Problem Set 9 - Siavash Davani's Website

Exercise 1 (Functions of operators in infinite-dimensional Hilbert space)

As discussed in the previous problem set, a function of an operator is an object that takes an operator as input and outputs another operator

\hat{A} \quad\longrightarrow\quad f(\hat{A}).

(1)

The term $f(\hat{A})$ is defined using the Taylor expansion of $f$ as

f(\hat{A}) = \sum_{n=0}^\infty c_n \, \hat{A}^n,

(2)

where $c_n$ are the coefficients taken from the power series expansion of the usual expansion of the function in calculus

f(x) = \sum_{n=0}^\infty c_n \, x^n.

(3)

In the last problem set, we mainly focused on operators in finite-dimensional vector spaces. However, many of the results transition into the infinite-dimensional case as we will see in this problem set. In this set, we take the vector space to be the Hilbert space of square-integrable functions $L_2$ .

In the infinite-dimensional case, it is not possible to represent vectors as a column of numbers and operators as matrices. Therefore, we shall be a bit more careful when we talk about functions of operators. The form written in Eq. 2 is actually very useful because it is an algebraic equation and does not depend on a particular representation of the operator $\hat{A}$ .

One of the main results from the finite-dimensional case was that, if $\ket{v}$ is an eigenvector to $\hat{A}$ with the eigenvalue λ, $\ket{v}$ will also be an eigenvector to any function of $\hat{A}$ as

f(\hat{A})\ket v = f(\lambda)\ket v.

(4)

This is a result that is valid also in the infinite-dimensional case as you will prove now!

We know that the position basis states $\ket{e_x}$ are eigenstates to the position operator as
$\hat{x}\ket{e_x} = x\ket{e_x}.$
(5)
Use this property and Eq. 2 to show that $\ket{e_x}$ will be an eigenstate to any function of $\hat{x}$ as
$f(\hat{x})\ket{e_x} = f(x)\ket{e_x}.$
(6)
Note that this is correct only if $f(x)$ is defined for all $x$ . This is always the case in the situations we are interested in, therefore it will not cause us any problem.
Solution
Applying the definition of the function of the operator, we have
$\begin{align} f(\hat{x})\ket{e_x} &= \sum_{n=0}^\infty c_n \, \hat{x}^n \ket{e_x} \\ &= \sum_{n=0}^\infty c_n \, \Big[ \hat{x}^n \ket{e_x} \Big] \\ &= \sum_{n=0}^\infty c_n \, \Big[ x^n \ket{e_x} \Big] \\ &= \sum_{n=0}^\infty c_n \, x^n \ket{e_x} \\ &= \Big[ \sum_{n=0}^\infty c_n \, x^n \Big] \ket{e_x} \\ &= f(x) \, \ket{e_x} \\ \end{align}$
(7)

Now that we know the action of the operator $f(\hat{x})$ on the position basis states, we can evaluate its action on any arbitrary state by simply decomposing our state in that basis.

Take a particular function of the operator $\hat{x}$ as
$g(\hat{x}) = e^{i\alpha\hat{x}}.$
(8)
When this operator acts on a state $\ket{\psi}$ , it outputs another state
$\ket{\psi'}=g(\hat{x})\ket{\psi}.$
(9)
Represent the state $\ket{\psi}$ using its corresponding wavefunction $\psi(x)=\braket{x}{\psi}$ in the position basis
$\ket{\psi} = \intinf \psi(x)\ket{e_x} \, dx$
(10)
to show that the wavefunction corresponding to the new state $\ket{\psi'}$ is
$\psi'(x) = \braket{e_x}{\psi'} = \psi(x) e^{i\alpha x}.$
(11)
Solution
First, we evaluate the following term
$\begin{align} \ket{\psi'} &= g(\hat{x})\ket{\psi} \\ &= g(\hat{x})\intinf \psi(x)\ket{e_x} \, dx \\ &= \intinf \psi(x) \, g(\hat{x}) \, \ket{e_x} \, dx \\ &= \intinf \psi(x) \, g(x) \,\ket{e_x} \, dx \\ &= \intinf \psi(x) \, e^{i\alpha x} \,\ket{e_x} \, dx \end{align}$
(12)
This immediately shows that the wavefunction corresponding to $\ket{\psi'}$ is $\psi(x) \, e^{i\alpha x}$ . But if we would like to see that explicitly, we should calculate
$\begin{align} \psi'(x) &= \braket{e_x}{\psi'} \\ &= \bra{e_x} \intinf \psi(x') \, g(x') \,\ket{e_{x'}} \, dx' \\ &= \intinf \psi(x') \, g(x') \, \braket{e_x}{e_{x'}} \, dx' \\ &= \intinf \psi(x') \, g(x') \, \delta(x-x') \, dx' \\ &= \psi(x) \, g(x) \\ &= \psi(x) \, e^{i\alpha x} \end{align}$
(13)

This is a very interesting result. The action of $f(\hat{x})$ on $\ket{\psi}$ will simply transform the position wavefunction by multiplying it by $f(x)$

\ket{\psi'}=f(\hat{x})\ket{\psi} \quad\longleftrightarrow\quad \psi'(x) = f(x)\psi(x)

(14)

We can see that this is very powerful because the basis $\ket{e_x}$ is complete, meaning that every arbitrary state $\ket \psi$ from the vector space can be fully decomposed in that basis. This gives us the possibility to evaluate the action of functions of $\hat{x}$ on any state $\ket \psi$ .

We have similar results for the momentum operator $\hat{p}$ and functions of it $h(\hat{p})$ . We only need to represent everything in terms of the momentum basis states.

Assume we are given a function of the momentum operator as $h(\hat{p})$ . When it acts on a state $\ket{\psi}$ , we have
$\ket{\psi'}=h(\hat{p})\ket{\psi}.$
(15)
If we define the original momentum wavefunction as $\phi(p)=\braket{e_p}{\psi}$ and the momentum wavefunction after the operation as $\phi'(p)=\braket{e_p}{\psi'}$ , show that we have
$\phi'(p) = h(p)\phi(p).$
(16)
Hint: Represent the state in the momentum basis states
$\ket{\psi} = \intinf \phi(p)\ket{e_p} \, dp,$
(17)
and put it in Eq. 15.
Solution
The derivations are identical to the previous case with the only difference that we have to represent everything in the momentum basis. We should evaluate
$\begin{align} \ket{\psi'} &= h(\hat{p})\ket{\psi} \\ &= h(\hat{p})\intinf \phi(p)\ket{e_p} \, dp \\ &=\intinf \phi(p) \, h(\hat{p}) \, \ket{e_p} \, dp \\ &=\intinf \phi(p) \, h(p) \, \ket{e_p} \, dp \\ \end{align}$
(18)
which is the representation of the $\ket{\psi'}$ state in the momentum basis meaning that we have
$\phi'(p) = \braket{e_p}{\psi'} = h(p)\phi(p)$
(19)

The results we proved so far show that when we want to evaluate the action of a function of operators like $\hat{x}$ or $\hat{p}$ on a state, it is always useful to represent everything in the corresponding basis of that operator because the action will become trivial.

We now know how to evaluate the action of $h(\hat p)$ on a state

\ket{\psi'}=h(\hat{p})\ket{\psi} \quad\longleftrightarrow\quad \phi'(p) = h(p)\phi(p).

(20)

Although this relation completely describes the action, it might not be intuitive in all cases because it is a relation between the momentum wavefunctions in the Fourier domain. But how can we know what will the final position wavefunction $\psi'(x)=\braket{e_x}{\psi'}$ be after a function of the momentum operator $h(\hat{p})$ acts on the state

\ket{\psi'}=h(\hat{p})\ket{\psi} \quad\longleftrightarrow\quad \psi'(x) = \,? \quad\quad\quad\,\,

(21)

The answer to this question is not always straightforward because we have to take an inverse Fourier transform of $\phi'(p)$ from Eq. 20 to get our answer

\psi'(x) = \mathcal{FT}^{-1} \Big\{ \phi'(p) \Big\}

(22)

But let us see a particular example of such operators in the following where we can explicitly find a closed form for the final position wavefunction.

Let $\hat{T}(x_0)$ be a function of the momentum operator and a scalar $x_0$ as
$\hat{T}(x_0) = e^{-i\hat{p} x_0 / \hbar}.$
(23)
Show that when this operator acts on position basis states, it shifts them by $x_0$ . That is
$\ket{e_{x+x_0}} = \hat{T}(x_0)\ket{e_{x}}.$
(24)
Because of this property, the $\hat{T}$ operator is called the translation operator.
Hint: Use the fact that the position basis states are represented in the momentum basis as
$\ket{e_x} = \frac{1}{\sqrt{2\pi\hbar}} \intinf e^{-ipx/\hbar} \ket{e_p} \, dp.$
(25)
Solution
The key idea here is that we do not know the action of the $\hat{T}(x_0)$ operator directly acting on the position basis states $\ket{e_x}$ but we know its action on the momentum basis states $\ket{e_p}$ because $\hat{T}(x_0)$ is a function of the momentum operator $\hat{p}$ . Therefore we expand the state $\ket{e_x}$ in terms of the momentum basis states to make the action of the operator trivial. We can thus write
$\begin{align} \hat{T}(x_0)\ket{e_{x}} &= \hat{T}(x_0) \frac{1}{\sqrt{2\pi\hbar}} \intinf e^{-ipx/\hbar} \ket{e_p} \, dp \\ &= \frac{1}{\sqrt{2\pi\hbar}} \intinf e^{-ipx/\hbar} \, \Big[ \hat{T}(x_0)\ket{e_p} \Big] \, dp \\ &= \frac{1}{\sqrt{2\pi\hbar}} \intinf e^{-ipx/\hbar} \, \Big[ e^{-i\hat{p} x_0 / \hbar} \, \ket{e_p} \Big] \, dp \\ &= \frac{1}{\sqrt{2\pi\hbar}} \intinf e^{-ipx/\hbar} \, \Big[ e^{-ip x_0 / \hbar} \, \ket{e_p} \Big] \, dp \\ &= \frac{1}{\sqrt{2\pi\hbar}} \intinf e^{-ipx/\hbar} \, e^{-ip x_0 / \hbar} \, \ket{e_p} \, dp \\ &= \frac{1}{\sqrt{2\pi\hbar}} \intinf e^{-ip(x+x_0)/\hbar} \, \ket{e_p} \, dp \\ &= \ket{e_{x+x_0}} \end{align}$
(26)
Now that we know the action of $\hat{T}(x_0)$ on the position basis states, given an arbitrary state
$\ket{\psi} = \intinf \psi(x)\ket{e_x} \, dx,$
(27)
show that the $\hat{T}(x_0)$ operator will shift the whole wavefunction by $x_0$ when it acts on $\ket{\psi}$ . That is,
$\ket{\psi'}=\hat{T}(x_0)\ket{\psi} \quad\longleftrightarrow\quad \psi'(x) = \psi(x-x_0).$
(28)
Solution
We just need to apply the result of the previous part.
$\begin{align} \ket{\psi'} &= \hat{T}(x_0)\ket{\psi} \\ &= \hat{T}(x_0) \intinf \psi(x)\ket{e_x} \, dx \\ &= \intinf \psi(x) \, \hat{T}(x_0) \, \ket{e_x} \, dx \\ &= \intinf \psi(x) \, \ket{e_{x+x_0}} \, dx \\ \intertext{using a change of variables $x'=x+x_0$, we have} &= \intinf \psi(x'-x_0) \, \ket{e_{x'}} \, dx' \end{align}$
(29)
which means
$\psi'(x) = \braket{e_x}{\psi'} = \psi(x-x_0)$
(30)

Exercise 2 (Time evolution operator)

Now that we are fully equipped with the notion of functions of operators, we are ready to introduce the time evolution operator.

Let us first review what we learned in the lectures. We know the dynamics of a wavefunction is governed by the time-dependent Schrodinger equation

i\hbar\pdv{\psi(x,t)}{t} = -\frac{\hbar^2}{2m}\ppdv{\psi(x,t)}{x} + V(x) \psi(x,t).

(31)

To solve this differential equation, we used the technique of separation of variables and looked for stationary solutions

\psi(x,t)=\psi_n(x)e^{-iE_nt/\hbar}.

(32)

If we put this form of the wavefunction in the time-dependent Schrodinger equation, we will derive the time-independent Schrodinger equation

E_n\psi_n(x) = -\frac{\hbar^2}{2m}\ppdv{\psi_n(x)}{x} + V(x) \psi_n(x)

(33)

which is an eigenvalue problem for the $\ket{\psi_n}$ state

\hat{H}\ket{\psi_n} = E_n\ket{\psi_n}.

(34)

where the Hamiltonian operator is $\hat{H}=\frac{\hat{p}^2}{2m} + V(\hat{x})$ , the wavefunction is defined as $\psi_n(x)=\braket{e_x}{\psi_n}$ , and $\ket{\psi_n}$ is the state associated with a particular energy $E_n$ .

The goal of solving the Schrodinger equation was to calculate $\psi(x,t)$ for any $t$ , given an initial wavefunction $\psi(x)=\psi(x,0)$

\psi(x) \quad\longrightarrow\quad \psi(x,t).

(35)

Knowing the energy eigenstates, this was done by first decomposing the initial wavefunction as a linear combination of the energy eigenstates

\psi(x,0)=\sum_n c_n\psi_n(x).

(36)

And then the time-evolved wavefunction was simply

\psi(x,t)=\sum_n c_n\psi_n(x) \, e^{-iE_nt/\hbar},

(37)

as if each $\psi_n(x)$ wavefunction is evolving independently according to Eq. 32.

Given this summary, let us reinterpret the goal in Eq. 35 using linear algebra and function spaces. Each of the wavefunctions in Eq. 35 correspond to a state in our Hilbert space. We can call the initial state $\ket{\psi}$ and the state after the time evolution $\ket{\psi(t)}$ . This means Eq. 35 will become

\ket{\psi} \quad\longrightarrow\quad \ket{\psi(t)}.

(38)

The purpose of this problem is to find an operator $\hat{U}(t)$ that relates these two states

\ket{\psi(t)} = \hat{U}(t) \ket{\psi},

(39)

which means the $\hat{U}(t)$ acts on the state $\ket{\psi}$ at time $t=0$ and outputs the evolved state at an arbitrary time $t$ . The operator is therefore called the time evolution operator.

Use the notion of functions of operators from previous problems to show that the relation in Eq. 32 can be rewritten as
$\ket{\psi(t)} = e^{-i\hat{H}t/\hbar} \ket{\psi_n}.$
(40)
Note that $\ket{\psi(t)}$ is the state corresponding to the time-evolved wavefunction $\psi(x,t)$ ; that is $\psi(x,t) = \braket{e_x}{\psi(t)}$ .
Hint: Use the fact that $\ket{\psi_n}$ is an eigenstate to the Hamiltonian operator $\hat{H}$ .
Solution
We find ourselves again using the fundamental lemma that when a function of an operator acts on the eigenstates of that operator, the action is trivial and it will only multiply the state with the corresponding scalar. In this case, the $\ket{\psi_n}$ state is an eigenstate to the Hamiltonian operator $\hat{H}$ with the eigenvalue of $E_n$ . Therefore, we have
$e^{-i\hat{H}t/\hbar} \ket{\psi_n} = e^{-i E_n t/\hbar} \ket{\psi_n},$
(41)
which results in the following equation
$\ket{\psi(t)} = e^{-i\hat{H}t/\hbar} \ket{\psi_n} = e^{-i E_n t/\hbar} \ket{\psi_n}.$
(42)
If we take the inner product of the both sides of this equation with $\ket{e_x}$ , we have
$\begin{align} \braket{e_x}{\psi(t)} &= e^{-i E_n t/\hbar} \braket{e_x}{\psi_n} \\ \Longrightarrow\quad \psi(x,t) &= e^{-i E_n t/\hbar} \psi_n(x), \end{align}$
(43)
which is exactly the time evolution equation for the wavefunctions we have in Eq. 32.
The equality $\ket{\psi(t)} = e^{-i\hat{H}t/\hbar} \ket{\psi_n}$ was only shown for energy eigenstates $\ket{\psi_n}$ in the previous part. Assuming we decompose the initial state $\ket{\psi}$ using the energy eigenstates
$\ket{\psi} = \sum_n c_n \ket{\psi_n},$
(44)
apply the operator $e^{-i\hat{H}t/\hbar}$ to it to get another state
$\ket{\psi'} = e^{-i\hat{H}t/\hbar} \ket{\psi}.$
(45)
Use the decomposition in Eq. 44 and the result of the previous part to show that $\ket{\psi'}$ is exactly the state corresponding to the time-evolved wavefunction in Eq. 37. That is,
$\braket{e_x}{\psi'} = \sum_n c_n\psi_n(x) \, e^{-iE_nt/\hbar}.$
(46)
Solution
With all the mathematical machinery we have developed so far, this does not seem very difficult. In the previous task, we applied the $e^{-i\hat{H}t/\hbar}$ operator to a particular eigenstate $\ket{\psi_n}$ , but there will be nothing special if we apply it to a superposition of all the eigenstates!
$\begin{align} \ket{\psi'} &= e^{-i\hat{H}t/\hbar} \ket{\psi} \\ &= e^{-i\hat{H}t/\hbar} \sum_n c_n \ket{\psi_n} \\ &= \sum_n c_n \, e^{-i\hat{H}t/\hbar} \, \ket{\psi_n} \\ \intertext{now the operator acts on the eigenstates $\ket{\psi_n}$ and we will have} \ket{\psi'} &= \sum_n c_n \, e^{-i E_n t/\hbar} \, \ket{\psi_n}, \end{align}$
(47)
which is another expression for the relation we have in Eq. 37 for the wavefunctions and we can derive that relation as
$\begin{align} \braket{e_x}{\psi'} &= \sum_n c_n \, e^{-i E_n t/\hbar} \, \braket{e_x}{\psi_n} \\ \Longrightarrow\quad \psi'(x) &= \sum_n c_n \, e^{-i E_n t/\hbar} \, \psi_n(x). \end{align}$
(48)
This shows that the state $\ket{\psi'}$ is exactly the state corresponding to the time-evolved wavefunction from the Schrodinger equation therefore we have
$\ket{\psi(t)} = \ket{\psi'} = e^{-i\hat{H}t/\hbar} \, \ket{\psi}.$
(49)

This is a fascinating result! It shows that the wavefunction corresponding to the state

e^{-i\hat{H}t/\hbar} \ket{\psi}

(50)

exactly matches the one we derived after solving the Schrodinger equation. This allows us to write

\ket{\psi(t)} = e^{-i\hat{H}t/\hbar} \ket{\psi}

(51)

for arbitrary states $\ket{\psi}$ , which means

\hat{U}(t) = e^{-i\hat{H}t/\hbar}

(52)

is the time evolution operator we were looking for.

And finally, let us show that the Schrodinger equation can be derived starting from the time evolution operator defined in Eq. 52.

The time evolution operator allows us to write

\ket{\psi(t+dt)} = \hat{U}(dt) \ket{\psi(t)}

(53)

which means starting from the $\ket{\psi(t)}$ state, we can evolve time for a short period of time $dt$ to get $\ket{\psi(t+dt)}$ .

If we assume $dt$ is small, we can approximate the time evolution operator using only the first order term in its power series

\hat{U}(dt) = e^{-i\hat{H} dt/\hbar} = \sum_{n=0}^\infty \frac{1}{n!} \Big( \frac{-i}{\hbar} \hat{H} dt \Big)^n \approx \hat{I} - \frac{i}{\hbar} \hat{H} dt.

(54)

This means Eq. 53 will become

\begin{align} \ket{\psi(t+dt)} &= \hat{U}(dt) \ket{\psi(t)} \\ &\approx \Big( \hat{I} - \frac{i}{\hbar} \hat{H} dt \Big) \ket{\psi(t)} \\ &= \ket{\psi(t)} - \frac{i}{\hbar} dt \,\, \hat{H}\ket{\psi(t)} \end{align}

(55)

Therefore we have

i\hbar \left[ \frac{\ket{\psi(t+dt)}-\ket{\psi(t)}}{dt} \right] = \hat{H}\ket{\psi(t)}.

(56)

Taking the limit $dt \to 0$ , we can write

i\hbar \pdv{}{t} \ket{\psi(t)} = \hat{H}\ket{\psi(t)}

(57)

which is exactly the Schrodinger equation (see Eq. 31) written in terms of states in the Hilbert space.