeying local realism (i.e. no hidden variable theory) can be the correct description of nature. (This does not prove QM is correct, but QM is consistent with all experiments, and there is no known alternative.)

6.1.Bell States

Many of the features of entanglement can be explored in the simplest bipartite system, where each subsystem is a single qubit. The complete system is then a 2-qubit system, so the Hilbert space has dimension 4. We can use an orthonormal basis of separable states

\[ \Big\{ \ket{x} \otimes \ket{y} \; : \; x, y \in \{0, 1\} \Big\} \; . \]

In general, linear combinations of these states will be entangled. Given a specific state, one way to check if it is separable is simply to try and write it in the form \(\ket{\psi} \otimes \ket{\phi} = a\ket{0} \otimes \ket{\phi} + b\ket{1} \otimes \ket{\phi}\) for some \(a, b \in \mathbb{C}\) and some state \(\ket{\phi}\). If this is not possible, the state is entangled. Another way is to check if the reduced density matrix (of either subsystem) gives a pure state, meaning that the state is separable, or a mixed state, meaning that the state is entangled. Recall that the state is pure iff. \(\Tr(\rho^2) = \Tr(\rho) = 1\).

In fact \(\Tr(\rho_A^2) = \Tr(\rho_B^2)\) gives a measure of the entanglement, with maximum value \(1\) for no entanglement (i.e. a separable state) to the minimum value \(1/2\) (for a single qubit subsystem) for maximally entangled states.

For maximally entangled states, the trace of the reduced density matrix of both sides is \(1/2\).

The four Bell states

\[ \ket{\beta_{xy}} = \frac{1}{\sqrt{2}} \Big( \ket{0} \otimes \ket{y} + (-1)^x \ket{1} \otimes \ket{\overline{y}} \Big) \]

where \(\overline{y} = NOT \; y\) (defined by \(\overline{0} = 1\) and \(\overline{1} = 0\)) are maximally entangled, and also form an orthonormal basis.

Note that in terms of a 2-qubit system the Bell state basis is related to the standard basis by a unitary transformation, so these are entirely equivalent choices of basis states. However, for the bipartite system the Bell states cannot be created from the separable states by any LOCC process. This is because the required unitary transformations are not of the form \(\hat{U}_A \otimes \hat{U}_B\) as that would only transform a separable state to a separable state. Also, measurement by Alice and/or Bob would also result in a separable state. Note that a measurement by Alice or Bob on a Bell state can result in a separable state such as \(\ket{0} \otimes \ket{0}\) but of course measurement is not a reversible transformation.

On the other hand, it is possible for Alice or Bob to individually transform any Bell state to any other Bell state. Specifically, the unitary operators \(\hat{U}_{xy} \otimes \hat{I}\) (which Alice can use) and \(\hat{I} \otimes \hat{U}_{xy}\) (which Bob can use) transform the Bell state \(\ket{\beta_{00}}\) to the Bell state \(\ket{\beta_{xy}}\), i.e.

\[ \hat{U}_{xy} \otimes \hat{I} \ket{\beta_{00}} = \hat{I} \otimes \hat{U}_{xy} \ket{\beta_{00}} = \ket{\beta_{xy}}\,, \]

where the unitary operators are given in the standard single qubit matrix representation by

\begin{eqnarray*} U_{00} = \mathbb{I}_2 = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) & , & U_{01} = \sigma_1 = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \; , \\ U_{10} = \sigma_3 = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) & , & U_{11} = i\sigma_2 = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \; . \end{eqnarray*}

Note that when dealing with qubit, or multi-qubits systems we will often drop the hat from operators and use, unless differently specified, the basis \(\ket{0},\ket{1}\) represented as the two standard basis vectors \((1,0)^T\,,(0,1)^T\), hence \(\hat{I}\) for a qubit will become \(\mathbb{I}_2\) the \(2\times 2\) identity matrix, and in this basis we will directly use the Pauli matrices instead of having to define three abstract operators (usually called spin operators \(\hat{S}_x\,,\hat{S}_y\,,\hat{S}_z\) or sometimes \(\hat{J}_x\,,\hat{J}_y\,,\hat{J}_z\)) represented by these three matrices in this given basis.

Consider \(\hat{U}_{11}=i\sigma_2\), then we have

\[ \hat{U}_{11} \ket{0} = -\ket{1}\,,\qquad\hat{U}_{11}\ket{1} = \ket{0}\,. \]

Then if we consider

\begin{align*} \hat{U}_{11}\otimes \hat{I} \ket{\beta_{00}} &= \hat{U}_{11}\otimes \hat{I} \left(\frac{1}{\sqrt{2}} \ket{0}\otimes \ket{0}+\frac{1}{\sqrt{2}} \ket{1}\otimes \ket{1}\right) \\ &= \left(-\frac{1}{\sqrt{2} }\ket{1}\otimes \ket{0}+\frac{1}{\sqrt{2}} \ket{0}\otimes \ket{1}\right)=\ket{\beta_{11}}\,, \end{align*}

as expected.

6.2.Superdense Coding

This process is called superdense because we can use one qubit to transmit two bits of information. However, this is only possible at the cost of using a resource of entanglement corresponding to one Bell state. I.e. we cannot convey more than one bit of information using a single qubit unless the two parties already share an entangled state.

To see this, first note that it is obvious that a qubit can be used in place of a bit. E.g. simply send the state \(\ket{0}\) corresponding to the bit \(0\), or the state \(\ket{1}\) corresponding to the bit \(1\). The recipient just measures, corresponding to the operator \(\frac{1}{2}\left(\mathbb{I}_2- \sigma_3\right)\) to distinguish, with probability \(1\), these two orthogonal states

\[ \frac{1}{2}\left(\mathbb{I}_2- \sigma_3 \right)\ket{0} = 0 \ket{0}\,,\qquad\frac{1}{2}\left(\mathbb{I}_2- \sigma_3\right)\ket{1} = 1 \ket{1}\,, \]

i.e. the outcome of the measurement is with probability \(1\) the classical bit Alice wanted to send.

Note as well that Alice can indeed prepare the initial qubit to represent in this way the classical bit she wants to send to Bob. Suppose she wants to send the classical bit \(x\in\{0,1\}\), she can just prepare a random state \(\ket{\psi}\) and measure on it \(\frac{1}{2}\left(\mathbb{I}_2- \sigma_3\right)\). The result of this measurement can only be \(0\) or \(1\), if it is precisely the bit \(x\) she wants to send then she knows she has now projected \(\ket{\psi}\) to the right state \(\ket{x}\) and she sends it straight away.

If the result of the measurement is not the same as the bit she wanted to send then she knows she has projected the state onto \(\ket{\bar{x}}\) and she just needs to perform the unitary evolution \(\sigma_1 \ket{\bar{x}} = \ket{x}\) and then she can send the state to Bob.

Now, although a qubit contains two real numbers (as it corresponds to a point on a sphere, as we saw in the Bloch sphere picture), the recipient can only make one measurement which will give one of two possible values, and then the state will be transformed to one of two orthogonal states. (Which states depend on the choice of measurement operator.)

Now, suppose Alice and Bob share the Bell state \(\ket{\beta_{00}}\) and Alice wants to send the 2-digit binary number \((xy)_2\) to Bob. She can do this by transforming the system by acting on her qubit with the unitary operator \(\hat{U}_{xy}\) which transforms the whole system to the state \(\ket{\beta_{xy}}\). Note that this does not transmit any information to Bob, his reduced density matrix is \(\hat{\rho}_B = \frac{1}{2}\hat{I}\) before and after Alice's transformation.

Superdense coding relies on the fact that Alice can transform \(\ket{\beta_{00}}\) to any \(\ket{\beta_{xy}}\) using only a local unitary transformation on her part of the Hilbert space.

Alice can then send her qubit to Bob. Note that this qubit also contains no information on its own. This means that if say Eve intercepts it, she just has a qubit with reduced density matrix \(\frac{1}{2}\hat{I}\) completely independent of \(x\) and \(y\). However, assuming Bob receives the qubit from Alice, he will have the full Bell state \(\ket{\beta_{xy}}\). Since the four Bell states are orthogonal, he can distinguish them (with probability \(1\)) with a suitable measurement. This means any measurement operator which has the four Bell states as eigenstates (with distinct eigenvalues.) E.g. Bob will definitely measure the result \((xy)_2\) if he measures

\[ \hat{B} = 0\ket{\beta_{00}}\bra{\beta_{00}}+1\,\ket{\beta_{01}}\bra{\beta_{01}} + 2\, \ket{\beta_{10}}\bra{\beta_{10}} + 3 \,\ket{\beta_{11}}\bra{\beta_{11}} \; . \]

6.3.No-Cloning Theorem

Crucial to the claim above that one qubit could only be used to transmit one bit of information (without use of entanglement) was that fact that the recipient could only perform one measurement without destroying the original state. From our knowledge of measurement, this is obvious. However, suppose Bob received a qubit and then cloned it. He could then measure on the many copies to deduce (to any desired accuracy by making more copies) the probability distribution of the results of any measurement(s). E.g. doing this for the measurement \(\frac{1}{2}\left(\mathbb{I}_2- \sigma_3\right)\) would give an estimate of \(\cos^2(\theta/2)\) (defined on the Bloch sphere) as the probability of getting the result corresponding to \(\ket{0}\) rather that \(\ket{1}\). This would give an arbitrary number of bits of information as the binary representation of this number, for example \(\cos^2(\theta/2)=(0.1100010101...)_2\). However, the No-Cloning Theorem states that this is not possible.

In QM it is impossible to clone an unknown state \(\ket{\psi}\). I.e. we cannot transform \(\ket{\psi} \otimes \ket{\Omega} \rightarrow \ket{\psi} \otimes \ket{\psi}\) for arbitrary unknown \(\ket{\psi}\), where \(\ket{\Omega}\) is a fixed initial state.

Note, the attempt here is to create a quantum photocopier, taking any input \(\ket{\psi}\) and a blank sheet of paper \(\ket{\Omega}\) and making a perfect copy. Note that it is possible to transform \(\ket{\psi} \otimes \ket{\Omega} \rightarrow \ket{\phi} \otimes \ket{\psi}\) for arbitrary \(\ket{\psi}\) but the state \(\ket{\phi}\) will not (in general) be the state \(\ket{\psi}\), or even depend on the state \(\ket{\psi}\).

Note that measurement cannot help. Any measurement of an unknown state will give a result dependent on the choice of measurement, and the final state will be an eigenstate of the measurement operator. We can only deduce that the original state was not orthogonal to this final state. This means that such a quantum copier must use unitary evolution. However, we can prove by contradiction that there is no such unitary transformation. Alternatively we can prove that there is no such linear transformation. (Either is sufficient to prove the theorem.)

Linearity: Take any two linearly independent states \(\ket{\psi_1}\) and \(\ket{\psi_2}\) and assume we have a quantum copier. Also, let \(\ket{\psi} = a\ket{\psi_1} + b\ket{\psi_2}\) for any non-zero \(a, b \in \mathbb{C}\). Then the copier acts as

\begin{eqnarray*} \ket{\psi_1} \otimes \ket{\Omega} & \rightarrow & \ket{\psi_1} \otimes \ket{\psi_1} \\ \ket{\psi_2} \otimes \ket{\Omega} & \rightarrow & \ket{\psi_2} \otimes \ket{\psi_2} \\ \ket{\psi} \otimes \ket{\Omega} & \rightarrow & \ket{\psi} \otimes \ket{\psi} \end{eqnarray*}

but by linearity we must also have

\[ \ket{\psi} \otimes \ket{\Omega} = a\ket{\psi_1} \otimes \ket{\Omega} + b\ket{\psi_2} \otimes \ket{\Omega} \rightarrow a\ket{\psi_1} \otimes \ket{\psi_1} + b\ket{\psi_2} \otimes \ket{\psi_2} \ne \ket{\psi} \otimes \ket{\psi} \; . \]

Hence we have a contradiction.

Unitarity: The argument is similar. Take all states to be normalised and consider two states \(\ket{\psi_1}\) and \(\ket{\psi_2}\). The copier must act as

but since it is a unitary operation, inner products are preserved. This means that

\[ \left( \bra{\psi_1} \otimes \bra{\Omega} \right) \left( \ket{\psi_2} \otimes \ket{\Omega} \right) = \left( \bra{\psi_1} \otimes \bra{\psi_1} \right) \left( \ket{\psi_2} \otimes \ket{\psi_2} \right) \]

but this means that

\[ \ip{\psi_1}{\psi_2} = \left( \ip{\psi_1}{\psi_2} \right)^2 \; . \]

This is only possible if \(\ip{\psi_1}{\psi_2} = 1\) so \(\ket{\psi_1} = \ket{\psi_2}\) or \(\ip{\psi_1}{\psi_2} = 0\) so the states are orthogonal. So, again we see it is impossible to copy arbitrary unknown states.

Note that the above does not mean that we cannot clone an arbitrary state \(\ket{\psi}\). It is just that we must choose the right unitary transformation, depending on \(\ket{\psi}\) and obviously we cannot do that if we don't know anything about the state.

It is possible to write down a unitary operator which transforms

\begin{equation} \label{e:not-a-copier} U\big( \ket{n}\otimes\ket{0} \big) \rightarrow \ket{n}\otimes\ket{n}\,. \end{equation}

In the matrix notation, this operator is given by

\begin{equation} U = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\,. \end{equation}

But despite the fact that \eqref{e:not-a-copier} looks like a generic copier, the states on which it acts are not the most general one-qubit states. After all, the most general such state is

\begin{equation} \ket{\psi} = \alpha\ket{0} + \beta\ket{1}\,. \end{equation}

If you act with the \(U\) given above on this state \(\ket{\psi}\), you will find (try it!) that it does not produce \(\ket{\psi}\otimes\ket{\psi}\). So the copier only works for the basis states, not for linear combinations of them.

6.4.Teleportation

Teleportation is the process of transferring a quantum state, say from Alice to Bob, without any quantum communication. Surprisingly, although it is not possible to clone, it is still possible to teleport an unknown quantum state using only classical communication, although this does require the resource of entanglement. It is in some sense complementary to the process of superdense coding, where entanglement is again the resource. In both cases two bits correspond to one qubit, and we require one pair of maximally entangled qubits as the resource to convert in either direction.

Consider the simplest example where Alice has one unknown qubit state \(\ket{\psi} = a \ket{0} + b \ket{1}\). Note that even if she knew what the state was, in general she would need to communicate two real numbers (e.g. the two angles giving the position of the Bloch sphere) to Bob so he could create a copy. Obviously, to any reasonable accuracy, this would require many bits of information, and certainly just 2 bits would not be enough. However, this is not what happens in teleportation.

In the process below, Alice and Bob do not need to know anything about the state \(\ket{\psi}\), they do not learn anything about it in the teleportation process, and Bob does not create a copy as at the end Alice no longer has the state \(\ket{\psi}\).

The starting point is that Alice and Bob must share an entangled state. Assume they share the Bell state \(\ket{\beta_{00}}\) so the state of the whole system is

\[ \ket{\psi} \otimes \ket{\beta_{00}} = \frac{1}{\sqrt{2}}\ket{\psi} \otimes \ket{0} \otimes \ket{0} + \frac{1}{\sqrt{2}}\ket{\psi} \otimes \ket{1} \otimes \ket{1} \]

where Alice has the first two qubits, and Bob has the third.

Now Alice and Bob can use LOCC to teleport the state \(\ket{\psi}\), i.e. so that the final state of the system is \(\ket{\Phi} \otimes \ket{\psi}\), and Alice's final 2-qubit state \(\ket{\Phi}\) does not depend on \(\ket{\psi}\). Alice then no longer has the state \(\ket{\psi}\), but Bob does – it has been teleported.

The process is as follows.

Alice entangles \(\ket{\psi}= a\ket{0}+b\ket{1}\) with the other two qubits. She can do this with a unitary transformation on her two qubits since the second is already entangled with Bob's qubit.

First we act with the controlled-NOT (CNOT) gate given by the unitary operator transforming
\[ \ket{00} \rightarrow \ket{00} \; , \;\; \ket{01} \rightarrow \ket{01} \; , \;\; \ket{10} \rightarrow \ket{11} \; , \;\; \ket{11} \rightarrow \ket{10} \]
on Alice's two qubits producing the state
\begin{align*} \ket{\psi}\otimes\ket{\beta_{00}}&= \frac{1}{\sqrt{2}} \Big(a\ket{000} +a \ket{011}+b \ket{100}+b\ket{111} \Big) \\ \text{CNOT}\Big\downarrow& \\ \hat{U}_{\text{CNOT}}\otimes \hat{I} \ket{\psi}\otimes\ket{\beta_{00}} &= \frac{1}{\sqrt{2}} \Big( a\ket{000} + a\ket{011} + b\ket{110} + b\ket{101} \Big) \; . \end{align*}

Passing to the standard basis representation the CNOT gate is given by the \(4\times 4\) matrix
\begin{equation} U_{\text{CNOT}}= \left(\begin{matrix}1 & 0 & 0 & 0\\ 0& 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{matrix}\right) = \left(\frac{\mathbb{I}_2+\sigma_3}{2}\right)\otimes \mathbb{I}_2 + \left(\frac{\mathbb{I}_2-\sigma_3}{2}\right)\otimes \sigma_1\,, \end{equation}
where we note that \(U_{\text{CNOT}}\) is not of the form \(A\otimes B\) (see discussion in section 5.3) which is the reason why the unitary operator \(U_{\text{CNOT}}\otimes \mathbb{I}_2\) entangles Alice two qubits and leaves Bob qubit invariant.
If she then measures her two qubits, Bob's qubit state will depend on the result of her measurement. This can be done so that Bob's qubit is a unitary transformation of the state \(\ket{\psi}\) that we want to teleport. (To clarify this, we present the measurement process as a unitary transformation followed by measurement to distinguish the standard basis states. Instead Alice could simply measure to distinguish the states \(\ket{\pm} \otimes \ket{y}\).)

First Alice acts on the first qubit with the unitary operator mapping \(\ket{0} \rightarrow \ket{+}\) and \(\ket{1} \rightarrow \ket{-}\) called the Hadamard gate. This is just given by the unitary operator (change of basis matrix)
\[ U_H = \frac{1}{\sqrt{2}} \left( \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix}\right)\,, \]
which clearly implements
\[ \hat{U}_H \ket{0} = \ket{+}\,,\qquad \hat{U}_H\ket{1} = \ket{-}\,. \]

So after this unitary evolution we have
\begin{align*} & \hat{U}_H\otimes \hat{I}\otimes \hat{I} \frac{1}{\sqrt{2}} \Big( a\ket{000} + a\ket{011} + b\ket{110} + b\ket{101} \Big) =\\ &= \frac{1}{2} \Big[ a \left(\ket{0}+\ket{1}\right)\otimes \left(\ket{00}+\ket{11}\right) + b\left( \ket{0}-\ket{1}\right) \otimes \left( \ket{10} + \ket{01} \right) \Big] =\\ &= \frac{1}{2} \Big[ \ket{00}\otimes \left(a \ket{0}+b\ket{1}\right) + \ket{01}\otimes \left(a \ket{1}+b\ket{0} \right)+\\ &\phantom{=22}+ \ket{10}\otimes \left(a \ket{0}-b\ket{1} \right)+ \ket{11}\otimes \left(a \ket{1}-b\ket{0} \right)\Big]\,, \end{align*}
where we used all linearities properties of tensor products.

We can rewrite this expression as the state
\[ \frac{1}{2} \sum_{x, y} \ket{x} \otimes \ket{y} \otimes \hat{U}_{xy}\ket{\psi} \; , \]
where \(\hat{U}_{xy}\) are the unitary transformations defined above when we introduced the Bell states
\[ \hat{U}_{00} = \mathbb{I}_2\,,\qquad \hat{U}_{01} =\sigma_1\,,\qquad\hat{U}_{10}=\sigma_3\,,\qquad\hat{U}_{11}=i\sigma_2\,. \]

Alice will now measure her two qubits (e.g. measuring \((\mathbb{I}_2-\sigma_3)/2\) for each of them) and she will get one of the four results \((xy)_2\) with \(x,y\in\{0,1\}\), which means that Bob's state will then be \(\hat{U}_{xy}\ket{\psi}\). After Alice measures \((\mathbb{I}_2-\sigma_3)/2\) for both her qubits with outcomes \(x,y\) entanglement is destroyed and we are left with the separable state \(\ket{x}\otimes \ket{y}\otimes \hat{U}_{xy} \ket{\psi}\).
Bob can then recover the state \(\ket{\psi}\) by a unitary transformation of his qubit, but which one depends on the result of Alice's measurement so she must communicate which of the four possible results she got – this requires 2 bits of classical communication. Therefore Alice sends the two bits of information giving the values of \(x\) and \(y\), and Bob then acts on his state \(\hat{U}_{xy}\ket{\psi}\) with \(\hat{U}_{xy}^{-1} = \hat{U}_{xy}^{\dagger}\) so his state is then \(\ket{\psi}\). I.e. we have teleported the state \(\ket{\psi}\) from Alice lab to Bob's at the price of using the Bell pair shared by Alice and Bob.

Note that we have not cloned \(\ket{\psi}\)! Alice does not have this state any longer, she is left with the two qubits \(\ket{x}\otimes \ket{y}\), which are independent of \(\ket{\psi}\). Note also that neither Alice nor Bob need to know what the teleported state \(\ket{\psi}\) is at all and with this protocol they have not gained any additional information about it.

Both with teleportation and superdense coding we have achieved something classically impossible, however, in both protocols we have expended some resources: for teleportation we have destroyed the Bell state by performing a measurement thus destroying entanglement and producing a separable state, while in superdense coding Alice had to give away her qubit and the Bell pair is not shared anymore.

6.5.Quantum Key Distribution

This is an application using QM to ensure that Alice and Bob can produce a secret shared random key. Such a key can then be used for absolutely secure communication. Specifically, QKD allows them to generate a random string of bits so that they will both know the value of each bit, and can be sure no one else does. Of course, they could do this without any quantum theory by meeting but this may not be convenient and they would have to know in advance how long the key should be, and they would have to keep the key secure until they needed to use it. Instead, QKD allows them to do this using quantum communication

Note that the purpose is secure classical communication, but QKD itself does not communicate any information. Instead, QKD allows Alice and Bob to share a key which Alice can use to encrypt a message. The encrypted message can then be transmitted (using classical communication) to Bob. He can then use the same key to decrypt the message. The point is that even if the transmission is intercepted, the encrypted message does not provide any information about the actual message without knowledge of the key.

To see how this works consider a message \(M\) represented as a binary number (note that this can be done in any other basis) of \(n\) bits and a random secret key \(K\) also \(n\) bits long shared by Alice and Bob. Since the encryption works independently on each bit, consider a bit \(m\) of the message and the corresponding bit \(k\) in the key. We do not make any assumptions about the value of \(m\), but we assume that \(k\) has values \(0\) and \(1\) with equal probability (and the value is known only by Alice and Bob.) Then the mechanism is as follows:

Alice produces the encrypted message \(C = M \oplus K\) where \(\oplus\) means bitwise addition modulo 2, i.e. for each bit \(c = m \oplus k\) which is just \(m + k\) except \(1 \oplus 1 = 0\). This is also sometimes noted as the XOR operation (or exclusive OR): \((True\, XOR\, False) = (False\, XOR \,\,True) = True\) while \((False\, XOR\, False) = (True\, XOR\,\, True) = False\).
Alice transmits the encrypted message \(C\) to Bob. If Eve intercepts this, she gains no information about \(M\) since for each bit \(c\) there is equal probability that this is the same as \(m\) or different from \(m\). Since there are only two possible values for each bit, \(C\) is completely random and independent of \(M\) without knowledge of the key \(K\).
Bob decodes the message since by calculating \(C \oplus K\) since by modular arithmetic
\[ C \oplus K = (M \oplus K) \oplus K = M \oplus (K \oplus K) = M \oplus 0 = M \; . \]

Note that it is vital that the key is as long as the message and it is not reused. E.g. if the same bit \(k\) is used for two bits \(m_1\) and \(m_2\) of the message the property that either \(m_1 = m_2\) or that \(m_1 \ne m_2\) is unchanged for \(c_1\) and \(c_2\). A key shorter than the message would make the encryption susceptible to attacks.

Note that for a key as long as the message this encryption method is not susceptible to any sort of attack. One cannot apply any frequency analysis because all the characters are equally likely.

However this method has an obvious drawback! Requiring such a new key is a significant practical obstacle to using this encryption method. This is why this method was called One-Time Pad (OTP) as the key can be used only once and then it has to be destroyed. Furthermore the key cannot be transmitted in clear and required to be physically transported (by a spy wearing a fake mustache carrying an inconspicuous briefcase handcuffed to their wrist) from the sender to the recipient, or the be agreed before-hand, for example by using a pre decided book or the daily newspaper.

Standard encryption methods (such as the ones you use daily on the internet) instead rely on trapdoor functions. These are functions where even with full knowledge of the function \(f\) itself, it is very difficult to find the inverse function \(f^{-1}\), but if you know \(f^{-1}\) it is easy to find \(f\). In this case Bob can choose such a function \(f^{-1}\) and then tell Alice what \(f\) is. She then sends \(C = f(M)\) and Bob decrypts it using \(f^{-1}(C) = M\). This is very convenient since there is no need to share a secret key. The disadvantage is that it is possible to find \(f^{-1}\) with knowledge of \(f\). Normally this is fine since either the information in the message is of no use after a certain time, or we assume that it is not worth the significant effort to find it. On the other hand, in some instances it may be important that the information is never found by a third party, and for such applications trapdoor functions are not secure.

So the main problem of OTP is the transmission of the encryption key, how does QM help exchange a secret random key without Alice and Bob having to meet? There are many different protocols, but one is the BB84 Protocol:

Alice makes a random choice \(0\) or \(1\) (with equal probability) and also a random choice \(X\) or \(Z\). According to her choices (which she records), she prepares a single qubit state
\[ (0, Z) \rightarrow \ket{0} \; , \;\; (1, Z) \rightarrow \ket{1} \; , \;\; (0, X) \rightarrow \ket{+} \; , \;\; (1, X) \rightarrow \ket{-} \; . \]
She sends this qubit to Bob (using quantum communication).
Bob receives the qubit and randomly chooses \(Z\), in which case he measures \(\frac{1}{2}( \mathbb{I}_2 - \sigma_3)\), or \(X\), in which case he measures \(\frac{1}{2}( \mathbb{I}_2 - \sigma_1)\) (remember what these operators mean on the Bloch sphere).
Alice and Bob repeat this process as often as required to generate a sufficiently long key.
Alice and Bob announce (publicly, so no secrecy is assumed) their choices of \(X\) or \(Z\) for each qubit and discard all results where they did not make the same choice. Where they made the same choice, Bob's measurement is guaranteed to agree with Alice's choice of \(0\) or \(1\). This generates the random shared key. (If it's not long enough, they can repeat to extend the key.)

Why is this secure? Suppose Eve intercepts a qubit. What can she do before forwarding it to Bob? To learn something she must measure. However, since the four possible states are not all orthogonal, Eve cannot make a measurement which will distinguish them with certainty. E.g. suppose Eve chose to measure \(\frac{1}{2}( \mathbb{I}_2 - \sigma_3)\). If Alice had chosen \(Z\), Eve would get the result \(0\) or \(1\) matching Alice's key, and Eve would forward the qubit unchanged to Bob. However, with equal probability Alice would have chosen \(X\), in which case for either \(\ket{\pm}\), Eve would get a random result \(0\) or \(1\) and forward correspondingly a random qubit \(\ket{0}\) or \(\ket{1}\) to Bob. If Bob chose to measure \(X\) this result would be discarded (as Alice chose \(Z\)) but if Bob chose \(Z\), he would get the random result of Eve's measurement, so half the time this would differ from Alice's key.

Alice choice	Eve Choice	Bob Choice	Results
Z	Z	Z	A and B and E measures all match: BAD
Z	Z	X	Discarded
Z	X	X	Discarded
Z	X	Z	In \(50\%\) of this case E measures match A.
			In \(50\%\) of this case B measures does not match A.

Figure 6.0: List of possible cases. Similar if Alice chooses X.

The above means that overall, for each bit of the shared key which Eve has intercepted and measured, there is a \(25\%\) chance that Alice and Bob will not have the same value for their keys as schematically depicted in Table tab:table1. If Alice and Bob do nothing, Eve will know \(75\%\) of the key which could be disastrous. However, if Alice and Bob compare a random subset of their keys, they can estimate the error rate. If it is too high, they will assume interference and discard the key, hoping to repeat the process. If they get a low enough error rate, they can assume the worst case scenario that all the errors are due to Eve. However, they can reduce the key in a way which makes it highly unlikely Eve will have any information about it.

Note that there are many other ways to exchange a key. E.g. if Alice and Bob share Bell states such as \(\ket{\beta_{00}}\) they could simply both measure \(\frac{1}{2}( \mathbb{I}_2 - \sigma_3)\), getting the same random result. This is sufficient if they already share enough entangled qubits. However, if Alice prepares the Bell states and sends one qubit to Bob, Eve could simply also measure \(\frac{1}{2}( \mathbb{I}_2 - \sigma_3)\) then forward the state to Bob. Eve would then learn the full key without being detected. To circumvent this, Alice and Bob can employ the same strategy as above, as provided they both choose \(X\) or both choose \(Z\), they will get the same results. Again Eve will be detected with probability \(25\%\) for each bit. Note that here, and above, it is essential that Alice and Bob do not announce any choice of \(X\) or \(Z\) before they are sure Bob has received the qubit.

6.6.Bell Inequalities

17 & 18

Einstein Podolsky and Rosen were not particularly happy with the non-local (“spooky action at a distance”) and probabilistic (“God does not play dice”) nature of quantum mechanics. So they postulated that quantum mechanics must be incomplete! Any “complete” theory must satisfy the postulate of local realism, i.e.

Locality: no faster than light influences;
Realism: measurements must be deterministic, i.e. the results tell us a property of the system.

This lead to the notion of hidden variables theories. According to EPR there must be some extra parameters not included in quantum mechanics that if we were to measure would give us a fully deterministic world. Quantum mechanics then only seems probabilistic because we lack the knowledge about these hidden variables.

Short Qiskit video on the Bell inequality published after the 2022 Nobel Prize was given to Clauser, Aspect and Zeilinger for their related work.

However Bell showed that any such theory with local realism must satisfy a certain type of inequality while quantum mechanics violate this! There are different inequalities which are all generally called Bell Inequalities. We consider a specific version known as the CHSH (Clauser, Horne, Shimony & Holt) Inequality.

In all cases the feature is that a result is derived which must be obeyed by any theory satisfying certain conditions such as locality (an effect at one point can be detected at another point only if something travels between these points) and some definition of realism meaning that measurements always reveal a property of the system (i.e. the system really had that property whether we measured or not.)

Roughly speaking, classical physics should obey these conditions whereas QM does not seem to. The question is to make this precise, and Bell inequalities allow us to demonstrate that QM does not satisfy these conditions, and so no classical hidden variable theory can reproduce all predictions of QM. The huge advantage of this approach is that we don't have to rule out candidate hidden variable theories one by one.

6.6.1.CHSH Bell-Inequality

Suppose we have a system with four observables \(Q\), \(R\), \(S\) and \(T\) which each can take only the values \(\pm 1\). The realism property tells us that any state of this system must have specific values for these four observables, i.e. \((q,r,s,t)\) .

Take a large number of states of this system (which can have different values of these observables) and measure the quantity \(QS + RS + QT - RT\) for each such state and calculate the average, i.e. the expectation value \(E(QS + RS + QT - RT)\). Now, due to the restricted values for each observable \(Q = \pm R\), so on each state when we measure we either have \(Q+R = 0\) and \(Q-R = \pm 2\) or \(Q+R = \pm 2\) and \(Q-R = 0\). Hence either \((Q+R)S = 0\) and \((Q-R)T = \pm 2\) or \((Q+R)S = \pm 2\) and \((Q-R)T = 0\), so \(QS + RS + QT - RT\) can only take the values \(\pm 2\). Obviously taking the average we must have

\begin{multline} -2 \le E(QS + RS + QT - RT) = \\[1ex] E(QS) + E(RS) + E(QT) - E(RT) \le 2 \; . \end{multline}

Note that the above argument used realism since we assumed that each state of this system had definite values of all 4 observables, i.e. we can really assign values to both \(Q\) and \(R\) for each state.

Now, consider the following EPR experiment: we have a system where Alice and Bob are separated (as far as we want) and Charlie is exactly in the middle.

Charlie will prepare lots and lots of Bell states \(\ket{\beta_{11}}\) and send one qubit of each simultaneously to Alice and Bob so that they receive them at exactly the same time.
On receiving each qubit Alice will make a random choice of \(Q\) or \(R\) and immediately measure the qubit, and similarly Bob will measure randomly either \(S\) or \(T\). Assuming locality, this setup excludes the possibility that Alice or Bob's measurement can affect the other via something travelling at finite speed ( Knowing the time taken to measure, the distance between Alice and Bob give a lower bound on the speed of any propagation. It is certainly possible to ensure that any such propagation must be faster than the speed of light, and so violate relativity.).
If our quantum mechanics were really a local realism theory we would have that Alice and Bob results are predetermined by some hidden variable that describe the Bell state sent out by Charlie.
For each qubit Alice and Bob record their choice of measurement and the result. This way they can later compare and calculate \(E(QS)\), \(E(RS)\), \(E(QT)\) and \(E(RT)\).

Note that each Bell state is only contributing to one of those expectation values, so Alice and Bob never really measure \(QS + RS + QT - RT\) for any individual state. However, by the assumption of realism, measuring say \(E(QS)\) on a random subset of states will give a good estimate (subject only to statistical errors) of \(E(QS)\) averaged over all states.

Specifically, we can take measurements using matrices (in the standard representation)

\[Q = \sigma_1 \otimes \mathbb{I}_2\,,\qquad R = \sigma_3 \otimes \mathbb{I}_2\]

which Alice can measure using only her qubit, and

\[S = \mathbb{I}_2 \otimes \frac{-1}{\sqrt{2}} (\sigma_1 + \sigma_3)\,,\qquad T = \mathbb{I}_2 \otimes \frac{-1}{\sqrt{2}} (\sigma_1 - \sigma_3) \]

which Bob can measure.

Note that since \([Q,R]\neq 0\) and \([S,T]\neq 0\) we cannot simultaneously measure both \(Q\) and \(R\), or similarly \(S\) and \(T\). Hower both Alice and Bob only measure one and then we just consider the average over many copies of the same state.

Using QM we predict that the outcome of each measurement is either \(\pm 1\) (Check the eigenvalues of all these operators) and that for the Bell state \(\ket{\beta_{11}}\) we can calculate

\begin{align*} E(QS) = \ev{QS} &= \ipop{\beta_{11}}{QS}{\beta_{11}} = \bra{\beta_{11}} \left(\sigma_1 \otimes \mathbb{I}_2\right)\left( \mathbb{I}_2 \otimes \frac{-1}{\sqrt{2}} (\sigma_1 + \sigma_3)\right) \ket{\beta_{11}}\\ &=-\frac{1}{2\sqrt{2}} \left(\bra{01}-\bra{10}\right) \sigma_1 \otimes (\sigma_1 + \sigma_3)\left(\ket{01}-\ket{10}\right)\\ & = -\frac{1}{2\sqrt{2}} \left(\bra{01}-\bra{10}\right) \mathbb{I}_2 \otimes (\sigma_1 + \sigma_3)\left(\ket{11}-\ket{00}\right)\\ & = -\frac{1}{2\sqrt{2}} \left(\bra{01}-\bra{10}\right)\left(\ket{10}-\ket{01}-\ket{11}-\ket{00}\right)\\ &=- \frac{1}{2\sqrt{2}} \left(-1-1\right) =\frac{1}{\sqrt{2}}\,, \end{align*}

and similarly (exercise)

\[ E(RS) = E(QT) = -E(RT) = \frac{1}{\sqrt{2}} \; . \]

However, the combination

\[ E(QS) + E(RS) + E(QT) - E(RT) = \frac{4}{\sqrt{2}} = 2\sqrt{2} \gt{} 2 \]

so this clearly violates the Bell (CHSH) inequality.

Experiments have confirmed this violation of the Bell inequality, so nature is not described by a theory obeying local realism. Nature is consistent with QM.

However there are loopholes to this argument. In particular there is something called super-determinism where everything is predetermined, including the “randomness” of Alice and Bob choices of the observable to measure. There is no such thing as free will and all our choices and actions are dictated by a fully deterministic world. Whenever one wants to understand the foundations of quantum mechanics inevitably the discussion will turn philosophical and for us it is a good point to stop.

6.7.Problems

2
Consider the operators
\[ \hat{N}_1 \equiv \sigma_1 \otimes \hat{I} \;\; \mathrm{and} \;\; \hat{N}_2 \equiv \sigma_3 \otimes \sigma_1 \]
acting on a 2-qubit system.
1. Write \(\hat{N}_1\) and \(\hat{N}_2\) as \(4 \times 4\) matrices in the representation where the standard basis states \(\ket{m} \otimes \ket{n}\) are written as 4-component column vectors with all components zero except a \(1\) in the row counted by the 2-digit binary number \((mn)_2\), e.g.
  \(\ket{1} \otimes \ket{0} \rightarrow (0 \; 0 \; 1 \; 0)^T\).
2. Write the operators \(\hat{N}_{+} \equiv \hat{N}_1 + \hat{N}_2\) and \(\hat{N}_{\times} \equiv \hat{N}_1 \hat{N}_2\) in matrix form. Explain why the structure of the \(4 \times 4\) matrices shows that \(\hat{N}_{\times}\) can be written in the form \(\hat{A} \otimes \hat{B}\) but \(\hat{N}_{+}\) cannot.
3. Show that \(\hat{U} \equiv \frac{1}{\sqrt{2}}\hat{N}_{+}\) and \(\hat{N}_{\times}\) are unitary operators.
4. Show that \(\hat{U}\) acting on the \(4\) basis states \(\ket{m} \otimes \ket{n}\) produces the \(4\) Bell states
  \[ \ket{\beta_{xy}} \equiv \frac{1}{\sqrt{2}} \left( \ket{0} \otimes \ket{y} + (-1)^x \ket{1} \otimes \ket{\overline{y}} \right) \]
  and show that none of the Bell states is a separable state.
5. Find the 4 states produced by \(\hat{U}\) acting on the \(4\) states \(\ket{\pm} \otimes \ket{\pm}\) where
  \(\ket{\pm} \equiv \frac{1}{\sqrt{2}} ( \ket{0} \pm \ket{1} )\). Are any of the resulting states separable?
  
  If any are separable, can you explain why? (Hint: look at \(\hat{N}_1\) and \(\hat{N}_2\) acting on these states.)
Solution:
▶
1. In this representation standard matrix multiplication shows that the columns of the matrices are (left to right) the results of the action of the basis vectors corresponding to the binary numbers \(00, 01, 10, 11\) in that order. Also recall that \(\sigma_1\) acts as \(\ket{0} \rightarrow \ket{1}\) and \(\ket{1} \rightarrow \ket{0}\) while \(\sigma_3\) acts as \(\ket{0} \rightarrow \ket{0}\) and \(\ket{1} \rightarrow -\ket{1}\). E.g. for \(\hat{N}_1\) we map \(\ket{0} \otimes \ket{0} \rightarrow \ket{1} \otimes \ket{0}\) so the first column of \(\hat{N}_1\) is given by the vector for \(\ket{1} \otimes \ket{0}\), i.e. \((0 \; 0 \; 1 \; 0)^T\). Repeating for the other basis states, and doing the same for \(\hat{N}_2\) we find
  \[ N_1 = \left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) \; , \;\; N_2 = \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{array} \right) \; . \]
  Note the structure is of \(2 \times 2\) blocks given by the form of the first operator in the tensor product (i.e. \(\sigma_1\) for \(\hat{N}_1\)) and the structure of each individual block (itself a \(2 \times 2\) matrix) is given by the second operator in the tensor product i.e. \(I\) for \(\hat{N}_1\)).
2. Just add and multiply the matrices to find
  \[ N_{+} = \left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \end{array} \right) \; , \;\; N_{\times} = \left( \begin{array}{cccc} 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right) \; . \]
  We see that the blocks in \(N_{+}\) are not of the same form (some are proportional to \(\sigma_1\), others to \(I\)) so it cannot be written as a single tensor product. For \(N_{\times}\) all the blocks are proportional to \(\sigma_1\) so it can be written as \(-i\sigma_2 \otimes \sigma_1\).
3. Just check that the corresponding matrices are unitary.
4. Just write the 4 Bell states in vector form and note that these are the 4 columns of the matrix \(N_{+}\). Note that the values of \(x\) and \(y\) are \(x=m\) and \(y=\overline{n}\) for \(m=0\) while \(y=n\) for \(m=1\).
  
  Any separable state can be written in the form \(\ket{\psi} \otimes \ket{\phi}\). Since we can always write \(\ket{\psi} = \ket{0} + b\ket{1}\) for a qubit system, we see that any separable state must be of the form \(a\ket{0} \otimes \ket{\phi} + b\ket{1} \otimes \ket{\phi}\) but in the Bell states, since \(\ket{y}\) and \(\ket{\overline{y}}\) are linearly independent, we clearly cannot have both \(a\ket{\phi} = \ket{y}\) and \(b\ket{\phi} = (-1)^x\ket{\overline{y}}\).
5. Either write these states as a linear combination of the standard basis states (and we know how \(\hat{U}\) acts on these states from the above parts) or just calculate the action of \(\hat{N}_1\) and \(\hat{N}_2\) directly. In particular note that \(\sigma_3 \ket{\pm} = \ket{\mp}\) while \(\sigma_1 \ket{\pm} = \pm \ket{\pm}\). This means that if we denote the \(\pm\) signs by \(s_1 = \pm 1\) and \(s_2 = \pm 1\) we have
  \[ \hat{N}_1 \ket{s_1} \otimes \ket{s_2} = s_1 \ket{s_1} \otimes \ket{s_2} \; , \;\; \hat{N}_2 \ket{s_1} \otimes \ket{s_2} = s_2 \ket{-s_1} \otimes \ket{s_2} \; . \]
  Obviously the second factor in the tensor product is the same in both cases so the linear combination corresponding to the action of \(\hat{U}\) is the separable state
  \[ \frac{1}{\sqrt{2}} \Big( s_1 \ket{s_1} + s_2 \ket{-s_1} \Big) \otimes \ket{s_2} \; . \]
  Note that if \(s_1 = s_2\) the first factor is proportional to \(\ket{0}\) while if \(s_1 = -s_2\) it is proportional to \(\ket{1}\).
3
For a 2-qubit system we define the 4 Bell states in terms of the standard basis states as:
\[ \ket{\beta_{xy}} \equiv \frac{1}{\sqrt{2}} \left( \ket{0} \otimes \ket{y} + (-1)^x \ket{1} \otimes \ket{\overline{y}} \right) . \]
1. Write the Bell states in the basis \(\big\{ \ket{++}, \ket{+-}, \ket{-+}, \ket{--} \big\}\).
2. Write the Bell states in the basis \(\big\{ \ket{LL}, \ket{LR}, \ket{RL}, \ket{RR} \big\}\).
3. If Alice and Bob share the Bell state \(\ket{\beta_{00}}\) (one qubit each) what are the possible outcomes and the associated probabilities, and what are Alice and Bob's final states in each case:
  1. Alice measures \(\sigma_3\).
  2. Bob measures \(\sigma_3\).
  3. Alice measures \(\sigma_1\).
  4. Alice measures \(\sigma_2\).
  5. Alice measures \(\sigma_1\) and then Bob measures \(\sigma_3\).
  6. Alice measures \(\sigma_1\) after Bob measures \(\sigma_3\).
  7. Alice measures \(\sigma_1\) and then Bob measures \(\sigma_1\).
Solution:
▶
1. Just replace the basis states using
  \[ \ket{0} = \frac{1}{\sqrt{2}} \Big( \ket{+} + \ket{-} \Big) \; , \;\; \ket{1} = \frac{1}{\sqrt{2}} \Big( \ket{+} - \ket{-} \Big) \; . \]
  For example, using notation \(\ket{01} = \ket{0} \otimes \ket{1}\) etc.
  \[ \ket{00} = \frac{1}{2} \Big( \ket{+} + \ket{-} \Big) \otimes \Big( \ket{+} + \ket{-} \Big) = \frac{1}{2} \Big( \ket{++} + \ket{+-} + \ket{-+} + \ket{--} \Big) \; . \]
  similarly we find
  \begin{eqnarray*} \ket{01} & = & \frac{1}{2} \Big( \ket{++} - \ket{+-} + \ket{-+} - \ket{--} \Big) \\ \ket{10} & = & \frac{1}{2} \Big( \ket{++} + \ket{+-} - \ket{-+} - \ket{--} \Big) \\ \ket{11} & = & \frac{1}{2} \Big( \ket{++} - \ket{+-} - \ket{-+} + \ket{--} \Big) \; . \end{eqnarray*}
  Then we just take the appropriate linear combinations to find
  \begin{eqnarray*} \ket{\beta_{00}} & = & \frac{1}{\sqrt{2}} \Big( \ket{++} + \ket{--} \Big) \\ \ket{\beta_{01}} & = & \frac{1}{\sqrt{2}} \Big( \ket{++} - \ket{--} \Big) \\ \ket{\beta_{10}} & = & \frac{1}{\sqrt{2}} \Big( \ket{+-} + \ket{-+} \Big) \\ \ket{\beta_{11}} & = & \frac{1}{\sqrt{2}} \Big( -\ket{+-} + \ket{-+} \Big) \; . \end{eqnarray*}
2. This part is essentially the same as above, using
  \[ \ket{0} = \frac{1}{\sqrt{2}} \Big( \ket{L} + \ket{R} \Big) \; , \;\; \ket{1} = \frac{-i}{\sqrt{2}} \Big( \ket{L} - \ket{R} \Big) \; . \]
  The result is
  \begin{eqnarray*} \ket{\beta_{00}} & = & \frac{1}{\sqrt{2}} \Big( \ket{LR} + \ket{RL} \Big) \\ \ket{\beta_{01}} & = & \frac{-i}{\sqrt{2}} \Big( \ket{LL} - \ket{RR} \Big) \\ \ket{\beta_{10}} & = & \frac{1}{\sqrt{2}} \Big( \ket{LL} + \ket{RR} \Big) \\ \ket{\beta_{11}} & = & \frac{-i}{\sqrt{2}} \Big( -\ket{LR} + \ket{RL} \Big) \; . \end{eqnarray*}
3. Noting that the eigenstates of \(\sigma_1\), \(\sigma_2\), \(\sigma_3\) are respectively \(\ket{+}\), \(\ket{L}\), \(\ket{0}\) with eigenvalue \(1\) and \(\ket{-}\), \(\ket{R}\), \(\ket{1}\) with eigenvalue \(-1\), the above results can be used:
  1. With probability \(1/2\) in each case, Alice measures \(1\) and the state becomes \(\ket{00}\) or \(-1\) and the state becomes \(\ket{11}\).
  2. Exactly as above, with probability \(1/2\) in each case, Bob measures \(1\) and the state becomes \(\ket{00}\) or \(-1\) and the state becomes \(\ket{11}\).
  3. Now noting that
    \[ \ket{\beta_{00}} = \frac{1}{\sqrt{2}} \Big( \ket{++} + \ket{--} \Big) \]
    again with probability \(1/2\) in each case, Alice measures \(1\) and the state becomes \(\ket{++}\) or \(-1\) and the state becomes \(\ket{--}\).
  4. This time noting that
    \[ \ket{\beta_{00}} = \frac{1}{\sqrt{2}} \Big( \ket{LR} + \ket{RL} \Big) \]
    again with probability \(1/2\) in each case, Alice measures \(1\) and the state becomes \(\ket{LR}\) or \(-1\) and the state becomes \(\ket{RL}\).
  For parts (v) and (vi) use the results from parts (iii) and (ii) respectively. Note that in both cases, after the first measurement the state is a separable state, so for (v) Bob's measurement does not change Alice's state (which is either \(\ket{+}\) or \(\ket{-}\)), and for (vi) Alice's measurement does not change Bob's state (\(\ket{0}\) or \(\ket{1}\)). Also, if Bob has the state \(\ket{+}\) or \(\ket{-}\) and measures \(\sigma_3\), in either case he will get results \(1\) or \(-1\) with probability \(1/2\) and his state will become \(\ket{0}\) or \(\ket{1}\) respectively. Similarly, Alice gets \(1\) or \(-1\), producing state \(\ket{+}\) or \(\ket{-}\) respectively, with probability \(1/2\) when measuring \(\sigma_1\) on state \(\ket{0}\) or \(\ket{1}\). The final result is that for (v) or (vi) the results are exactly the same (as should be expected since the measurements correspond to the commuting operators \(\sigma_1 \otimes I\) and \(I \otimes \sigma_3\)). With probability \(1/4\) we have the following results and final states:
  \[ (1, 1) \; , \; \ket{+0} \; ; \;\; (1, -1) \; , \; \ket{+1} \; ; \;\; (-1, 1) \; , \; \ket{-0} \; ; \;\; (-1, -1) \; , \; -\ket{-1} \; . \]
  
  Part (vii) follows immediately from part (iii). Bob will always get the same result as Alice so with probability \(1/2\) the combined results and final states are :
  \[ (1, 1) \; , \; \ket{++} \; ; \;\; (-1, -1) \; , \; \ket{--} \; . \]
Charlie knows that Alice and Bob are meeting for lunch on Saturday. He wants to send them a message with a surprise announcement that they can read together at that time, and not before. The problem is, he will be travelling and unable to communicate with them that day. However, he will see Alice on Thursday and Bob on Friday before he leaves, so he decides to give them each half the message with instructions to combine the information over lunch on Saturday. Suppose Charlie needs 100 bits for his message.
1. At first Charlie decides to give Alice half the bits, say the odd ones (i.e.
  the first, the third, the fifth, etc.) and to give Bob the even ones. However, he soon realises that this won't work since Alice and Bob will probably be able to guess some of the message from their 50 bits. After some thought he realises that he can avoid this problem by giving them each 100 bits. How might this work?
2. Just before Alice arrives on Thursday, Charlie realises the flaw in his plan. Alice and Bob are both impatient so will just send each other the information before Saturday! Then (what luck!) he remembers he has 50 Bell states. He checks his 100-bit message and performs some unitary transformations of the Bell states. When Alice leaves he gives her 50 qubits and some instructions, and the next day gives the remaining 50 qubits to Bob.
  
  Assuming Alice and Bob cannot meet before Saturday, and they do not have a quantum communication channel, explain how Charlie's plan might work.
Solution:
▶
1. One way is that the message can be found by combining Alice and Bob's 100 bits using bitwise addition modulo \(2\). I.e. each bit of the message will be a \(0\) if Alice and Bob's corresponding bits are the same, and a \(1\) if they are different. Obviously Alice or Bob alone cannot deduce anything about the message since for each bit they have it is equally likely that the other will have the same or different value for that bit.
2. He can use the \(50\) Bell states to encode \(50\) pairs of bits. As in superdense coding, if Alice and Bob have one qubit of each Bell pair, they cannot determine anything about which of the four Bell states it is. Of course, when they meet they can combine the qubits and measure which Bell states they have, generating the \(100 = 50 \times 2\) bit message.
  
  Note that this is not necessarily a flawless mechanism to keep the message secret. E.g. if Alice and Bob each measure \(\sigma_3\) and discuss their results (using classical communication), they can determine half the message. This is because they will get the same result for \(\ket{\beta_{00}}\) or \(\ket{\beta_{10}}\), but different results for \(\ket{\beta_{01}}\) or \(\ket{\beta_{11}}\). Of course, once they have done this they will never be able to recover the full message, so it is not quite the same as the case where they each have 50 classical bits of the message.
Alice needs to urgently send a secret message to Bob. For simplicity, assume she can do this with just 2 bits. They would like to do this using superdense coding, but they don't share any entangled states and to make matters worse someone ordered the wrong equipment – Alice's quantum entangler is broken, as is her quantum receiver and Bob's quantum transmitter. Fortunately, they each share a Bell state with Charlie whom they trust. So, their plan is for Alice to send the message to Charlie using superdense coding, and for Charlie to read it and transmit it to Bob, again using superdense coding. Then, they get news that Eve is spying on Charlie (it is vital that she does not see the message) and also blocking all his quantum communications. All seems lost, but actually there is a way for Alice to send the message to Bob using superdense coding with Charlie's help. Explain how this can be done in a way that does not require Charlie to use quantum communication, and so that Eve gains no information about the message from eavesdropping on any classical communications, or watching what Charlie does.

Solution:
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You can picture this in terms of a line connecting Alice to Charlie, and another line connecting Bob to Charlie. Each line represents a shared Bell state. Effectively, since there is a line from Alice to Charlie to Bob, it is possible for Charlie to manipulate the two qubits he has so that Alice and Bob share a Bell state. This can be done using teleportation. E.g. Alice can transform the Bell pair she shares according to her 2-bit message (the first part of superdense coding), Charlie can teleport the qubit he shares with Alice to Bob (using classical communication which does not reveal any details of the state being teleported), Alice can send her qubit to Bob (quantum communication). Finally Bob will have the Bell state chosen by Alice so can measure to read the 2-bit message. Of course Alice must transform the Bell state before she sends her qubit to Bob, but otherwise the order of the three steps does not matter.

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1.	Introduction
2.	Quantum mechanics essentials
3.	Measurement and uncertainty
4.	Qubits and the Bloch sphere
5.	Bipartite systems
6.	Entanglement applications
6.1.	Bell States
6.2.	Superdense Coding
6.3.	No-Cloning Theorem
6.4.	Teleportation
6.5.	Quantum Key Distribution
6.6.	Bell Inequalities
6.7.	Problems
7.	Information theory
8.	Changelog
9.	Bibliography