We may also allow Alice and Bob to communicate through classical
channels, i.e. Classical Communications (CC). This means that
they can send classical bits to each other. This includes speaking on
the phone or sending texts and emails, but we often want to quantify
the amount of information exchanged so we count the number of bits
transferred. (This counting does not include any pre-agreed protocol
which is required to interpret the data transferred. I.e. if Alice
sends a single bit to Bob with value \(1\), we assume he knows whether
this means “I measured X and got value 1 rather than \(0\)” or “I
decided to perform unitary transformation \(U\) rather than \(V\)”.)
Together if Alice and Bob can both perform LO and communicate
classically, we say they can perform LOCC operations.
An additional possibility is that Alice and Bob can communicate
through quantum channels. This means that they can send qubits to each
other. This is more powerful than classical communication. In fact
with unlimited capacity for quantum communication, they can perform
arbitrary operations on the whole system. A simple argument is that
Alice could just send her whole quantum system to Bob. He could then
do anything on the full system and then send subsystem A back to
Alice. However, we can focus on specific issues such as what can be
done by sending a single qubit compared to a single bit.
5.1.Tensor Products
When we say that we consider a system with Hilbert space \(\mathcal{H}\)
which can be separated (partitioned) into two (or more) subsystems in
mathematical language we are saying that our Hilbert space can be
written as a tensor product of two (or more) Hilbert spaces and
we will write it as \(\mathcal{H} = \mathcal{H}_A\otimes
\mathcal{H}_B\).
The tensor product of two vector spaces is a new vector space and this
is a construction that could be discussed in a purely linear algebra
setup without ever mentioning quantum mechanics similar to what you
have seen when discussing the direct sum \(\mathcal{H}_A \oplus
\mathcal{H}_B\).
In this module we will not be needing the most general definition of
tensor product so we will just start from the vectors of
\(\mathcal{H}_A \) and \(\mathcal{H}_B\) to construct the vectors of this
new vector space \(\mathcal{H}_A \otimes \mathcal{H}_B\) and understand
how the usual linearity properties work in this extended space.
With this in mind let us denote \(\ket{\psi_1} \in \mathcal{H}_A\) and
\(\ket{\psi_2} \in \mathcal{H}_B\) then the vector \(\ket{\psi_1} \otimes
\ket{\psi_2} \in \mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B\).
We can write a basis \(\{ \ket{i} \otimes \ket{m} \}\) for \(\mathcal{H}\)
in terms of a basis \(\{\ket{i}\}\) for \(\mathcal{H}_A\) and a basis
\(\{\ket{m}\}\) for \(\mathcal{H}_B\). Just by counting how many basis
elements we have if \(\mathcal{H}_A\) has dimension \(N_A\) and
\(\mathcal{H}_B\) has dimension \(N_B\) we have that \(\mathcal{H}\) has
dimension \(N_A N_B\).
The tensor product has various linear properties
\(c\, \Big(\! \ket{\psi } \otimes \ket{\phi} \Big) = \Big(
c\ket{\psi } \Big) \otimes \ket{\phi} = \ket{\psi} \otimes \Big(
c\ket{\phi} \Big)\)
\(a \ket{\psi_1} \otimes \ket{\phi} + b \ket{\psi_2} \otimes
\ket{\phi} = \Big( a \ket{\psi_1} + b \ket{\psi_2} \Big) \otimes
\ket{\phi}\)
\(a \ket{\psi } \otimes \ket{\phi_1} + b \ket{\psi} \otimes
\ket{\phi_2} = \ket{\psi} \otimes \Big( a \ket{\phi_1} + b
\ket{\phi_2} \Big) \)
Note in particular that in order to simplify the sum of two tensors we must have that either the first vector is the same or the second one is, if we find an expression of the form \( \ket{\psi_1 } \otimes \ket{\phi_1}+ \ket{\psi_2 } \otimes \ket{\phi_2}\) we have to leave it like that.
The inner products of \(\mathcal{H}_A\) and \(\mathcal{H}_B\) induce an inner product on the tensor product space so that the inner product of \(\ket{\psi_1} \otimes \ket{\phi_1}\) with
\(\ket{\psi_2} \otimes \ket{\phi_2}\) in \(\mathcal{H}\) is defined by the inner products in
\(\mathcal{H}_A\) and \(\mathcal{H}_B\) as:
\[ \Big( \bra{\psi_1} \otimes \bra{\phi_1} \Big) \Big( \ket{\psi_2} \otimes \ket{\phi_2} \Big) = \ip{\psi_1}{\psi_2} \,\ip{\phi_1}{\phi_2} . \]
States of a bipartite system which can be written as
\(\ket{\psi}\otimes\ket{\phi}\) are
separable, all others are
entangled.
If a pure state can be written in the form \(\ket{\psi} \otimes
\ket{\phi}\) we say it is a separable state. However, a typical
state is a linear combination of such states and is not separable. We
say a non-separable state is entangled.
For example suppose that the basis \(\{\ket{i}\}\) for \(\mathcal{H}_A\)
is orthonormal as well as the basis \(\{\ket{m}\}\) for \(\mathcal{H}_B\),
then the basis \(\{ \ket{i} \otimes \ket{m} \}\) for \(\mathcal{H}\) will
also be orthonormal. We can then write a separable vector
\[
\ket{\psi}\otimes \ket{\phi} = \left(\sum_i a_i \ket{i}\right) \otimes \left(\sum_m b_m \ket{m}\right) = \sum_{i,m} a_i b_m \ket{i}\otimes \ket{m}\,.
\]
However the most general vector \(\ket{\Psi}\in \mathcal{H}\) will be written as
The components of a bipartite state \(\ket{\Psi}\) on a basis
\(\{ \ket{i}\otimes\ket{m}\}\) are two-index coefficients \(c_{im}\) (tensor components).
\[\ket{\Psi} = \sum_{i,m} c_{im} \ket{i}\otimes \ket{m}\,,
\]
for some complex numbers \( c_{im}\in\mathbb{C}\) which generically cannot be written as \( a_i b_m\). We will see later on a quantity that will tell us immediately if a state is entangled or separable.
If we have the orthonormal basis \(\{ \ket{i} \otimes \ket{m} \}\) the
inner product between two general states can then be computed as
\begin{align*}
\ip{\Phi}{\Psi} &= \left( \sum_{j,n} d_{jn}^* \bra{j}\otimes \bra{n}\right) \left( \sum_{i,m} c_{im} \ket{i}\otimes \ket{m}\right)\\
& = \sum_{i,j,m,n} d_{jn}^* c_{im} \ip{j}{i}\ip{n}{m} = \sum_{i,m} d_{im}^* c_{im}\,,
\end{align*}
where we used the orthonormality of the \(\ket{i}\) and \(\ket{m}\).
The inner product between two states of a bipartite system, expressed in components, equals \(\sum_{i,m} d_{im}^* c_{im}\).
We can construct mixed ensemble of separable and entangled states however we need first to understand how linear operators work on tensor product spaces (see next section).
Once we have the tensor product of two vector spaces \(\mathcal{H}_A\otimes \mathcal{H}_B\) we can consider more and more tensors, i.e. \(\mathcal{H}_A\otimes \mathcal{H}_B \otimes \mathcal{H}_C\).
For example the Hilbert space of a \(N\)-qubits system is the \(2^N\) dimensional Hilbert space
\[
H_{\text{N-qubits}} = \mathcal{H}_q^{\otimes ^N}\,,
\]
where we simply mean the tensor product \( \mathcal{H}_q \otimes \mathcal{H}_q \otimes ...\otimes \mathcal{H}_q\) of \(N\) copies of a single qubit system.
Quantum computing will be the analysis of algorithms performed on the space of \(N\)-qubits seen as quantum circuits built out of gates (unitary operators) acting on a single qubit state or on pairs of qubits, analogues of the classical logical gates NOT, AND, OR, XOR acting on usual strings of bits, e.g. \(01101000\,01101001\),... .
Consider the space of \(3\)-qubits
\[\mathcal{H} = \mathcal{H}_q\otimes\mathcal{H}_q\otimes\mathcal{H}_q = \mbox{span} \{ \ket{000},\ket{001},\ket{010},\ket{011},\ket{100},\ket{101},\ket{110},\ket{111}\}\]
where we used the shorthand notation \( \ket{000} = \ket{0}\otimes \ket{0}\otimes \ket{0}\) and so on.
The operator \(\hat{I}\otimes \sigma_1\otimes \hat{I}\) will act on the second qubit and leave the first two qubits invariant, furthermore we know that \( \sigma_1\ket{0} = \ket{1}\) and \(\sigma_1\ket{1} =\ket{0}\) hence \(\sigma_1\) acts on this basis as the usual NOT logical gate where \(\mbox{NOT} \,0 = \bar{0}=1\) and \(\mbox{NOT} \,1 =\bar{1} = 0\).
We can then write the action of this operator on the general \(3\)-qubit state \(\ket{xyz}\) with \(x,y,z\in\{0,1\}\) as
\[
\hat{I}\otimes \sigma_1\otimes \hat{I} \ket{xyz} = \ket{x\bar{y}z}\,.
\]
Similarly an operator of the form \(\sum_J \hat{A}_J \otimes \hat{I}\otimes \hat{B}_J\) will act both on the first and third qubit while leaving the second qubit unchanged.
5.2.Linear Operators and Local Unitary Operations
Linear operators on \(\mathcal{H}\) can be written as linear combinations of
operators of the form \(\hat{A} \otimes \hat{B}\). (As for separable states, a
general linear operator cannot be written in this way as a single tensor product.) Such operators act on separable
states as
\[ \hat{A} \otimes \hat{B} \ket{\psi} \otimes \ket{\phi} = \left( \hat{A} \ket{\psi} \right) \otimes \left( \hat{B} \ket{\phi} \right) . \]
By linearity this defines the action of a general linear operator on an arbitrary state in \(\mathcal{H}\).
As for addition of tensor product of vectors, also for operators we cannot simplify \(\hat{A}\otimes \hat{B} + \hat{C}\otimes \hat{D}\) in general, however if either the first factors are the same, or the second factors are, we have
\[
\hat{A}\otimes \hat{B}+\hat{C}\otimes \hat{B} = ( \hat{A}+\hat{C})\otimes \hat{B}\,,\qquad \hat{A}\otimes \hat{B}+\hat{A}\otimes \hat{D} = \hat{A}\otimes (\hat{B}+\hat{D})\,.
\]
Note that standard operation or properties of the operators \(\hat{A}\) and \(\hat{B}\) descend to their tensor product, for example
\begin{align*}
(\hat{A}\otimes \hat{B}) ^\dagger &= \hat{A}^\dagger \otimes \hat{B}^\dagger\,,\\
(\hat{A}\otimes \hat{B})(\hat{C}\otimes \hat{D}) &= (\hat{A}\hat{C}\otimes \hat{B}\hat{D})\,,\\
\Tr_{\mathcal{H}_A\otimes \mathcal{H}_B} (\hat{A}\otimes \hat{B}) &= \Tr_{\mathcal{H}_A} ( \hat{A})\,\Tr_{\mathcal{H}_B} ( \hat{B})\,,
\end{align*}
or
the tensor product of two unitary operators is unitary;
the tensor product of two hermitian operators is hermitian;
the tensor product of two positive operators is positive;
the tensor product ot two projectors is a projector.
If we have a bipartite system and consider only local unitary operations then
Alice and Bob can each perform only very restricted unitary transformations of
the form \(\hat{U}_A \otimes \hat{I}\) for Alice and \(\hat{I} \otimes \hat{U}_B\)
for Bob. Note that these two operators commute
\begin{align*}
[\hat{U}_A \otimes \hat{I}, \hat{I} \otimes \hat{U}_B] &=\left(\hat{U}_A \otimes \hat{I}\right)\left( \hat{I} \otimes \hat{U}_B\right)- \left( \hat{I} \otimes \hat{U}_B\right)\left(\hat{U}_A \otimes \hat{I}\right)\\
&= \left( \hat{U}_A \hat{I}\right) \otimes\left( \hat{I} \,\hat{U}_B\right)-\left(\hat{I}\, \hat{U}_A \right) \otimes\left( \hat{U}_B\hat{I} \right) \\
&=\hat{U}_A\otimes \hat{U}_B -\hat{U}_A\otimes \hat{U}_B = 0\,,
\end{align*}
Local operations are of the form \(\hat{U}_A\otimes \hat{U}_B\), while general unitary operators on the bipartite Hilbert space cannot be simplified to this form.
so Alice and Bob independently
transform their own systems, and their product is
\(\hat{U}_A \otimes \hat{U}_B\) which can be seen easily from
\begin{align*}
(\hat{I} \otimes \hat{U}_B)(\hat{U}_A \otimes \hat{I})\ket{\psi}\otimes \ket{\phi} &= (\hat{I} \otimes \hat{U}_B) (\hat{U}_A \ket{\psi})\otimes (\hat{I}\ket{\phi}) = (\hat{I} \hat{U}_A \ket{\psi}) \otimes (\hat{U}_B \hat{I} \ket{\phi})\\
&=(\hat{U}_A \ket{\psi}) \otimes (\hat{U}_B \ket{\phi }) = (\hat{U}_A \otimes \hat{U}_B) \ket{\psi}\otimes \ket{\phi}\,,
\end{align*}
so we see that together Alice and Bob can only
transform the system by unitary operators of this restricted form.
Local operations cannot turn a separable state into an
entangled state.
One very important result is that if Alice and Bob start with a
separable state \(\ket{\psi} \otimes \ket{\phi}\) then the most general
unitary transformation they can perform using LO will produce another
separable state, \(\hat{U}_A \ket{\psi} \otimes \hat{U}_B
\ket{\phi}\). I.e. they cannot create an entangled state from a
separable state.
Let us compute the exponential of the operators \(\hat{A}\otimes
\hat{I}\) and \(\hat{I}\otimes \hat{B}\). We have
\begin{align*}
e^{\hat{A}\otimes \hat{I}} &=\sum_{n=0}^\infty \frac{(\hat{A}\otimes \hat{I})^n}{n!} = \sum_{n=0}^\infty \frac{ \hat{A}^n \otimes \hat{I}^n}{n!} = e^{\hat{A}}\otimes \hat{I}\,,\\
e^{\hat{I}\otimes \hat{B} } &=\sum_{n=0}^\infty \frac{(\hat{I}\otimes \hat{B} )^n}{n!} = \sum_{n=0}^\infty \frac{ \hat{I}^n \otimes \hat{B}^n }{n!} = \hat{I}\otimes e^{\hat{B}}\,.
\end{align*}
Hence we have
\[
e^{\hat{A}\otimes \hat{I}} e^{\hat{I}\otimes \hat{B} } = \left(e^{\hat{A}}\otimes \hat{I}\right)\,\left(\hat{I}\otimes e^{\hat{B}}\right)=e^{\hat{A}}\otimes e^{\hat{B}}\,,
\]
in particular notice that for general operators \(\hat{A}\,,\hat{B}\) the exponential of \(\hat{A}\otimes \hat{B}\), i.e. \(e^{\hat{A}\otimes \hat{B}}\) is different from \(e^{\hat{A}} \otimes e^{\hat{B}}\) since
\begin{align*}
e^{\hat{A}\otimes \hat{B}} &= \sum_{n=0}^\infty \frac{(\hat{A}\otimes \hat{B})^n}{n!}= \sum_{n=0}^\infty \frac{\hat{A}^n \otimes \hat{B}^n}{n!}\,,\\
e^{\hat{A}}\otimes e^{\hat{B}} &= \left(\sum_{n_1=0}^\infty \frac{\hat{A}^{n_1}}{n_1!}\right) \left(\sum_{n_2=0}^\infty \frac{\hat{B}^{n_2}}{n_2!}\right) = \sum_{n_1,n_2=0}^\infty \frac{\hat{A}^{n_1}\otimes \hat{B}^{n_2}}{n_1!\,n_2!}\,.
\end{align*}
Compute the commutator of \(\hat{A_1}\otimes \hat{B}\) with \(\hat{A_2}\otimes \hat{I}\). We have
\begin{align*}
\left(\hat{A}_1\otimes \hat{B}\right) \left( \hat{A}_2\otimes \hat{I}\right) &= \left(\hat{A}_1\hat{A}_2\right)\otimes \hat{B}\,,\\
\left(\hat{A}_2\otimes \hat{I}\right) \left(\hat{A}_1\otimes \hat{B}\right) &= \left(\hat{A}_2\hat{A}_1\right)\otimes \hat{B}\,,\\
\left[ \hat{A}_1\otimes \hat{B},\hat{A}_2\otimes \hat{I}\right] &=\left(\hat{A}_1\otimes \hat{B}\right) \left(\hat{A}_2\otimes \hat{I}\right)-\left(\hat{A}_2\otimes \hat{I}\right) \left(\hat{A}_1\otimes \hat{B}\right)\\
& = \left(\hat{A}_1\hat{A}_2\right)\otimes \hat{B}-\left(\hat{A}_2\hat{A}_1\right)\otimes \hat{B} \\
&= \left[\hat{A}_1,\hat{A}_2\right] \otimes \hat{B}\,.
\end{align*}
Similarly we can easily obtain
\[
\left[\hat{A}\otimes \hat{B}_1,\hat{I}\otimes \hat{B}_2\right] = \hat{A}\otimes \left[\hat{B}_1,\hat{B}_2\right]\,.
\]
Note in particular that \(\left[\hat{A}_1\otimes \hat{B}_1,\hat{A}_2\otimes \hat{B}_2 \right]\) is in general different from \( \left[\hat{A}_1,\hat{A}_2 \right]\otimes \left[\hat{B}_1,\hat{B}_2\right]\) as we can see by expanding everything out
\begin{align*}
\left[\hat{A}_1\otimes \hat{B}_1,\hat{A}_2\otimes \hat{B}_2 \right] & =\left(\hat{A}_1\otimes \hat{B}_1\right) \left(\hat{A}_2\otimes \hat{B}_2\right) -\left(\hat{A}_2\otimes \hat{B}_2\right) \left(\hat{A}_1\otimes \hat{B}_1\right)
\\
&=\hat{A}_1 \hat{A}_2\otimes \hat{B}_1\hat{B}_2-\hat{A}_2 \hat{A}_1\otimes \hat{B}_2\hat{B}_1\,,\\
\left[\hat{A}_1,\hat{A}_2 \right]\otimes \left[\hat{B}_1,\hat{B}_2\right] &= \left(\hat{A}_1\hat{A}_2-\hat{A}_2 \hat{A}_1\right)\otimes\left(\hat{B}_1\hat{B}_2-\hat{B}_2 \hat{B}_1\right)\\
&=\hat{A}_1\hat{A}_2\otimes \hat{B}_1\hat{B}_2-\hat{A}_1\hat{A}_2\otimes \hat{B}_2\hat{B}_1-\hat{A}_2\hat{A}_1\otimes \hat{B}_1\hat{B}_2+\hat{A}_2\hat{A}_1\otimes \hat{B}_2\hat{B}_1\,.
\end{align*}
5.2.1.Density matrix for tensor products
We can now consider the density matrix associated to a separable state
\(\ket{\Psi} = \ket{\psi} \otimes \ket{\phi}\) which is simply
\[
\hat{\rho} = \ket{\Psi}\bra{\Psi} = \Big(\!\ket{\psi}\bra{\psi}\Big)\otimes \Big(\!\ket{\phi}\bra{\phi}\Big) =\hat{\rho}_A\otimes \hat{\rho}_B\,,
\]
where \(\hat{\rho}_A = \ket{\psi}\bra{\psi}\) is just the density
matrix for system \(A\) and \(\hat{\rho}_B = \ket{\phi}\bra{\phi}\) the
density matrix for system \(B\).
A mixed ensemble of separable states is then given by
\[
\hat{\rho} = \sum_n p_n \, \hat{\rho}^{(n)}_{A}\otimes \hat{\rho}^{(n)}_{B}\,,
\]
where \(\{ \hat{\rho}^{(n)}_{A}\}\) are mixed or pure states of system A, while \(\{ \hat{\rho}^{(n)}_{B}\}\) are mixed or pure states of system B and as always \(p_n\geq 0\) such that \(\sum_n p_n=1\).
We will say that a mixed state is separable if and only if it is an
ensemble of separable states, entangled otherwise. We will not be
focusing particularly on mixed states in bipartite systems because
they can be seen as arising from a pure state in a larger system via
the process of reduced density matrix and partial trace (see later).
5.3.Matrix Representation
Since \(\mathcal{H} =\mathcal{H}_A\otimes \mathcal{H}_B\) is once again
a vector space with a complex inner product, we have that states and
operators in \(\mathcal{H} =\mathcal{H}_A\otimes \mathcal{H}_B\) can be
expressed as vectors and matrices. To express the tensor product of
two column vectors or matrices we use the convention that the first
term gives the block structure while the second specifies the detail
of the individual blocks up to multiplication by the appropriate
constant from the first. This is simplest to explain with examples
\[ \cvec{1 \\ 2 \\ 5} \otimes \cvec{3 \\ 4} \rightarrow
\cvec{1\cvec{3 \\ 4} \\ 2\cvec{3 \\ 4} \\ 5\cvec{3 \\ 4}} =
\cvec{3 \\ 4 \\ 6 \\ 8 \\ 15 \\ 20} \]
where the middle expression is just given to indicate the structure.
This process can be understood by ordering the basis elements of \(\mathcal{H} = \mathcal{H}_A\otimes \mathcal{H}_B\) starting from the basis of \(\mathcal{H}_A\) and \(\mathcal{H}_B\) with the following order
\begin{align*}
\mathcal{H} =\mbox{span} &\left\lbrace \cvec{1 \\ 0 \\ 0} \otimes \cvec{1 \\ 0},\cvec{1 \\ 0 \\ 0} \otimes \cvec{0 \\ 1}, \cvec{0 \\ 1 \\ 0} \otimes \cvec{1 \\ 0},\right.\\
&\phantom{=}\left.\cvec{0 \\ 1 \\ 0} \otimes \cvec{0 \\ 1},\cvec{0 \\ 0 \\ 1} \otimes \cvec{1 \\ 0},\cvec{0 \\ 0 \\ 1} \otimes \cvec{0 \\ 1} \right\rbrace\,,
\end{align*}
and now we represent this basis with the standard basis in the order given, i.e.
\[
\cvec{1 \\ 0 \\ 0} \otimes \cvec{1 \\ 0}\to \cvec{1\\0\\0\\0\\0\\0}\,,\,\,\,...\,\,\,,\, \cvec{0 \\ 0 \\ 1} \otimes \cvec{0 \\ 1}\to \cvec{0\\0\\0\\0\\0\\1}\,,
\]
which produces exactly the procedure outlined above with an example.
Similarly
for matrices
\[ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & -3 & 0 \end{array} \right) \otimes \left( \begin{array}{cc} 0 & 1 \\ -1 & 2 \end{array} \right)
\rightarrow \left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & -2 & 4 \\ 0 & 0 & 0 & -3 & 0 & 0 \\ 0 & 0 & 3 & -6 & 0 & 0 \end{array} \right) \; . \]
In the above examples the first Hilbert space has dimension \(3\) while the
second has dimension \(2\) but obviously similar relations hold for arbitrary
(finite) dimensional Hilbert spaces.
Assuming we use orthonormal basis for the original Hilbert spaces, these vector
and matrix representations correspond to an orthonormal basis for the tensor
product Hilbert space.
Note that both the most general vector and the most general linear operator are not of the simple form \(\ket{\psi}\otimes \ket{\phi}\) and \(\hat{A}\otimes \hat{B}\) but rather \(\sum_{a,b} \ket{\psi_a}\otimes \ket{\phi_b}\) and \(\sum_{I,J} \hat{A}_I\otimes \hat{B}_J\).
Let \(\mathcal{H} = \mathcal{H}_q\otimes \mathcal{H}_q\) be the Hilbert
space of two qubits, i.e. a four dimensional Hilbert space. If
possible write the linear operator on \(\mathcal{H}\)
\[ O = \left(\begin{matrix}1 & 0 & 0 & 2\\ 0& 1 & 2 & 0\\ 0 & 2 & 1 &
0\\ 2 & 0 & 0 & 1
\end{matrix}\right)
\]
as \(A\otimes B\) where \(A,B\) are linear operators on \(\mathcal{H}_q\), i.e. \(2\times 2\) matrices.
First we notice that
\[O = \mathbb{I}_4 + \left(\begin{matrix}0 & 0 & 0 & 2\\
0& 0 & 2 & 0\\
0 & 2 & 0 & 0\\
2 & 0 & 0 & 0
\end{matrix}\right) = \mathbb{I}_2\otimes \mathbb{I}_2 + \left(\begin{matrix}0 & 2 \\ 2 & 0\end{matrix}\right)\otimes\left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right) = \mathbb{I}_2\otimes \mathbb{I}_2 + 2 \,\sigma_1 \otimes \sigma_1\,,
\]
which cannot be simplified any further so \(O \neq A\otimes B\).
Let us consider again \(\mathcal{H} = \mathcal{H}_q\otimes \mathcal{H}_q\) and define
\[
U = \left(\begin{matrix}1 & 0 & 0 & 0\\
0& 1 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0
\end{matrix}\right)\,.
\]
First check that \(U\) is a unitary operator. Let us write it as \(A_1 \otimes B_1+A_2 \otimes B_2\), to do this we write
\begin{align*}
U &= \left(\begin{matrix}1 & 0 & 0 & 0\\
0& 1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{matrix}\right)+ \left(\begin{matrix}0 & 0 & 0 & 0\\
0& 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0
\end{matrix}\right)\\
&= \frac{\left(\begin{matrix} 1 & 0\\0 & 0 \end{matrix}\right)}{2}\otimes \mathbb{I}_2+ \frac{\left(\begin{matrix} 0 & 0\\0 & 1 \end{matrix}\right)}{2}\otimes \sigma_1 = \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{align*}
We can now understand why this operator implements the quantum logical
gate called
Controlled NOT (CNOT) where the first qubit
controls what we do to the second qubit, if the first qubit is
\(\ket{0}\) we leave the second qubit invariant, otherwise if the first
qubit is \(\ket{1}\) we take the NOT of the second qubit \(\ket{\bar{y}}
= \sigma_1 \ket{y}\).
As always to understand what an operator does we just need to check
what \(U\) does on the four basis states \(\mathcal{H} =
\mbox{span}\{\ket{00},\,\ket{01},\,\ket{10},\,\ket{11}\}\).
First of all check that
\[ \left(\frac{\mathbb{I}_2+\sigma_3}{2}\right) \ket{0} = \ket{0}\,,\qquad \left(\frac{\mathbb{I}_2+\sigma_3}{2}\right)\ket{1}= 0\,,\]
and similarly
\[ \left(\frac{\mathbb{I}_2-\sigma_3}{2}\right) \ket{0} =0 \,,\qquad \left(\frac{\mathbb{I}_2-\sigma_3}{2}\right)\ket{1}= \ket{1}\,,\]
so the first operators appearing in \(U\) acting on the first qubit behave as a
control:
-If the first qubit is \(\ket{0}\) we only need to consider \(\left(\frac{\mathbb{I}_2+\sigma_3}{2}\right)\otimes \mathbb{I}_2\), which acts as the identity on the second qubit;
-while if the first qubit is \(\ket{1}\) we only need to consider \(\left(\frac{\mathbb{I}_2-\sigma_3}{2}\right)\otimes \sigma_1\) which acts as the logical gate NOT on the second qubit and we have
\begin{align*}
U \ket{00} &= \left(\frac{\mathbb{I}_2+\sigma_3}{2}\right)\ket{0} \otimes \mathbb{I}_2\ket{0} + \phantom{\left(\frac{\mathbb{I}_2-\sigma_3}{2}\right) \ket{0} \otimes \sigma_1\ket{0}} = \ket{00}\,,\\
U \ket{01} &= \left(\frac{\mathbb{I}_2+\sigma_3}{2}\right)\ket{0} \otimes \mathbb{I}_2\ket{1} + \phantom{\left(\frac{\mathbb{I}_2-\sigma_3}{2}\right) \ket{0} \otimes \sigma_1\ket{1}} = \ket{01}\,,\\
U \ket{10} &= \phantom{\left(\frac{\mathbb{I}_2+\sigma_3}{2}\right)\ket{1} \otimes \mathbb{I}_2\ket{0}} + \left(\frac{\mathbb{I}_2-\sigma_3}{2}\right) \ket{1} \otimes \sigma_1\ket{0} = \ket{11}\,,\\
U \ket{11} &= \phantom{\left(\frac{\mathbb{I}_2+\sigma_3}{2}\right)\ket{1} \otimes \mathbb{I}_2\ket{1} }+ \left(\frac{\mathbb{I}_2-\sigma_3}{2}\right) \ket{1} \otimes \sigma_1\ket{1} = \ket{10}\,.
\end{align*}
Check that this is exactly the same action as the \(4\times 4\) matrix written above has on the four standard basis vectors \(\{ (1,0,0,0)^T,..., (0,0,0,1)^T \}\).
5.4.Local Measurements
If Alice and Bob perform measurements on their own systems, they can do so using
self-adjoint operators of the form \(\hat{F} = \hat{F}_A \otimes \hat{I}\) for Alice and
\(\hat{G} = \hat{I} \otimes \hat{G}_B\) for Bob. Assuming for simplicity no degeneracy for
the spectrum of \(\hat{F}_A\) or \(\hat{G}_B\) within each subsystem, these operators
have projection operators \(\hat{F}_{Ai} = \ket{i}\bra{i}\) and
\(\hat{G}_{Bm} = \ket{m}\bra{m}\) within Alice's and Bob's subsystems respectively.
In the full system the operators are degenerate, with degeneracy given by the
dimension of the other subsystem, i.e. by the dimension of \(\mathcal{H}_B\) for Alice's observable, and by that of
\(\mathcal{H}_A\) for Bob's. The corresponding projection operators in the full
system are \(\hat{F}_i = \hat{F}_{Ai} \otimes \hat{I}\) and
\(\hat{G}_{m} = \hat{I} \otimes \hat{G}_{Bm}\).
Using the spectral decomposition of the identity we can write these two projectors as
\[
\hat{F}_i = \sum_n \ket{i}\bra{i} \otimes \ket{n}\bra{n}\,,\qquad\hat{G}_{m} =\sum_j \ket{j}\bra{j} \otimes \ket{m}\bra{m}\,.
\]
It is clear then that the eigenspace for the operator \(\hat{F}_A \) which was spanned by \(\ket{i}\) in \(\mathcal{H}_A\) is now becoming degenerate for the operator \(\hat{F}=\hat{F}_A\otimes \hat{I} \) in \(\mathcal{H}_A \otimes \mathcal{H}_B\) spanned by all the vectors of the form \(\ket{i}\otimes \ket{\phi}\) with \(\ket{\phi}\in\mathcal{H}_B\).
Since \(\com{F}{G} = 0\) the measurements are compatible so Alice and Bob can both
measure and the final state will be in a simultaneous eigenstate of \(\hat{F}\)
and \(\hat{G}\). The outcomes of the measurements will be some pair \((f_i, g_m)\)
and the probability of this outcome is independent of whether Alice or Bob
measures first – in fact they could also measure simultaneously (or with spacelike separation so that no signal could travel between them to allow one measurement to potentially affect the other.) In this sense, as for local unitary transformations, they can consider their systems to be isolated from each other.
Suppose in fact that the state is in a pure state and let us assume for now separable as well
\[
\ket{\Psi} = \ket{\psi}\otimes \ket{\phi} = \sum_{i\,,m} \alpha_i \beta_m \ket{i}\otimes \ket{m}\,,
\]
where the coefficients \(\alpha_{i},\beta_{m}\in\mathbb{C}\) and \(\{\ket{i}\}\,,\,\{ \ket{m}\}\) form an orthonormal basis for \(\mathcal{H}_A\) and \(\mathcal{H}_B\) respectively.
As always the states \( \ket{\psi}\) and \(\ket{\phi}\) are normalised which means \(\sum_i \vert \alpha_i\vert^2 = \sum_m \vert \beta_m\vert^2 = 1\), this implies that if we define the combination \(\alpha_i \beta_m = \gamma_{im}\) then the condition \(\ip{\Psi}{\Psi}=1\) imposes \(\sum_{i\,,m} \vert \gamma_{im} \vert^2= 1\).
If Alice measures \(\hat{F}\) she will obtain outcome \(f_j\) with probability \(\vert \alpha_j\vert^2 = \sum_{m} \vert \gamma_{jm}\vert^2\) and the system will then collapse to the state
\[
\sum_m \beta_m \ket{j}\otimes \ket{m} = \ket{j}\otimes \ket{\phi}\,.
\]
If Bob then measures \(\hat{G}\) and obtains outcome \(g_n\) with probability \(\vert \beta_n \vert^2 = \sum_i \vert \gamma_{im}\vert^2\) we have that the final state becomes \(\ket{j}\otimes \ket{m}\).
The probability would be exactly the same if Bob had measured first, except the intermediate state would have been \(\ket{\psi}\otimes \ket {n}\).
Overall the combined measurements of \(\hat{F}\) and \(\hat{G}\) with outcomes \((f_j,g_n)\) have probability \(\vert \gamma_{jn}\vert^2\) which is the product of the two probabilities.
We can also rewrite everything in operator formalism.
We just need to remember that the operators \(\hat{F}_{Ai} = \ket{i}\bra{i}\) are mutually orthogonal projectors onto the eigenspace \(\mbox{span}\{\ket{i}\}\subseteq \mathcal{H}_A\).
The probability for Alice to measure \(f_i\) is
\[
\vert \ip{i}{\psi}\vert^2 = \Tr\left(\hat{\rho}_A\, \hat{F}_{Ai} \right)\,,
\]
and after measuring \(\hat{F}_A\) and finding \(f_i\) the state of Alice has collapsed to the normalised state
\[
\ket{\psi} \to \ket{i} = \frac{1}{\vert \ip{i}{\psi} \vert} \hat{F}_{Ai} \ket{\psi} = \frac{1}{ \sqrt{\Tr \left( \hat{\rho}_A\,\hat{F}_{Ai}\right)} }\hat{F}_{Ai}\ket{\psi}\,,
\]
or equivalently
\[
\hat{\rho}_A \to \frac{1}{\Tr \left( \hat{\rho}_A\, \hat{F}_{Ai} \right)} \hat{F}_{Ai}\, \hat{\rho}_A \hat{F}_{Ai}\,,
\]
where we have intensively used the fact that \( \hat{F}_{Ai}\) is a projector, i.e. \( \hat{F}_{Ai}^\dagger = \hat{F}_{Ai}\) and \( \hat{F}_{Ai}^2 = \hat{F}_{Ai}\).
For a bipartite system we just need to repeat this analysis while
carrying along the road Bob's system. Clearly when Alice measures the
observable \(\hat{F}_{A}\) on her system she does not perform any
operation on Bob's (LO) so we define
\[
\hat{F}_i = \hat{F}_{Ai}\otimes \hat{I}\,.
\]
If we have prepared the separable state \(\ket{\Psi}= \ket{\psi}\otimes \ket{\phi}\in\mathcal{H}_A\otimes \mathcal{H}_B\) then we can use the density matrix
\[
\hat{\rho} = \ket{\Psi} \bra{\Psi} = \ket{\psi}\bra{\psi}\otimes \ket{\phi}\bra{\phi} = \hat{\rho}_A\otimes \hat{\rho}_B\,.
\]
Now if Alice measures \(\hat{F} = \hat{F}_A\otimes \hat{I} \) she will obtain outcome \(f_i\) with probability
\[
\mbox{Tr}_{A\otimes B} \left( \hat{\rho}\, \hat{F}_i\right) = \mbox{Tr}_A \left(\hat{\rho}_A \,\hat{F}_{Ai}\right)\,,
\]
and after that her wave function will have collapsed while Bob' state will be unchanged
\[
\hat{\rho} \to \frac{1}{\Tr_A \left(\hat{\rho}_A \,\hat{F}_{Ai}\right)} \hat{F}_{Ai} \,\hat{\rho}_A \hat{F}_{Ai} \otimes \hat{\rho}_B = \frac{1}{\Tr_{A\otimes B} \left(\hat{\rho} \,\hat{F}_{i}\right)} \hat{F}_{i} \,\hat{\rho} \,\hat{F}_{i}\,,
\]
where we used the projector \(\hat{F}_i\) defined on the whole bipartite system.
Note in particular that
\[
\hat{F}_i = \hat{F}_{Ai}\otimes \hat{I} = \sum_{m} \left(\ket{i}\otimes \ket{m}\right)\otimes \left(\bra{i}\otimes \bra{m}\right)\,,
\]
where we have used the spectral decomposition of \(\hat{I}\) for Bob system to make manifest the fact that although the eigenspace corresponding to the eigenvalue \(f_i\) was non-degenerate in \(\mathcal{H}_A\), as soon as we consider a bipartite system it immediately becomes degenerate since any vector of the form \(\ket{i}\otimes \ket{\phi}\) with \(\ket{\phi}\in\mathcal{H}_B\) is an eigenvector of \(\hat{F}\otimes \hat{I}\) with same eigenvalue \(f_i\) (see comment at the beginning of this section).
Now similarly if Bob measures \(\hat{G}_B\) with spectrum \(\{ g_n,\ket{n}\}\) he will obtain outcome \(g_m\) with probability
\[
\Tr_B\left( \hat{\rho}_B \,\hat{G}_{Bm} \right) = \Tr_{A\otimes B} \left( \hat{\rho} \,\hat{G}_{m} \right)\,,
\]
where we defined \( \hat{G}_{Bm} = \ket{m}\bra{m} \) and \( \hat{G}_m = \hat{I}\otimes \hat{G}_{Bm}\).
After measurement the wave function will collapse to
\[
\hat{\rho} \to \frac{1}{\Tr_{A\otimes B} \left(\hat{\rho}\, \hat{G}_m\right) }\hat{G}_m\,\hat{\rho}\, \hat{G}_m\,.
\]
As above it does not matter who measures first (only the intermediate state will change), if we measure \(\hat{F} = \hat{F}_A\otimes \hat{I}\) and \(\hat{G} = \hat{I}\otimes \hat{G}_B\) the outcome \((f_i,g_m)\) will be measured with probability
\[
\Tr_{A\otimes B} \left( \hat{\rho} \,\hat{P}_{im} \right)\,,
\]
with \(\hat{P}_{im} = \hat{F}_{Ai}\otimes \hat{G}_{Bm} = \ket{i}\bra{i}\otimes \ket{m}\bra{m}\).
The state will then collapse to
\[
\hat{\rho} \to \frac{1}{\Tr_{A\otimes B}\left(\hat{\rho}\,\hat{P}_{im}\right)} \hat{P}_{im}\,\hat{\rho}\,\hat{P}_{im} = \ket{i}\otimes \ket{m}\,.
\]
Let us now repeat the same analysis but for an entangled state.
Suppose the bipartite system \(\mathcal{H}= \mathcal{H}_A\otimes \mathcal{H}_B\) is prepared in the state
\[
\ket{\Psi} = \sum_{i,m} \gamma_{im} \ket{i}\otimes \ket{m}\,,
\]
where the normalised coefficients \( \gamma_{im} \in\mathbb{C}\) are not of the form \(\gamma_{im} \neq \alpha_i \beta_m\) as above, i.e. the state is an entangled one.
Let us still define the coefficients
\[
\alpha_j =\sqrt{\sum_{m} \vert \gamma_{jm}\vert^2}\,,\qquad\beta_n =\sqrt{\sum_{i} \vert \gamma_{in}\vert^2}\,,
\]
allowing us to define the two auxiliary normalised states
\begin{align*}
\ket{\psi_n} &= \frac{1}{\beta_n} \sum_i \gamma_{in}\ket{i} \in\mathcal{H}_A\,,\\
\ket{\phi_j} &= \frac{1}{\alpha_j} \sum_m \gamma_{jm}\ket{m} \in\mathcal{H}_B\,,
\end{align*}
excluding values of \(n\) and \(j\) for which \(\beta_n=0\) or \(\alpha_j=0\).
(Little exercise for you check that these states are indeed normalised.)
We can then rewrite the state \(\ket{\Psi}\) as
\begin{align*}
\ket{\Psi} &= \sum_{i,m} \gamma_{im} \ket{i}\otimes \ket{m}\,,\\
&= \sum_i \alpha_i \ket{i}\otimes \ket{\phi_i}\,,\\
&=\sum_n \beta_m \ket{\psi_m} \otimes \ket{m}\,.\end{align*}
If Alice measures \(\hat{F}\) she will have outcome \(f_i\) with probability \(\vert \alpha_i\vert^2\) and after the measurement the state will have collapsed to
\[
\ket{\Psi}\to \hat{F}_i \ket{\Psi} = (\hat{F}_{Ai}\otimes \hat{I}) \ket{\Psi} \sim \ket{i} \otimes \ket{\phi_i}\,.
\]
The key difference from the previous, separable case is that starting from an entangled state after Alice measurement we obtain a separable one with \(\ket{i}\in\mathcal{H}_A\) and \(\ket{\phi_i}\in\mathcal{H}_B\) but Bob' state depends on the result of Alice measurement!
Local measurements can have a non-local effect. If Alice
performs a local measurement on an entangled state, the system will
collapse to a separable one, and Bob's state will depend on the
result of Alice's measurement.
This is the novelty of quantum mechanics: we say that quantum mechanics is
non-local! Local measurements, for example Alice measuring a spin in her laboratory in New York, can have non-local effects, i.e. changing Bob' state who lives on Mars.
We will shortly see that this “spooky action at a distance”, as Einstein used to call it, will not allow us to communicate faster than the speed of light, i.e. it will not violate causality.
The key point is that if both Alice and Bob know the full initial state \(\ket{\Psi}\), then after measuring \(\hat{F}\) and finding \(f_i\) Alice knows that Bob' state is \(\ket{\phi_i}\) however Bob does not, unless Alice tells him (we will say they
communicate classically) the result of her measurement.
For the moment let us forget about Bob and let us try and assign a density matrix \(\hat{\rho}_A\) to describe only Alice system and accommodate for this lack of (classical) knowledge regarding Bob, this will introduce the concept of
Reduced Density Matrix.
5.5.Reduced Density Matrix
Given a bipartite (or multi-partite) system, we can define the partial
trace over a subsystem to be a trace in that subsystem only. I.e. the
partial trace over system B (or Hilbert space \(\mathcal{H}_B\)) is
defined by
\[ \Tr_B \left( \hat{C} \otimes \hat{D} \right) = \Tr(\hat{D})\, \hat{C} \]
and all other properties follow from linearity.
Note that the partial trace \(\Tr_B\) maps linear operators acting on \(\mathcal{H}_A \otimes \mathcal{H}_B\) to linear operators acting only on \(\mathcal{H}_A\), i.e. the result of a partial trace is NOT a number but rather an operator on the remaining Hilbert space.
Similarly the partial trace over \(A\)
\[ \Tr_A \left( \hat{C} \otimes \hat{D} \right) = \Tr(\hat{C}) \,\hat{D} \,,\]
produces an operator on \(\mathcal{H}_B\).
We define the
reduced density matrix for a subsystem to be the partial trace of the
density matrix over the other subsystem(s). I.e. for a bipartite system
\[\hat{\rho}_A \equiv \Tr_B(\hat{\rho})\]
and
\[\hat{\rho}_B \equiv \Tr_A(\hat{\rho})\,.\]
Generically the reduced density matrices will describe mixed states even if
the full system is in a pure state.
This is reminiscent of our discussion regarding mixed states: if we “forget” about Bob system, i.e. if we consider the partial trace over \(B\), Alice will have some lack of classical knowledge about her system which means that her state will not be a pure one but rather a mixed state, i.e. the mixed state described by the reduced density matrix.
Let us consider the two-qubit system \(\mathcal{H}= \mathcal{H}_q\otimes \mathcal{H}_q\) in the pure, entangled state
\[
\ket{\beta_{00}} =\frac{1}{\sqrt{2}} \left(\ket{0}\otimes \ket{0}+\ket{1}\otimes \ket{1}\right)\,.
\]
(This will be the prototypical example of entangled states called a Bell state or EPR pair)
Firstly let us compute the density matrix in operator form
\[
\hat{\rho} = \ket{\Psi}\bra{\Psi} = \frac{1}{2} \left( \ket{0}\bra{0}\otimes \ket{0}\bra{0}+ \ket{0}\bra{1}\otimes \ket{0}\bra{1}+ \ket{1}\bra{0}\otimes \ket{1}\bra{0}+ \ket{1}\bra{1}\otimes \ket{1}\bra{1}\right)\,.
\]
We can compute the reduced density
\begin{align*}
\hat{\rho}_A &=\Tr_B \hat{\rho} = \frac{1}{2} \left[ \ket{0}\bra{0} \mbox{Tr}_B \left( \ket{0}\bra{0}\right)+\ket{0}\bra{1} \mbox{Tr}_B \left( \ket{0}\bra{1}\right)+\phantom{\frac{1}{2}}\right.\\
&\left.\phantom{====\Tr_B \hat{\rho} \frac{1}{2}}+\ket{1}\bra{0} \mbox{Tr}_B \left( \ket{1}\bra{0}\right)+ \ket{1}\bra{1} \mbox{Tr}_B \left( \ket{1}\bra{1}\right)\right] \\
&= \frac{1}{2}\left(\ket{0}\bra{0}+\ket{1}\bra{1}\right) =\frac{\hat{I}}{2}\,.
\end{align*}
We will see that these Bell states will be maximally entangled, for the moment we have just seen that the reduced density matrix over the second (or first) qubit produces the most mixed qubit state, i.e. half of the identity, i.e. the centre of the Bloch sphere.
We can also obtain this result by first constructing the density matrix as a \(4\times 4\) matrix using the standard basis for the usual orthonormal basis states \(\{\ket{00},\,\ket{01},\,\ket{10},\,\ket{11}\}\).
We first rewrite the state \(\ket{\beta_{00}}\) as the \(4\)-dimensional vector
\[
\ket{\beta_{00}} = \frac{1}{\sqrt{2}} \left(\ket{0}\otimes \ket{0}+\ket{1}\otimes \ket{1}\right) \to \vec{v} = \frac{1}{\sqrt{2}} \left( \begin{matrix}1 \\ 0 \\0 \\1 \end{matrix}\right) \,,
\]
then as always the density matrix becomes
\[
\hat{\rho} = \ket{\beta_{00}}\bra{\beta_{00}} \to \rho = \vec{v}\,\vec{v}^\dagger =\frac{1}{2}\left( \begin{matrix}1 & 0 & 0 & 1\\
0& 0& 0& 0\\ 0& 0& 0& 0 \\ 1& 0 & 0 & 1\end{matrix}\right) \,,
\]
which we rewrite in the sum of tensor operators as
\[
\rho = \frac{1}{2} \left(\begin{matrix}1 & 0 \\ 0& 0 \end{matrix}\right)\otimes \left(\begin{matrix}1 & 0 \\ 0& 0\end{matrix}\right)+\frac{1}{2} \left(\begin{matrix}0 & 1 \\ 0& 0 \end{matrix}\right)\otimes \left(\begin{matrix}0 & 1 \\ 0& 0\end{matrix}\right)+\frac{1}{2} \left(\begin{matrix}0 & 0 \\ 1& 0 \end{matrix}\right)\otimes \left(\begin{matrix}0 & 0 \\ 1& 0\end{matrix}\right)+\frac{1}{2} \left(\begin{matrix}0 & 0 \\ 0& 1 \end{matrix}\right)\otimes \left(\begin{matrix}0 & 0 \\ 0& 1\end{matrix}\right)\,,
\]
and finally perform the partial trace over the second operator
\begin{align*}
\rho_A & = \frac{1}{2} \left(\begin{matrix}1 & 0 \\ 0& 0 \end{matrix}\right) \,\Tr \left(\begin{matrix}1 & 0 \\ 0& 0\end{matrix}\right)+\frac{1}{2} \left(\begin{matrix}0 & 1 \\ 0& 0 \end{matrix}\right)\,\Tr \left(\begin{matrix}0 & 1 \\ 0& 0\end{matrix}\right)+\\
&\phantom{=}+\frac{1}{2} \left(\begin{matrix}0 & 0 \\ 1& 0 \end{matrix}\right)\,\Tr \left(\begin{matrix}0 & 0 \\ 1& 0\end{matrix}\right)+\frac{1}{2} \left(\begin{matrix}0 & 0 \\ 0& 1 \end{matrix}\right)\,\Tr \left(\begin{matrix}0 & 0 \\ 0& 1\end{matrix}\right)\\
&=\frac{1}{2} \left(\begin{matrix}1 & 0 \\ 0& 0 \end{matrix}\right)+ \frac{1}{2} \left(\begin{matrix}0 & 0 \\ 0& 1 \end{matrix}\right) = \frac{\mathbb{I}_2}{2}\,,
\end{align*}
as already obtained above.
The reduced density matrix \(\hat{\rho}_A\) is a density matrix which describes
Alice's subsystem when she has no knowledge of Bob's system. The following
properties illustrate this:
\(\hat{\rho}_A\) is invariant under all LO in system B. For unitary transformations this is a simple consequence of cyclicity of the trace. For measurements in B this is true provided Alice does not know the results of any measurements.
This explains why Alice typically sees a mixed state (before she performs any measurements.)
Under unitary transformations in system A, \(\hat{\rho}_A\) transforms as
expected for a density matrix.
Local measurements in system A can be described in terms of operators
acting on \(\mathcal{H}_A\) and \(\hat{\rho}_A\).
For the point about measurement, the main result is that the probability and
final state can be calculated using \(\hat{F}_i\) and \(\hat{\rho}\), or using
\(\hat{F}_{Ai}\) and \(\hat{\rho}_A\), and either method gives the same predictions
and results. Specifically
\[ \Tr_B \left( \hat{F}_i \, \hat{\rho}\, \hat{F}_i \right) = \hat{F}_{Ai} \,\hat{\rho}_A \,\hat{F}_{Ai} . \]
The reduced density matrix captures a lot of information regarding the nature of the state in consideration.
In particular we have the following theorem.
If the system \(\mathcal{H}_A\otimes \mathcal{H}_B\) is in a pure state
\(\ket{\Psi}\) then the reduced density matrix \(\hat{\rho}_A =
\mbox{Tr}_B \hat{\rho}\) is pure if and only if \(\ket{\Psi}\) is
separable, i.e. \(\ket{\Psi} = \ket{\psi}\otimes \ket{\phi}\) with
\(\ket{\psi}\in\mathcal{H}_A\) and \(\ket{\phi}\in\mathcal{H}_B\).
(\(\Leftarrow\)) Let us start with the separable pure state \(\ket{\Psi}
= \ket{\psi}\otimes \ket{\phi}\). Then \(\hat{\rho} =
\ket{\Psi}\bra{\Psi} = \ket{\psi}\bra{\psi}\otimes
\ket{\phi}\bra{\phi}\), so for the reduced density matrix we have
\begin{equation}
\begin{aligned}
\hat{\rho}_A = \mbox{Tr}_B \left(\hat{\rho}\right)
&= \mbox{Tr}_B \left(\ket{\psi}\bra{\psi}\otimes \ket{\phi}\bra{\phi}\right)\\[1ex]
&= \ket{\psi}\bra{\psi} \mbox{Tr}_B \left( \ket{\phi}\bra{\phi}\right)=\ket{\psi}\bra{\psi} \,,
\end{aligned}
\end{equation}
since the state \( \ket{\phi}\) is normalised, i.e. \(\mbox{Tr}_B \left(
\ket{\phi}\bra{\phi}\right)=\ip{\phi}{\phi}=1\). So if the starting
pure state \(\hat{\rho}\) is separable then the reduced density matrix
is pure \(\hat{\rho}_A=\ket{\psi}\bra{\psi} \).
(\(\Rightarrow\)) Conversely let us assume the reduced density matrix
\(\hat{\rho}_A = \ket{\psi}\bra{\psi}\) is given by a pure state. Let us
complete the vector \( \ket{\psi}\) to an orthonormal basis for
\(\mathcal{H}_A\), i.e. \(\mathcal{H}_A = \mbox{span}\{\ket{\psi},
\ket{\psi^\perp_i}\}\) with \(\ip{\psi}{\psi^\perp_i} = 0\) and
\(\ip{\psi^\perp_i}{\psi^\perp_j} = \delta_{ij}\).
The most general state in \(\mathcal{H}_A\otimes \mathcal{H}_B\) can be
written as
\[
\ket{\Psi} = c \ket{\psi}\otimes \ket{\phi} +\sum_i c_i \ket{\psi^\perp_i}\otimes \ket{\phi_i}\,,
\]
for some \(\ket{\phi},\ket{\phi_i}\in\mathcal{H}_B\) and \(c,c_i \in\mathbb{C}\).
If Alice measures the observable
\(\ket{\psi_j^\perp}\bra{\psi_j^\perp}\otimes \hat{I}\) on \(\ket{\Psi}\)
she has the expectation value
\[
\mbox{Tr}_{A\otimes B} \left[ \ket{\Psi}\bra{\Psi} \left(\ket{\psi_j^\perp}\bra{\psi_j^\perp}\otimes \hat{I} \right) \right] = \vert c_j \vert^2\,,
\]
using the orthonormality properties of the vectors \(\ket{\psi_i^\perp}\). But this must be equal to
\[
\mbox{Tr}_A \left(\hat{\rho}_A \ket{\psi_j^\perp}\bra{\psi_j^\perp} \right) = \ip{\psi}{\psi_j^\perp} \ip{\psi_j^\perp}{\psi} = 0\,,
\]
hence \(c_j=0\) for all \(j\) which means that the state \(\ket{\Psi} = c \ket{\psi}\otimes \ket{\phi}\) is indeed separable.
Measurement destroys entanglement.
If the spectrum of \(\hat{F}_A\) is non-degenerate then measuring
\(\hat{F}_A\) in the system \(\mathcal{H}_A\) produces a separable state
on the system \(\mathcal{H}_A\otimes \mathcal{H}_B\). In other words,
measurement destroys entanglement.
We know that if we measure \(\hat{F}_A\) on the state \(\hat{\rho}_A\) and
we find the outcome \(f_i\) we must collapse the wave function to the
one dimensional (because the spectrum is non-degenerate) eigenspace
spanned by the corresponding vector \(\ket{i}\). Hence we have that the
reduced density matrix \(\hat{\rho}_A\) must go to
\[
\hat{\rho}_A \to \hat{\rho}'_A= \ket{i}\bra{i}\,,
\]
but since the new density matrix \(\hat{\rho}'_A\) is clearly pure we
must have from the theorem above that the state \(\hat{\rho}\) in
\(\mathcal{H}_A\otimes \mathcal{H}_B\) has collapsed to a separable
state
\[
\hat{\rho} \to \hat{\rho}'_A\otimes \hat{\rho}'_B = \ket{i}\bra{i} \otimes \ket{\phi_i}\bra{\phi_i}\,.
\]
With this new concept of reduced density matrix we can also understand
why entanglement, although non-local in nature, does not violate
causality. Let us start again with our favourite entangled pure state
\begin{align*}
\ket{\Psi} &= \sum_{i,m} \gamma_{im} \ket{i}\otimes \ket{m}\,,\\
&= \sum_i \alpha_i \ket{i}\otimes \ket{\phi_i}\,,\\
&=\sum_n \beta_m \ket{\psi_m} \otimes \ket{m}\,,\end{align*}
where the various coefficients have been defined as above.
We know that Alice measuring \(\hat{F}= \hat{F}_A\otimes \hat{I}\) will
collapse the state \(\hat{\rho}\to \hat{\rho}_A'\otimes \hat{\rho}_B'\).
If Bob could detect this we would have that Alice measurement would
result in instantaneous communication which violates the causality of
physics, i.e. any signal should travel from Alice to Bob no faster
than the speed of light.
However how could Bob detect this? Well the only thing he can do is
perform a measurement for his favourite observable \(\hat{G} =
\hat{I}\otimes \hat{G}_B\), so what we are really trying to understand
is whether the probabilities of outcomes for Bob have changed before
and after Alice's measurement.
Before Alice measures we know that Bob has the reduced density matrix
\[
\hat{\rho}_B = \Tr_A \left( \ket{\Psi}\bra{\Psi}\right) =\sum_{i,j} \alpha_i \alpha_j^* \ket{\phi_i}\bra{\phi_j} \Tr_A \left(\ket{i}\bra{j}\right)= \sum_i \vert \alpha_i \vert^2 \ket{\phi_i}\bra{\phi_i}\,,
\]
i.e. Bob has the ensemble \(\{(|\alpha_i|^2, \ket{\phi_i}\}\).
Now Alice measure \(\hat{F}\) and finds outcome \(f_i\) so she knows that
\(\ket{\Psi} \to \ket{i}\otimes \ket{\phi_i}\) which means
\[
\hat{\rho}\to \hat{\rho}' = \ket{i}\bra{i}\otimes \ket{\phi_i}\bra{\phi_i} = \hat{\rho}_A'\otimes \hat{\rho}_B'
\]
where the now collapsed density matrix \(\hat{\rho}_B'\) is now
different from the reduced density matrix \(\hat{\rho}_B\) compute
above, i.e. \(\hat{\rho}_B' = \ket{\phi_i}\bra{\phi_i} \neq
\hat{\rho}_B\).
However Bob does not know that Alice has measured and found \(f_i\)
unless Alice communicates this information, Bob only knows that he has
the state \( \ket{\phi_i} \) with probability \(|\alpha_i|^2\). Although
Alice has instantaneously changed Bob system, Bob cannot detect this,
there has not been any instantaneous transmission of information
between Alice and Bob. Said differently if Alice and Bob have
prepared 100 copies of the same state \(\ket{\Psi}\) and for a hundred
times Alice has measure \(\hat{F}\), unless she classically communicates
in which instances she has found outcome \(f_i\), Bob cannot detect any
difference in his probability distributions because he cannot possibly
know which one are the instances for which Alice has found outcome
\(f_i\) or outcome say \(f_1\)!
To summarise: Bob
cannot
Instantaneously detect the result of Alice's measurement: we
have already seen that the order of measurements is irrelevant, the
probability of outcome \((f_i,g_m)\) is \(\vert \gamma_{im}\vert^2\);
Know what Alice has measured: Suppose that Alice chooses a
different observable \(\hat{F}'\) this will just select a different
orthonormal basis \(\{\ket{\tilde{i}}\}\) of eigenstates of \(\hat{F}'\)
but still we have that
\[
\ket{\Psi} = \sum_i \alpha_i \ket{i}\otimes \ket{\phi_i} = \sum_i \tilde{\alpha}_i \ket{\tilde{i}}\otimes \ket{\tilde{\phi}_i}\,,
\]
for some new normalised coefficients \(\tilde{\alpha}_i \) and states
\(\ket{\tilde{\phi}_i}\in\mathcal{H}_B\). So Bob's ensemble can be
thought as \(\{ ( \, | \alpha_i|^2, \ket{\phi_i})\}\) or
\(\{(\,|\tilde{\alpha}_i|^2, \ket{\tilde{\phi}_i})\}\) but still the
mixed state is described by the same reduced density matrix
\(\hat{\rho}_B = \Tr_A \hat{\rho}\).
Even know that Alice has made a measurement: otherwise this
would mean that Bob's result could depend on whether or not Alice
made a measurement at any point in time, even in the future! Clearly
not possible in quantum mechanics.
However so far we have assumed that Alice and Bob do not communicate
at all. The non-locality of quantum mechanics cannot be detected with
local operations only (LO) but as soon as we add classical
communications (CC) between Alice and Bob everything changes. If
Alice calls Bob to tell him the result of her measurements Bob can
indeed detect the change in his state and the story gets interesting.
5.7.Quantum Communication
This means that we allow Alice and Bob to send quantum states to each
other. If we allow arbitrary quantum communication, we don't really
have a bipartite system. E.g. Bob could send his whole system to
Alice. She could then perform arbitrary operations on the complete
system, and then send Bob's part back to him. Obviously there is no
sense in which the two parts of the system were separated or
non-interacting. Instead, if we allow quantum communication at all we
typically impose specific restrictions, such as Alice can send one
qubit to Bob. In this way we can explore questions such as what could
be done with a single qubit compared to a single classical bit of
communication.
In applications concerned with security, again we assume Eve can
intercept any transmissions. However, unlike classical communications,
Eve cannot make a copy of the qubits (due to the no-cloning
theorem). Also, unless she already knows that they are eigenstates of
a specific measurement operator, she cannot measure them without some
non-zero probability that she disturbs the state through the
measurement process. This means that if Eve gains any information
about the qubits before forwarding them, this can be detected with
non-zero probability. The result is that quantum communication can be
used to ensure security in a way which is not possible with classical
communication. See specifically the discussion of Quantum Key
Distribution.
5.7.1.Hardware
Although in this module we will entirely focus on the “software”
part, to better understand quantum communication is perhaps useful to
see the “hardware” of quantum computers, i.e. the physical
realization of quantum computers, or even just two-level systems,
i.e. the qubit.
In a real world scenario we have three important quantities: the
decoherence time \(\tau_Q\), i.e. the time it takes the environment to
corrupt our quantum system, the operation time \(\tau_{op}\), i.e. the
time it takes to perform unitary transformations, and maximum number
of operations \(n_{op} = \tau_Q/\tau_{op}\), roughly how many operations
we can do to our system before it is destroyed.
I will not present here the pros and cons for various physical
realisation but if you are interested Chapter 7 of Nielsen and Chuang
discusses all these topics.
Perhaps the simplest apparatus are optical photons. A single photon
state can be produced in a lab by attenuating the output of a laser,
furthermore we can act on photons using mirrors, phase shifters,
beamsplitters and can be made interacting using nonlinear optical
media. This to say that we can indeed produce various states,
entangled or not, and act on them with unitary transformation. We can
take an electromagnetic cavity (think about two very good mirrors at
distance \(\lambda\)) and produce a quantum superposition of zero and
one photon of wave-length \(\lambda\) bouncing back and forth, i.e. a
qubit \(a \ket{0}+b\ket{1}\). Alice and Bob can then produce an
entangled state and each one of them store their qubit
(photon/no-photon superposition) in two separate cavities. Alice will
then perform some unitary transformation on her qubit and then (very
carefully) give the cavity to Bob who has now access to the whole
two-qubit system and can perform his favourite \(4 \times 4\)
dimensional unitary transformation.
Another candidate for real world qubits are single-atom cavities. We
can think of a single atom standing between two mirrors (called a
Fabry-Perot cavity). For simplicity let us assume that this atom
energy levels are just two, the lowest called ground state with energy
\(E_0=0\), and an excited state with energy \(E_1\gt{}E_0\). We can then have
a single photon with energy \(E_2-E_1\) interacting with this single
atom since it's resonant, i.e. the atom in the ground state can
“eat” the photon and go to the excited state. Quantum information
can then be treated in various ways: for example we can use photon
states (quantum superposition of \(0\) and \(1\) photons) and the cavity
with atom provides the non-linear interactions between them, or
represented by the atom (quantum superposition of ground and excited
state) with photons communicating between different atoms.
The final example uses the spin of particles, being that the orbitals
of atoms (ion traps) or the nuclear spin states (NMR the same as the
medical device!). The key point is that now we can use magnetic fields
to interact with spin systems. For example we can start with a
spin-1/2 particle, say for example an electron (although these are not
the particles used in real systems) that can be described by
specifying the component of its spin by convention along the \(z\)-axis,
hence our qubit is \(a \ket{\uparrow}+ b\ket{\downarrow}\) where
\(\ket{\uparrow}\) would be a state rotating counter-clockwise along the
\(z\)-axis and \(\ket{\downarrow}\) clockwise. Couplings between
different electrons (or chemical bonds between neighbouring atoms the
real world realisation) provide the interactions to produce entangled
states. So the only Alice and Bob can share two entangled states, each
one of them act with some magnetic field on their on spin system and
then send it to the other person.
Recently Google has announced that their new quantum processor:
“Sycamore” had reached quantum supremacy. They have a two
dimensional array of 54 of what they are called transmons qubits
where each qubit is coupled to the four nearest neighbours. Their
qubits are obtained from superconducting circuits for which the
conduction electrons condense into Cooper pairs (a macroscopic quantum
effect), these two superconducting islands are coupled via two
Josephson junctions which are just two superconducting regions
separated by a barrier. These Cooper pairs are pseudo-particles formed
by two electrons paired together and they are the fundamental objects
in the theory of superconductors. The qubit is now realised by the
quantum superposition of a Cooper pair transferred between these two
islands.
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