2.1.States and wave functions
Let us reminder ourselves of the quantum mechanics of a single
particle on the real line \(x\in\mathbb{R}\). In the formulation of
quantum mechanics using wave functions, this system is described by a
complex-valued wave function of space and time, \(\psi(x, t)\). The
probability of finding a particle in a region \(x\in[a,b]\) at some moment
in time \(t\) is given by
\begin{equation}
P(a,b; t) = \int_{a}^{b} \big|\psi(x, t)\big|^2\, {\rm d}x\,.
\end{equation}
This probability is clearly \(\geq 0\) because of the properties of the
norm of a complex number. We normalise the wave function such that
\begin{equation}
\label{e:normalisation}
P(-\infty, +\infty; t) = 1 \,,
\end{equation}
that is to say, the probability of finding the particle somewhere is
one.
The time-evolution of the wave function is given by the
Schrödinger equation,
\begin{equation}
i\hbar \frac{\partial\psi(x,t)}{\partial t}
= -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)
+ V(x) \psi(x,t)\,.
\end{equation}
The right-hand side equals the Hamiltonian operator acting on the wave
function, \(\hat{H} \psi(x,t) = \hat{K} \psi(x,t) + \hat{V} \psi(x,t)\)
where \(K,V\) are the kinetic and potential energy respectively. If we
impose \eqref{e:normalisation} for a single moment in time \(t=t_0\),
then the Schrödinger equation guarantees that it will remain valid for
arbitrary other times.
The function \(\psi(x)\) is said to describe the
state of the
system. Rather than knowing exactly where a particle is, as we do
classically, we only know the probability density \(P(x)\) (or more
precisely, the amplitude \(\psi(x)\)) of finding the particle somewhere
on the real line. Because the Schrödinger equation is
linear,
any linear superposition of wave functions is also a solution. So you
can have a wave function strongly peaked on earth, and another one
strongly peaked at the moon, and the linear combination is still
a valid quantum mechanical wave function,
\begin{equation}
\psi(x) = \psi_{\text{earth}}(x) + \psi_{\text{moon}}(x)\,.
\end{equation}
In quantum mechanics, only measurement will force the system into one
of the two classical configurations.
2.2.The Dirac notation (bra-ket)
Because the space of wave functions is linear (wave functions can be
superposed) and because we have a norm, we can view each wave function
as a
vector in a (complex) vector space:
Hilbert
space. Typically, this vector space will be infinite-dimensional,
because there is an infinite, or even a continuum, of possible
classical configurations. But this is not necessary, and in fact we
will consider in this module mainly systems for which we have only a
finite (but possibly large) number of classical configurations.
In Dirac notation, the ket \(|\psi\rangle \in {\cal H}\)
corresponds to the physical state. The dual bra vectors
\(\langle\phi|\) live in \({\cal H}^*\) and together they can form the
inner product \(\langle\phi|\psi\rangle\).
The Dirac notation
consists in writing \(|\psi\rangle\) for a vector in Hilbert space
corresponding to the wave function \(\psi(x)\). And instead of calling
it a vector, we call it a
ket, for reasons that will become
clear shortly. The fact that we no longer write the \(x\) label is
significant. Compare the situation in linear algebra. There, we can
have a physical, arrow-like object which we call a vectors (let's say
\(\vec{v}\)). To write down concretely which vector we mean, we choose a
basis of unit vectors, and then write down the
components of
the vector on that basis, e.g. \(\vec{v} = (2, 3)\). But changing the
basis does not change the vector itself, only its components. With
wave functions a similar thing happens. The representation \(\psi(x)\)
refers to the “basis” of position eigenstates labelled by the
position \(x\). But it is perfectly possible to write down the wave
function in a different basis, for instance the basis of momentum
eigenstates.
So we use \(|\psi\rangle\) from now on, as a more abstract way of expressing
the vector in Hilbert space. For any two such vectors, we have a
positive definite inner product for the corresponding wave fuctions,
\begin{equation}
\text{inner product}(\phi, \psi\big)
= \int_{-\infty}^{\infty} \phi^*(x,t) \psi(x,t)\, {\rm d}x
=: \langle \phi | \psi \rangle\,.
\end{equation}
On the right-hand side we have introduced the inner product
in the Dirac notation, \(\langle \phi | \psi\rangle\). It requires that
we have access to the
dual vector \(\langle \phi|\), which as you
can see from the explicit integral representation, is simply related
to the complex conjugate of the wave function \(\phi\). This new object
\(\langle \phi|\) is called a
bra, so that the inner product
(or bracket) between two states reads bra-ket.
In this module we will almost always consider finite-dimensional
Hilbert spaces. That means that we can define the Hilbert space by
specifying a finite set of basis states. Of course, there is not a
unique choice of basis states, and it will often be useful to consider
different choices (related by changes of basis matrices). Commonly we
will consider orthonormal bases, i.e. ones where all the basis states
have norm \(1\) and are mutually orthogonal.
To better understand the bra symbols we need to introduce the concept
of
dual of a vector space \(V\). Formally, the dual \(V^*\) of a
vector space \(V\) is the vector space of linear functionals from \(V\)
into \(\mathbb{C}\). Or in formulas,
\begin{multline}
V^* = \{\Phi \,:\,V\to \mathbb{C}\,\,\mbox{s.t.} \,\, \Phi(a
{\mathbf{z}} + b{\mathbf{w}}) = a \,\Phi({\mathbf{z}}) +
b\,\Phi({\mathbf{w}}), \\[1ex] \,\forall
\,a\,,b\in\mathbb{C}\,,\,\mbox{and}\,\forall\,
{\mathbf{z}}\,,{\mathbf{w}}\in V\}\,.
\end{multline}
\begin{multline}
V^* = \{\Phi \,:\,V\to \mathbb{C}\,\,\mbox{s.t.} \,\, \Phi(a
{\mathbf{z}} + b{\mathbf{w}}) \\ = a \,\Phi({\mathbf{z}}) +
b\,\Phi({\mathbf{w}}), \\[1ex] \,\forall
\,a\,,b\in\mathbb{C}\,,\,\mbox{and}\,\forall\,
{\mathbf{z}}\,,{\mathbf{w}}\in V\}\,.
\end{multline}
Show that this is indeed a vector space over the complex numbers and
show that if \(V\) has dimension \(n\) so does \(V^*\).
But this dual space is nothing mysterious. If \(V=\mathbb{C}^n\),
vectors in the standard basis are simple \(n\) dimensional column
vectors, i.e. \(n\times 1\) matrices if you wish, similarly you can
think of \(V^*\) as the vector space of \(1\times n\) matrices, i.e. row
vectors. Furthermore whenever \(V\) is endowed with a complex inner
product \(\langle\cdot,\cdot\rangle\), precisely as our Hilbert space of
states \(\mathcal{H}\), for each vector \(\mathbf{z}\in V\) we can
associate an element \(\Phi_\mathbf{z} \in V^*\) schematically as
\(\Phi_\mathbf{z} = \langle \mathbf{z},\cdot\rangle\)
\[
\Phi_\mathbf{z} ( \mathbf{w}) \doteq \langle \mathbf{z},\mathbf{w}\rangle
\]
for all \(\mathbf{w} \in V\). It is easy to check (just write down
what it means) that \(\Phi_\mathbf{z}(\cdot)\) just defined is indeed a
linear functional from \(V\) into \(\mathbb{C}\), hence \(\Phi_\mathbf{z}
\in V^*\). This means that as soon as our vector space \(V\) has a
complex inner product we have immediately an isomorphism between \(V\)
and \(V^*\)
\[
V\ni \mathbf{z} \mapsto \Phi_{\mathbf{z}} (\cdot) = \langle \mathbf{z},\cdot\rangle \,\in V^*\,.
\]
A ket \(|\psi\rangle\) can be thought of as a column-vector, and
then the bra \(\langle \phi|\) is a row vector, so that their inner
product is a scalar.
In quantum mechanics we have vectors, i.e. elements of \(\mathcal{H}\),
denoted by ket vectors \(\ket{\psi}\), and thanks to the complex inner
product for each vector \(\ket{\psi}\) we can consider the corresponding
element in the dual space \(\bra{\psi} \in \mathcal{H}^*\). A bra vector
\(\bra{\psi}\) applied to a ket vector \(\ket{\phi}\) gives precisely the
inner product \(\ip{\psi}{\phi}\).
Note in particular that it does
NOT make any sense
to consider \(\ket{\psi} + \bra{\phi}\) since one is a column vector
that cannot be added to the other object which is a row vector!
Suppose our quantum mechanical system is described by a
three-dimensional Hilbert space \(\mathcal{H}\) written in terms of the
orthonormal basis \(\ket{0},\ket{1},\ket{2}\), i.e. \(\mathcal{H} =
\mbox{span}\{\ket{0},\ket{1},\ket{2}\}\). We can represent any vector
in \(\mathcal{H}\) as a three dimensional column vector using the
standard basis
\begin{align*}
&\ket{0}\mapsto \left(\begin{matrix}1 \\ 0 \\0
\end{matrix}\right)\,, \ket{1}\mapsto \left(\begin{matrix}0 \\ 1 \\0
\end{matrix}\right)\,, \ket{2}\mapsto \left(\begin{matrix}0 \\ 0 \\1
\end{matrix}\right)\,,\\[1ex]
&\ket{\psi} = a \ket{0}+b\ket{1}+c\ket{2} \mapsto \left(\begin{matrix}a \\ b \\c
\end{matrix}\right)\,,\\
&\bra{\psi} = a^* \bra{0}+b^* \bra{1}+c^* \bra{2} \\
&\quad \mapsto \left(\begin{matrix}a^* & b^* &c^*
\end{matrix}\right) = \left(\begin{matrix}a \\ b \\c
\end{matrix}\right)^\dagger\,.
\end{align*}
To understand the second line let us remember that we are told that
three basis vectors are orthonormal hence we know that the matrix that
represents the inner product in this basis is given by the identity
matrix. At this point it is very simple to compute the inner product
between two states, say the inner product of \(\ket{\phi} = d
\ket{0}+e\ket{1}+f\ket{2}\) with \(\ket{\psi} = a
\ket{0}+b\ket{1}+c\ket{2}\)
\[
\ip{\phi}{\psi} = \left(\begin{matrix} d^* & e^* & f^* \end{matrix}\right)
\left(\begin{matrix}a\\b\\c\end{matrix}\right) = \left(\begin{matrix}d\\e\\f\end{matrix}\right)^\dagger\left(\begin{matrix}a\\b\\c\end{matrix}\right)\,,
\]
where remember \(A^\dagger = (A^*)^T\) is the transpose complex conjugate.
In particular we also see that if the ket \(\ket{\psi} = a
\ket{0}+b\ket{1}+c\ket{2} \) is represented by the column vector
\(\left( \begin{matrix} a\\b\\c\end{matrix}\right)\), then the bra
\(\bra{\psi}\) can really be thought of as \(\bra{\psi} =
(\ket{\psi})^\dagger\) and represented by the row vector
\((a,b,c)^\dagger = (a^*,b^*,c^*)\).
2.3.Hilbert space formalities
You can now write this all up in formal language if you want. A
quantum mechanical system is described by a ket \(|\psi\rangle\) in
Hilbert space \(\mathcal{H}\). A Hilbert space is a (complex) vector
space with Hermitian inner product. This means that for any
\(\ket{\psi} \in \mathcal{H}\) and \(\ket{\phi} \in \mathcal{H}\):
For any complex numbers \(a\) and \(b\), \((a\ket{\psi} + b\ket{\phi}) \in \mathcal{H}\).
(
linear combinations of vectors = quantum superposition)
The inner product of \(\ket{\psi}\) with \(\ket{\phi}\) is a complex number
denoted
\[\ip{\psi}{\phi} = \ip{\phi}{\psi}^* \in\mathbb{C}.\]
The inner product is Hermitean,
\[\ip{\psi}{\phi} = \big(\ip{\phi}{\psi}\big)^*.\]
The inner product is linear in the second state (and so anti-linear in
the first state). I.e. if \(\ket{\phi} = c_1\ket{\phi_1} + c_2\ket{\phi_2}\) then
\begin{eqnarray*}
\ip{\psi}{\phi} & = & c_1\ip{\psi}{\phi_1} + c_2\ip{\psi}{\phi_2} \,,\\[1ex]
\ip{\phi}{\psi} & = & c_1^*\ip{\phi_1}{\psi} + c_2^*\ip{\phi_2}{\psi} .
\end{eqnarray*}
In other words, the inner product is linear in the
second
factor, and anti-linear in the
first; it is
sesquilinear.
Note that in your linear algebra module you might have seen a slightly
different definition for an hermitian inner product which is linear in
the
first term! This is just a convention and in this module
we will keep the inner product to be linear in the second
term. Combining linearity in the second term with hermiticity tells us
that the inner product is not quite linear in the first term
This inner product is real, \(\ip{\psi}{\psi} \in
\mathbb{R}\). However, we also have a physical state condition
(and we will only consider such states in this module):
\(\ip{\psi}{\psi} \ge 0\) and \(\ip{\psi}{\psi} = 0 \iff \ket{\psi} =
0\). We will use the notation \(\norm{\psi} \equiv
\sqrt{\ip{\psi}{\psi}}\) for the norm of \(\ket{\psi}\).
Finally, states which differ only by a
normalisation factor are physically equivalent, i.e.
\[\ket{\psi} \sim c \ket{\psi}\]
for any non-zero \(c \in
\mathbb{C}\). There are two ways to work with this equivalence
relation. One is to ignore the normalisation but then include
appropriate factors of the norms of states in formulae. The other,
which we will usually assume, is to always work with
normalised
states, i.e. unless indicated otherwise a state \(\ket{\psi}\) will
be assumed to be have \(\norm{\psi} = 1\). If you have a state which is
not normalised, just divide it by its norm to get a normalised
state.
Physical states are
rays in Hilbert space, as the
normalisation is irrelevant (and often tuned so the norm of the
vector is one).
Note that normalisation does not fix a unique representative of
the equivalence class of states since multiplying by a phase \(\exp(i
\theta)\) for any real phase \(\theta\) does not change the norm,
i.e. \(\norm{\psi} = 1\) if and only if \(\vert\vert e^{i\theta}
\ket{\psi} \vert \vert = 1\).
Sometimes (pure) quantum mechanical states are called rays in
the Hilbert space because of the equivalence \(\ket{\psi} \sim c
\ket{\psi}\) with \(c\in\mathbb{C}\) non-zero.
NOTE: Obviously the zero state cannot be normalised but that
is OK as it does not describe the state of a physical system, and
there is no physical process to transform a non-zero state to the zero
state (
Do not confuse the zero state (meaning the
unique state with norm zero) with a state labelled by
zero, i.e. \(\ket{0} \ne 0\)! The norm of \(\ket{0}\) is always non-zero
\(|| \ket{0}|| =1\), while the norm of the \(0\) vector is always
vanishing \(||0||=0\).).
2.4.Operators
In quantum mechanics we work with linear operators acting on the
states in a Hilbert space. Such operators are used to describe the
time-evolution of the system and to describe measurements. If
\(\hat{A}\) is a linear operator then acting on linear combinations of
states we have
\[ \hat{A} (a\ket{\psi} + b \ket{\phi}) = a(\hat{A}\ket{\psi}) + b(\hat{A}\ket{\phi})\,, \]
i.e. it is linear.
Also, products and linear combinations of linear operators, are again linear
operators.
The
adjoint (also commonly called the Hermitian conjugate) of \(\hat{A}\) is denoted \(\hat{A}^{\dagger}\) and defined by
\[ \left\langle \psi \vert \left( \hat{A}^{\dagger} \vert \phi\right\rangle \right)= \left[ \left\langle \phi \vert \left( \hat{A} \vert \psi \right\rangle \right)\right]^* \]
for all states \(\ket{\psi}\) and \(\ket{\phi}\).
In quantum mechanics, we usually focus on two types of linear operators:
- self-adjoint operators
(or Hermitian) meaning \(\hat{H}^{\dagger} = \hat{H}\). Self-adjoint operators correspond
to observables, i.e. quantities which can be measured. E.g. \(\hat{X}\) position operator, \(\hat{P}\) momentum operator, \(\hat{H}\) Hamiltonian operator, \(\hat{S}\) spin operator.
The reason for that comes from the fact that hermitian operators have real eigenvalues.
- unitary operators
meaning \(\hat{U}^{\dagger}\hat{U} = \hat{I}\).
Unitary operators are used to describe time-evolution in quantum mechanics.
Show that the eigenvalues of a self-adjoint operator \(\hat{H}\) must be
real. Show that eigenstates of a self-adjoint operator corresponding
to different eigenvalues are automatically orthogonal to each others.
Note: when we pass to matrices: self-adjoint operators become
hermitean matrices, i.e. \(H^\dagger = (H^T)^\star = H\) while unitary operators become
unitary matrices, i.e. \(U^\dagger U = U U^\dagger =I\). Do not confuse these two properties!
Given the matrices
\begin{align*}
A & = \left( \begin{matrix} 1 & i \\ -i & 2 \end{matrix} \right)\,,\\
B & = \left( \begin{matrix} \cos(\sqrt{5}) + \frac{2i \sin(\sqrt{5})}{\sqrt{5}} & \frac{\sin(\sqrt{5})}{\sqrt{5}} \\ - \frac{\sin(\sqrt{5})}{\sqrt{5}} & \cos(\sqrt{5}) - \frac{2i \sin(\sqrt{5})}{\sqrt{5}} \end{matrix} \right)\,,\\
C & = \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)\,,\\
D &= \left(\begin{matrix} 1 & i \\ 1+i & 3 \end{matrix} \right)\,,
\end{align*}
check that \(A\) is hermitean, \(B\) is unitary, \(C\) is both unitary and hermitean, \(D\) is neither unitary nor hermitean.
Find a basis for the vector space (check that it is indeed a vector
space!) of the \(2\times 2\) Hermitean matrices.
We also define the commutator of two operators \(\hat{A}\) with \(\hat{B}\) as:
\[ \com{A}{B} \equiv \hat{A}\hat{B} - \hat{B}\hat{A} . \]
There is also a similar definition of the anti-commutator
\[ \acom{A}{B} \equiv \hat{A}\hat{B} + \hat{B}\hat{A} . \]
The expectation value of an operator gives the average
outcome of the measurement of the corresponding observable, if we
start with the same system many times over.
The
expectation value of an observable \(\hat{A}\) on a state
\(\psi\) denoted by \(\langle A\rangle_\psi\) is given by
\[
\langle A\rangle_\psi = \bra{\psi} \hat{A} \ket{\psi}\,.
\]
In most cases if there is no confusion regarding which state we are considering we will drop the subscript and simply write \(\langle A\rangle\).
This expectation value can really be interpreted as the average outcome of many measurements of the same observable \(\hat{A}\) on the same state \(\ket{\psi}\), i.e. prepare \(1000\) times the same state \(\ket{\psi}\), measure \(1000\) times the same observable \(\hat{A}\) and then take the average.
Note that the expectation value is clearly a real number
\begin{equation}
\begin{aligned}
\langle A\rangle_\psi^* &= (\bra{\psi} \hat{A} \ket{\psi})^* = (\bra{\psi} \hat{A}^\dagger \ket{\psi}) \\[1ex]
& = (\bra{\psi} \hat{A} \ket{\psi})=\langle A\rangle_\psi\,,
\end{aligned}
\end{equation}
where the hermiticity of the inner product, i.e. \({\ip{\psi}{\phi}}^* = \ip{\phi}{\psi}\), and the hermiticity of \(\hat{A}\), \(\hat{A}^\dagger= \hat{A}\), both play a crucial role.
2.5.Matrix representation
As already mentioned above, for an \(N\)-dimensional Hilbert space, we
can represent states using ket vectors or complex \(N\)-component column
vectors, similarly for bra vectors we need \(N\)-component row vectors.
A standard choice is to represent the basis states by column vectors
with all components zero, except a single ‘one’. Then, provided we are
using an orthonormal basis (so that the matrix that represents the
inner product is given just by the identity matrix), the inner product
of two states represented by column vectors \(\vec{u}\) and \(\vec{v}\) is
given by standard matrix multiplication as \(\vec{u}^{\dagger}
\vec{v}\). Linear operators are then represented by \(N \times N\)
matrices.
By writing out the action of an operator on each of the
basis vectors in Hilbert space, we can construct its
matrix
representation.
In such a representation self-adjoint operators
\(\hat{H}^\dagger=\hat{H}\) are indeed Hermitian matrices \(H^\dagger=
H\), and unitary operators \(\hat{U}^\dagger \hat{U} = \hat{I}\) are
unitary matrices \(U^\dagger U = I\) where \(I\) is the identity matrix of
the appropriate dimension. Note that in most cases we will keep the
hat symbol \(\hat{\phantom{A}}\), as in \(\hat{A}\), to denote the
abstract operator without having having picked any particular basis to
be represented as the standard one, once we choose a particular
orthonormal basis to be represented via the standard one we will refer
to the matrix representing the operator \(\hat{A}\) in this basis with
the same letter but without the hat, i.e. just \(A\). Note that we will
be playing a lot with different basis so although the abstract
operator is one \(\hat{A}\) it might be represented by different
matrices \(A_1\,,\,A_2\,,...\) according to which basis we choose!
However we should remember from Linear Algebra I that if we change
basis, say from \(\{\vec{v}_1,...\vec{v}_2\}\) to
\(\{\vec{w}_1,...,\vec{w}_n\}\), then the matrix \(A_2\) representing the
linear transformation \(\hat{A}\) in the new basis is related to the
matrix \(A_1\) representing the same linear transformation \(\hat{A}\) but
in the old basis via
\[
A_2 =S^{-1} A_1 S\,,
\]
where \(S\) is the change of basis matrix to go from the new basis to the old one.
Let \(\mathcal{H} = \mbox{span}\{ \ket{0},\ket{1}\}\) be a \(2\)-dimensional Hilbert space of states, and assume that the basis vector are orthonormal.
We are given a linear operator \(\hat{A}\) defined on this basis
\begin{multline}
\hat{A} \ket{0} = a \ket{0} + b\ket{1}\,, \qquad\hat{A}\ket{1} = c \ket{0}+ d \ket{1}\,,
\end{multline}
\begin{multline}
\hat{A} \ket{0} = a \ket{0} + b\ket{1}\,,\\ \qquad\hat{A}\ket{1} = c \ket{0}+ d \ket{1}\,,
\end{multline}
with \(a,b,c,d\in\mathbb{C}\).
First of all we can pass to the vector/matrix representation by using
coordinates. We chose to represent the first basis vector \(\ket{0}
\to \left(\begin{matrix} 1\\0\end{matrix}\right)\) and the second one
as \(\ket{1} \to \left(\begin{matrix} 0\\1\end{matrix}\right)\). Once
we make this choice of basis the operator \(\hat{A} \) can be
represented as the \(2\times 2\) matrix
\[
A= \left(\begin{matrix} a & c \\b & d\end{matrix}\right)\,,
\]
and the abstract form \(\hat{A} \ket{0} = a \ket{0} + b\ket{1}\) can be simply stated in matrix language as
\(A\left(\begin{matrix} 1\\0\end{matrix}\right) = \left(\begin{matrix} a & c \\b & d\end{matrix}\right)\left(\begin{matrix} 1\\0\end{matrix}\right) = \left(\begin{matrix} a\\b\end{matrix}\right)\), and similarly for the other basis vector.
Let us now compute the adjoint \(\hat{A}^\dagger\) of \(\hat{A}\) and find which conditions we have to impose on the coefficients \(a,b,c,d\) such that \(\hat{A}\) becomes self-adjoint.
If we want to compute \(\hat{A}^\dagger\) we need to know what this operator does on a basis, i.e. we need to compute \(\hat{A}^\dagger \ket{0}\) and \(\hat{A}^\dagger\ket{1}\).
So we can write
\[
\hat{A}^\dagger \ket{0} = \alpha \ket{0} +\beta\ket{1}\,,\qquad \hat{A}^\dagger \ket{1} = \gamma \ket{0}+\delta\ket{1}\,,
\]
for some yet undetermined \(\alpha,\beta,\gamma,\delta\in\mathbb{C}\).
To fix these coefficients we need to remember that \(\ket{0},\ket{1}\) form an orthonormal basis so we have
\begin{align*}
&\bra{0} \hat{A}^\dagger \ket{0} =\bra{0}\left( \alpha \ket{0} +\beta\ket{1}\right) = \alpha\,,\\
&\bra{1} \hat{A}^\dagger \ket{0} =\bra{1}\left( \alpha \ket{0} +\beta\ket{1}\right) = \beta\,,\\
&\bra{0} \hat{A}^\dagger \ket{1} =\bra{0}\left( \gamma \ket{0} +\delta\ket{1}\right) = \gamma\,,\\
&\bra{1} \hat{A}^\dagger \ket{1} =\bra{1}\left( \gamma \ket{0} +\delta\ket{1}\right) = \delta\,.
\end{align*}
Finally we need to remember the definition \(\ipop{\psi}{\hat{A}^{\dagger}}{\phi} = \ipop{\phi}{\hat{A}}{\psi}^* \)
so we have
\begin{align*}
&\alpha = \bra{0} \hat{A}^\dagger \ket{0} =( \bra{0} \hat{A} \ket{0} )^*=a^* \,,\\
&\beta=\bra{1} \hat{A}^\dagger \ket{0} ( \bra{0} \hat{A} \ket{1} )^* = c^*\,,\\
&\gamma= \bra{0} \hat{A}^\dagger \ket{1}=( \bra{1} \hat{A} \ket{0} )^* = b^* \,,\\
&\delta=\bra{1} \hat{A}^\dagger \ket{1}=( \bra{1} \hat{A} \ket{1} )^*=d^*\,.
\end{align*}
Note the order of the vectors being flipped!
We have then
\[
\hat{A}^\dagger \ket{0} = a^* \ket{0} + c^* \ket{1}\,,\qquad \hat{A}^\dagger \ket{1} = b^* \ket{0}+d^*\ket{1}\,.
\]
Using the same basis as above this operator can be represented as the \(2\times 2\) matrix
\[
A^\dagger= \left(\begin{matrix} a^* & b^* \\c^* & d^*\end{matrix}\right)\,.
\]
When passing to coordinates the matrix representing the adjoint operator \(\hat{A}^\dagger \) is exactly \(A^\dagger = (A^*)^T\), i.e. the transpose complex conjugate of the matrix \(A\) representing the operator \(\hat{A}\).
Finally if we want the operator to be self-adjoint we must have \(\hat{A}^\dagger =\hat{A}\) which imposes \(a=a^*,\,b=c^*,\,d=d^*\). These conditions are identical to imposing that the matrix representing \(\hat{A}\) is an hermitian matrix, i.e. \(A=A^\dagger\).
Once we realise we are just doing linear algebra we can easily understand what happens when we change basis.
Suppose for example we are given the operator
\[
\hat{B} \ket{0} = 2i \ket{0} + 5 \ket{1}\,,\qquad\hat{B}\ket{1} = -3 \ket{0}+ (1+i) \ket{1}\,,
\]
which can be represented in the same basis as above by the matrix \(B= \left(\begin{matrix}2i & -3\\5& 1+i
\end{matrix}\right)\).
We want to use now a new orthonormal basis defined by \(\ket{\pm} = \frac{1}{\sqrt{2}} (\ket{0}\pm\ket{1})\)
(check that this is indeed an orthonormal basis) or equivalently \(\ket{0} = \frac{1}{\sqrt{2}}(\ket{+}+\ket{-}),\,\ket{1} = \frac{1}{\sqrt{2}}(\ket{+}-\ket{-})\).
We can proceed in two ways.
One possibility is to rewrite the action of \(\hat{B}\) in this new basis
\begin{align*}
\hat{B} \ket{+} &= \frac{1}{\sqrt{2}} \hat{B}( \ket{0}+\ket{1}) \\
& = \frac{1}{\sqrt{2}}((-3+2i) \ket{0} +(6+i)\ket{1})\\
& = \frac{(-3+2i)}{2}(\ket{+}+\ket{-}) + \frac{(6+i)}{2} (\ket{+}-\ket{-}) \\
& = \frac{3+3i}{2}\ket{+}+\frac{-9+i}{2}\ket{-}\,,\\
\hat{B} \ket{-} &= \frac{1}{\sqrt{2}} \hat{B}( \ket{0}-\ket{1}) \\
& = \frac{1}{\sqrt{2}}((3+2i) \ket{0} +(4-i)\ket{1})\\
& = \frac{(3+2i)}{2}(\ket{+}+\ket{-}) + \frac{(4-i)}{2} (\ket{+}-\ket{-}) \\
& = \frac{7+i}{2}\ket{+}+\frac{-1+3i}{2}\ket{-}\,.
\end{align*}
Hence in the new basis represented by \(\ket{+} \to \left(\begin{matrix} 1\\0\end{matrix}\right),\,\ket{-} \to \left(\begin{matrix} 0\\1\end{matrix}\right)\) the new matrix \(\tilde{B}\) representing the
same operator \(\hat{B}\) now takes the form
\[
\tilde{B} = \frac{1}{2}\left(\begin{matrix}3+3i & 7+i\\
-9+i & -1+3i \end{matrix}\right)\,.
\]
We could have reached the same conclusion noting that we are just making a change of basis from \(\{\ket{0},\ket{1}\}\) to \(\{\ket{+},\ket{-}\}\) and the change of basis matrix is simply given by
\[
S= \frac{1}{\sqrt{2}} \left(\begin{matrix} 1 & 1 \\
1 & -1\end{matrix}\right)\,,
\]
hence the matrix representation \(\tilde{B}\) of the same operator \(\hat{B}\) but in the new basis, is simply given by \(S^{-1} B S\) where \(B\) is the matrix representation of \(\hat{B}\) in the old basis. This matrix multiplication produces exactly the same matrix \(\tilde{B}\) just computed above.
Change of basis will play a crucial role in the discussion of qubits.
We will have a set of privileged operators and we will keep on
changing from a basis of eigenvectors for one such operator to a basis
of eigenvectors for another of these operators. Every time we change
basis the matrix representing these operators will change according to
a change of basis transformation, i.e. \(S^{-1} B S\).
2.6.Time-evolution
In quantum mechanics the time-evolution of the system is governed by a self-adjoint
operator called the Hamiltonian \(\hat{H}\). In this module we work in
the Schrödinger picture, so a state in the system evolves with time,
and we can consider a state \(\ket{\psi(t)}\) (think of it as a
time-dependent vector). The time-evolution is described by the
Schrödinger equation
\[ i\hbar\frac{{\rm d}}{{\rm d} t}\ket{\psi(t)} = \hat{H} \ket{\psi(t)} . \]
This can also be written in an integrated form to define the state in terms of
some initial state, say at time \(t\gt{}0\) in terms of the state at \(t=0\):
\[ \ket{\psi(t)} = \hat{U}_t \ket{\psi(0)} \]
where \(\hat{U}_t\) is a unitary operator. In the case where the Hamiltonian
operator is not time-dependent we have
\[ \hat{U}_t = \exp \left( -\frac{i}{\hbar} t \,\hat{H} \right) . \]
In quantum information we usually assume complete control over a
quantum system or subsystem. This means that we can interact with the
system in a arbitrary way, e.g. by rotating it, applying electric or
magnetic fields etc. In terms of time evolution this means that we
assume we have the ability to transform the state of the system
\[ \ket{\psi} \rightarrow \hat{U} \ket{\psi} \]
using any unitary operator \(\hat{U}\) we want. As such we will usually talk
about transformations by a unitary operator \(\hat{U}\), rather than in terms
of a Hamiltonian operator with evolution for some specific period of time.
2.6.1.Exponential of operators
In this Section we defined the time evolution operator in terms of the exponential of the Hamiltonian operator.
This is a general concept:
Def:
The exponential of a matrix \(A\), or more generally of an operator \(\hat{A}\), is defined by the Taylor series
\[
\exp( \hat{A}) = \sum_{n=0}^\infty \frac{\hat{A}^n}{n!} = \hat{I} + \frac{\hat{A}}{1!}+ \frac{\hat{A}^2}{2!} +...\,,
\]
where \(\hat{I}\) denotes the identity operator. Note that for the operators we will consider this series will always converge.
Suppose \(A = \mbox{diag}(\lambda_1,...\lambda_N)\) be a diagonal \(N\times N\) matrix with \(\lambda_i \in\mathbb{C}\).
Let us compute \(e^A\). To this end we need to compute \(A^n\) which is \(A \cdot A \cdot...\cdot A\) n-times.
In general this is a difficult task but for \(A\) diagonal it is actually very simple \(A^n = \mbox{diag}(\lambda_1^n,...\lambda_N^n)\) hence
\begin{align*}
e^A &= \sum_{n=0}^\infty \frac{A^n}{n!} = \sum_{n=0}^\infty \frac{1}{n!} \mbox{diag}(\lambda_1^n,...\lambda_N^n) \\
&= \mbox{diag}( \sum_{n=0}^\infty \frac{\lambda_1^n}{n!},\sum_{n=0}^\infty \frac{\lambda_2^n}{n!},...,\sum_{n=0}^\infty \frac{\lambda_N^n}{n!}) \\
&= \mbox{diag}( e^{\lambda_1},...,e^{\lambda_N})\,.
\end{align*}
So \(e^A\) is once again diagonal with diagonal elements simply given by the exponential of the diagonal elements of \(A\).
Consider
\[N= \left(\begin{matrix} 0 & 1 &0 \\ 0 & 0 & 1\\ 0& 0& 0\end{matrix}\right)\]
and compute \(e^{t N}\) with \(t\in\mathbb{R}\).
First we notice that \(N\) is a nilpotent matrix, i.e. we can find \(m\in\mathbb{N}\) such that \(N^m = 0\), in particular in this case
\[N^2 = \left(\begin{matrix} 0 & 0 &1 \\ 0 & 0 & 0\\ 0& 0& 0\end{matrix}\right)\]
and \(N^3 = 0\), i.e. the \(3\times 3\) zero matrix.
In this case the exponential series truncates after finitely many terms
\[
e^{t N} = \sum_{n=0}^\infty \frac{t^n N^n}{n!} = \mathbb{I}_3+ t N +\frac{t^2 N^2}{2!} = \left(\begin{matrix} 1 & t &\frac{t^2}{2} \\ 0 & 1 & t\\ 0& 0& 1\end{matrix}\right)\,.
\]
Note in particular that \(e^{t N}\) is
NOT simply given by the exponential of each entries of \(t N\)!
Let \(\sigma_2 = \left(\begin{matrix} 0 & -i \\ i & 0\end{matrix}\right)\), we want to compute
\(U_2(\alpha) = e^{i \alpha \sigma_2}\) and show that this is a unitary matrix for every \(\alpha\in\mathbb{R}\).
First we notice that \((i \sigma_2)^2= -\mathbb{I}_2\) hence \((i \sigma_2)^{2n} = (-1)^n \mathbb{I}_2\) while
\((i \sigma_2)^{2n+1} = (-1)^n (i \sigma_2)\).
This alternating pattern between even and odd powers allows us to evalute the exponential
\begin{align*}
U_2(\alpha) &= e^{i \alpha \sigma_2} = \sum_{n=0}^\infty \frac{\alpha^n (i \sigma_2)^n}{n!} \\
&= \sum_{n\,\text{even}} +\sum_{ n\,\text{odd}} = \sum_{n=0}^\infty \frac{(-1)^n \alpha^{2n}}{(2n)!} \mathbb{I}_2 + \sum_{n=0}^\infty \frac{(-1)^n \alpha^{2n+1}}{(2n+1)!} (i\sigma_2) \\
&= \cos(\alpha) \mathbb{I}_2+ i\sigma_2 \sin(\alpha) = \left(\begin{matrix}\cos(\alpha) & \sin(\alpha)\\ -\sin(\alpha) & \cos(\alpha) \end{matrix}\right)\,.
\end{align*}
In the last line we have used the Taylor expansion for the sine and cosine functions
\begin{align*}
\cos(\alpha) &= \sum_{n=0}^\infty \frac{(-1)^n \alpha^{2n}}{(2n)!}\,,\\
\sin(\alpha) &= \sum_{n=0}^\infty \frac{(-1)^n \alpha^{2n+1}}{(2n+1)!}\,.\\
\end{align*}
Finally it is easy to check that \(U_2(\alpha)^\dagger U_2(\alpha) = U_2(\alpha) U_2(\alpha)^\dagger = \mathbb{I}_2\) so the matrix \(U_2(\alpha)\) is indeed a unitary matrix and hence a possible time evolution operator of a \(2\)-dimensional system. We will see that this type of unitary time evolution will be crucial for the study of the qubit systems later on.
A final comment: From these examples it should be clear
that in general to compute the exponential of a matrix you
CANNOT simply compute the exponential of each entry! For more
on exponentials of matrices and the various ways to compute them, see
e.g. [
4].
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