Month: August 2015

Quantum Virial Theorem

August 31, 2015 phy1520 No comments , , , , ,

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Question: Quantum Virial Theorem ([1] pr. 2.7)

Consider a particle with Hamiltonian

\begin{equation}\label{eqn:qmVirialTheorem:20}
H = \frac{\Bp^2}{2 m} + V(\Bx),
\end{equation}

By calculating the time evolution of \( \antisymmetric{\Bx \cdot \Bp}{H} \), identify the quantum virial theorem and show the conditions where it is satisfied.

Answer

\begin{equation}\label{eqn:qmVirialTheorem:40}
\begin{aligned}
\antisymmetric{\Bx \cdot \Bp}{H}
&=
\inv{2 m} \antisymmetric{\Bx \cdot \Bp}{\Bp^2} + \antisymmetric{\Bx \cdot \Bp}{V(\Bx)} \\
&=
\inv{2 m} \lr{ x_r p_r \Bp^2 – \Bp^2 x_r p_r}
+
\lr{ x_r p_r V(\Bx) – V(\Bx) x_r p_r } \\
&=
\inv{2 m} \antisymmetric{ x_r }{\Bp^2} p_r
+
x_r \antisymmetric{ p_r}{ V(\Bx)},
\end{aligned}
\end{equation}

Evaluating those commutators separately, gives

\begin{equation}\label{eqn:qmVirialTheorem:60}
\begin{aligned}
\antisymmetric{ x_r }{\Bp^2}
&=
\antisymmetric{ x_r }{p_r^2}\qquad \text{no sum} \\
&=
2 i \Hbar p_r,
\end{aligned}
\end{equation}

and

\begin{equation}\label{eqn:qmVirialTheorem:80}
\antisymmetric{ p_r}{ V(\Bx)}
= -i \Hbar \PD{x_r}{V(\Bx)},
\end{equation}

so
\begin{equation}\label{eqn:qmVirialTheorem:100}
\begin{aligned}
\ddt{}\lr{\Bx \cdot \Bp}
&=
\inv{i \Hbar}
\antisymmetric{\Bx \cdot \Bp}{H} \\
&=
\inv{2 m} 2 p_r p_r – x_r \PD{x_r}{V(\Bx)} \\
&=
\frac{\Bp^2}{m} – \Bx \cdot \spacegrad V(\Bx).
\end{aligned}
\end{equation}

Taking expectation values, assuming that the states are independent of time, we have

\begin{equation}\label{eqn:qmVirialTheorem:120}
\begin{aligned}
0
&= \ddt{} \expectation{ \Bx \cdot \Bp } \\
&= \expectation{\frac{\Bp^2}{m}} – \expectation{\Bx \cdot \spacegrad V(\Bx)}.
\end{aligned}
\end{equation}

Note that taking the expectation with respect to stationary states was required to reverse the order of the time derivative with the expectation operation.

The right hand side is the quantum equivalent of the virial theorem, relating the average kinetic energy to the potential

\begin{equation}\label{eqn:qmVirialTheorem:140}
2 \expectation{T} = \expectation{\Bx \cdot \spacegrad V(\Bx)}
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

A symmetric real Hamiltonian

August 31, 2015 phy1520 No comments , ,

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Question: A symmetric real Hamiltonian ([1] pr. 2.9)

Find the time evolution for the state \( \ket{a’} \) for a Hamiltian of the form

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:20}
H = \delta \lr{ \ket{a’}\bra{a’} + \ket{a”}\bra{a”} }
\end{equation}

Answer

This Hamiltonian has the matrix representation

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:40}
H =
\begin{bmatrix}
0 & \delta \\
\delta & 0
\end{bmatrix},
\end{equation}

which has a characteristic equation of

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:60}
\lambda^2 -\delta^2 = 0,
\end{equation}

so the energy eigenvalues are \( \pm \delta \).

The diagonal basis states are respectively

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:80}
\ket{\pm\delta} =
\inv{\sqrt{2}}
\begin{bmatrix}
\pm 1 \\
1
\end{bmatrix}.
\end{equation}

The time evolution operator is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:100}
\begin{aligned}
U
&= e^{-i H t/\Hbar} \\
&=
e^{-i \delta t/\Hbar} \ket{+\delta}\bra{+\delta}
+ e^{i \delta t/\Hbar} \ket{-\delta}\bra{-\delta} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
+ \frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
-1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 \\
1
\end{bmatrix} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}
+\frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & -1 \\
-1 & 1
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}.
\end{aligned}
\end{equation}

The desired time evolution in the original basis is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:140}
\begin{aligned}
\ket{a’, t}
&=
e^{-i H t/\Hbar}
\ket{a’, 0} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar)
\end{bmatrix} \\
&=
\cos(\delta t/\Hbar) \ket{a’,0} -i \sin(\delta t/\Hbar) \ket{a”,0}.
\end{aligned}
\end{equation}

This evolution has the same structure as left circularly polarized light.

The probability of finding the system in state \( \ket{a”} \) given an initial state of \( \ket{a’,0} \) is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:160}
P
=
\Abs{\braket{a”}{a’,t}}^2
=
\sin^2 \lr{ \delta t/\Hbar }.
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Operator matrix element

August 29, 2015 phy1520 No comments , , , , , ,

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0dc1b8c5-232f-492d-b520-bcec41e45c88

Weird dreams

I woke up today having a dream still in my head from the night, but it was a strange one. I was expanding out the Dirac notation representation of an operator in matrix form, but the symbols in the kets were elaborate pictures of Disney princesses that I was drawing with forestry scenery in the background, including little bears. At the point that I woke up from the dream, I noticed that I’d gotten the proportion of the bears wrong in one of the pictures, and they looked like they were ready to eat one of the princess characters.

Guts

As a side effect of this weird dream I actually started thinking about matrix element representation of operators.

When forming the matrix element of an operator using Dirac notation the elements are of the form \( \bra{\textrm{row}} A \ket{\textrm{column}} \). I’ve gotten that mixed up a couple of times, so I thought it would be helpful to write this out explicitly for a \( 2 \times 2 \) operator representation for clarity.

To start, consider a change of basis for a single matrix element from basis \( \setlr{\ket{q}, \ket{r} } \), to basis \( \setlr{\ket{a}, \ket{b} } \)

\begin{equation}\label{eqn:operatorMatrixElement:20}
\begin{aligned}
\bra{q} A \ket{r}
&=
\braket{q}{a} \bra{a} A \ket{r}
+
\braket{q}{b} \bra{b} A \ket{r} \\
&=
\braket{q}{a} \bra{a} A \ket{a}\braket{a}{r}
+ \braket{q}{a} \bra{a} A \ket{b}\braket{b}{r} \\
&+ \braket{q}{b} \bra{b} A \ket{a}\braket{a}{r}
+ \braket{q}{b} \bra{b} A \ket{b}\braket{b}{r} \\
&=
\braket{q}{a}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix}
+
\braket{q}{b}
\begin{bmatrix}
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix} \\
&=
\begin{bmatrix}
\braket{q}{a} &
\braket{q}{b}
\end{bmatrix}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix}.
\end{aligned}
\end{equation}

Suppose the matrix representation of \( \ket{q}, \ket{r} \) are respectively

\begin{equation}\label{eqn:operatorMatrixElement:40}
\begin{aligned}
\ket{q} &\sim
\begin{bmatrix}
\braket{a}{q} \\
\braket{b}{q} \\
\end{bmatrix} \\
\ket{r} &\sim
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r} \\
\end{bmatrix} \\
\end{aligned},
\end{equation}

then

\begin{equation}\label{eqn:operatorMatrixElement:60}
\bra{q} \sim
{\begin{bmatrix}
\braket{a}{q} \\
\braket{b}{q} \\
\end{bmatrix}}^\dagger
=
\begin{bmatrix}
\braket{q}{a} &
\braket{q}{b}
\end{bmatrix}.
\end{equation}

The matrix element is then

\begin{equation}\label{eqn:operatorMatrixElement:80}
\bra{q} A \ket{r}
\sim
\bra{q}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\ket{r},
\end{equation}

and the corresponding matrix representation of the operator is

\begin{equation}\label{eqn:operatorMatrixElement:100}
A \sim
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}.
\end{equation}

Hermite polynomial normalization constant

August 21, 2015 phy1520 No comments , , , ,

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Question: Hermite polynomial normalization constant ([1] pr. 2.21)

Derive the normalization constant \( c_n \) for the Harmonic oscillator solution

\begin{equation}\label{eqn:hermiteOrtho:20}
u_n(x) = c_n H_n\lr{ x \sqrt{\frac{m\omega}{\Hbar}} } e^{-m \omega x^2/2 \Hbar},
\end{equation}

by deriving the orthogonality relationship using generating functions

\begin{equation}\label{eqn:hermiteOrtho:40}
g(x,t) = e^{-t^2 + 2 t x} = \sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}.
\end{equation}

Start by working out the integral

\begin{equation}\label{eqn:hermiteOrtho:60}
I = \int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx,
\end{equation}

consider the integral twice with each side definition of the generating function.

Answer

First using the exponential definition of the generating function

\begin{equation}\label{eqn:hermiteOrtho:80}
\begin{aligned}
\int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx
&=
\int_{-\infty}^\infty
e^{-t^2 + 2 t x}
e^{-s^2 + 2 s x} e^{-x^2} dx \\
&=
e^{-t^2 -s^2}
\int_{-\infty}^\infty
e^{-(x^2- 2 t x – 2 s x)} dx \\
&=
e^{-t^2 -s^2 + (s + t)^2}
\int_{-\infty}^\infty
e^{-(x – t – s)^2} dx \\
&=
e^{2 st}
\int_{-\infty}^\infty
e^{-u^2} du \\
&= \sqrt{\pi} e^{2 st}.
\end{aligned}
\end{equation}

With the Hermite polynomial definition of the generating function, this integral is

\begin{equation}\label{eqn:hermiteOrtho:100}
\begin{aligned}
\int_{-\infty}^\infty g(x, t) g(x, s) e^{-x^2} dx
&=
\int_{-\infty}^\infty
\sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}
\sum_{m=0}^\infty H_m(x) \frac{s^m}{m!}
e^{-x^2} dx \\
&=
\sum_{n=0}^\infty \frac{t^n}{n!}
\sum_{m=0}^\infty \frac{s^m}{m!}
\int_{-\infty}^\infty H_n(x) H_m(x) e^{-x^2} dx.
\end{aligned}
\end{equation}

Let

\begin{equation}\label{eqn:hermiteOrtho:120}
\alpha_{n m} = \int_{-\infty}^\infty H_n(x) H_m(x) e^{-x^2} dx,
\end{equation}

and equate the two expansions of this integral

\begin{equation}\label{eqn:hermiteOrtho:140}
\sqrt{\pi} \sum_{n=0}^\infty \frac{(2st)^n}{n!}
=
\sum_{n=0}^\infty \frac{t^n}{n!}
\sum_{m=0}^\infty \frac{s^m}{m!}
\alpha_{n m},
\end{equation}

or, after equating powers of \( t^n \)

\begin{equation}\label{eqn:hermiteOrtho:160}
\sqrt{\pi} (2 s)^n =
\sum_{m=0}^\infty \frac{s^m}{m!} \alpha_{n m}.
\end{equation}

This requires \( \alpha_{n m} \) to be zero for \( n \ne m \), so

\begin{equation}\label{eqn:hermiteOrtho:180}
\sqrt{\pi} 2^n = \frac{1}{n!} \alpha_{n n},
\end{equation}

and

\begin{equation}\label{eqn:hermiteOrtho:200}
\int_{-\infty}^\infty H_n(x) H_m(x) e^{-x^2} dx = \delta_{n m} \sqrt{\pi} 2^n n!.
\end{equation}

The SHO normalization is fixed by

\begin{equation}\label{eqn:hermiteOrtho:220}
\int_{-\infty}^\infty u_n^2(x) dx
= c_n^2
\int_{-\infty}^\infty H_n^2(x/x_0) e^{-(x/x_0)^2} dx
= c_n^2 x_0 \sqrt{\pi} 2^n n!,
\end{equation}

or

\begin{equation}\label{eqn:hermiteOrtho:240}
\begin{aligned}
c_n
&= \inv{\sqrt{ \sqrt{\pi} 2^n n! \sqrt{\frac{\Hbar}{m \omega}}}} \\
&= \lr{ \frac{m \omega}{\Hbar \pi} }^{1/4} 2^{-n/2} \inv{\sqrt{n!}}
\end{aligned}
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Relation of probability flux to momentum

August 19, 2015 phy1520 No comments ,

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In [1] it is mentioned that the probability flux

\begin{equation}\label{eqn:fluxAndMomentum:20}
\Bj(\Bx, t) = -\frac{i\Hbar}{2 m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj },
\end{equation}

is related to the momentum expectation at a given time by the integral of the flux over all space

\begin{equation}\label{eqn:fluxAndMomentum:40}
\int d^3 x \Bj(\Bx, t) = \frac{\expectation{\Bp}_t}{m}.
\end{equation}

That wasn’t obvious to me at a glance, however, this can be seen by recasting the integral in bra-ket form. Let

\begin{equation}\label{eqn:fluxAndMomentum:60}
\psi(\Bx, t) = \braket{\Bx}{\psi(t)},
\end{equation}

and note that the momentum portions of the flux can be written as

\begin{equation}\label{eqn:fluxAndMomentum:80}
-i \Hbar \spacegrad \psi(\Bx, t) = \bra{\Bx} \Bp \ket{\psi(t)}.
\end{equation}

The current is therefore

\begin{equation}\label{eqn:fluxAndMomentum:100}
\begin{aligned}
\Bj(\Bx, t)
&= \frac{1}{2 m}
\lr{
\psi^\conj \bra{\Bx} \Bp \ket{\psi(t)}
+\psi {\bra{\Bx} \Bp \ket{\psi(t)} }^\conj
} \\
&= \frac{1}{2 m}
\lr{
{\braket{\Bx}{\psi(t)}}^\conj \bra{\Bx} \Bp \ket{\psi(t)}
+ \braket{\Bx}{\psi(t)} {\bra{\Bx} \Bp \ket{\psi(t)} }^\conj
} \\
&= \frac{1}{2 m}
\lr{
\braket{\psi(t)}{\Bx} \bra{\Bx} \Bp \ket{\psi(t)}
+
\bra{\psi(t)} \Bp \ket{\Bx} \braket{\Bx}{\psi(t)}
}.
\end{aligned}
\end{equation}

Integrating and noting that the spatial identity is \( 1 = \int d^3 x \ket{\Bx}\bra{\Bx} \), we have

\begin{equation}\label{eqn:fluxAndMomentum:n}
\int d^3 x \Bj(\Bx, t)
=
\bra{\psi(t)} \Bp \ket{\psi(t)},
\end{equation}

This is just the expectation of \( \Bp \) with respect to a specific time-instance state.

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Quantum SHO ladder operators as a diagonal change of basis for the Heisenberg EOMs

August 19, 2015 phy1520 No comments , , , , , , , , , , , , , , , , , , ,

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Many authors pull the definitions of the raising and lowering (or ladder) operators out of their butt with no attempt at motivation. This is pointed out nicely in [1] by Eli along with one justification based on factoring the Hamiltonian.

In [2] is a small exception to the usual presentation. In that text, these operators are defined as usual with no motivation. However, after the utility of these operators has been shown, the raising and lowering operators show up in a context that does provide that missing motivation as a side effect.
It doesn’t look like the author was trying to provide a motivation, but it can be interpreted that way.

When seeking the time evolution of Heisenberg-picture position and momentum operators, we will see that those solutions can be trivially expressed using the raising and lowering operators. No special tools nor black magic is required to find the structure of these operators. Unfortunately, we must first switch to both the Heisenberg picture representation of the position and momentum operators, and also employ the Heisenberg equations of motion. Neither of these last two fit into standard narrative of most introductory quantum mechanics treatments. We will also see that these raising and lowering “operators” could also be introduced in classical mechanics, provided we were attempting to solve the SHO system using the Hamiltonian equations of motion.

I’ll outline this route to finding the structure of the ladder operators below. Because these are encountered trying to solve the time evolution problem, I’ll first show a simpler way to solve that problem. Because that simpler method depends a bit on lucky observation and is somewhat unstructured, I’ll then outline a more structured procedure that leads to the ladder operators directly, also providing the solution to the time evolution problem as a side effect.

The starting point is the Heisenberg equations of motion. For a time independent Hamiltonian \( H \), and a Heisenberg operator \( A^{(H)} \), those equations are

\begin{equation}\label{eqn:harmonicOscDiagonalize:20}
\ddt{A^{(H)}} = \inv{i \Hbar} \antisymmetric{A^{(H)}}{H}.
\end{equation}

Here the Heisenberg operator \( A^{(H)} \) is related to the Schrodinger operator \( A^{(S)} \) by

\begin{equation}\label{eqn:harmonicOscDiagonalize:60}
A^{(H)} = U^\dagger A^{(S)} U,
\end{equation}

where \( U \) is the time evolution operator. For this discussion, we need only know that \( U \) commutes with \( H \), and do not need to know the specific structure of that operator. In particular, the Heisenberg equations of motion take the form

\begin{equation}\label{eqn:harmonicOscDiagonalize:80}
\begin{aligned}
\ddt{A^{(H)}}
&= \inv{i \Hbar}
\antisymmetric{A^{(H)}}{H} \\
&= \inv{i \Hbar}
\antisymmetric{U^\dagger A^{(S)} U}{H} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} U H
– H U^\dagger A^{(S)} U
} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} H U
– U^\dagger H A^{(S)} U
} \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{A^{(S)}}{H} U.
\end{aligned}
\end{equation}

The Hamiltonian for the harmonic oscillator, with Schrodinger-picture position and momentum operators \( x, p \) is

\begin{equation}\label{eqn:harmonicOscDiagonalize:40}
H = \frac{p^2}{2m} + \inv{2} m \omega^2 x^2,
\end{equation}

so the equations of motions are

\begin{equation}\label{eqn:harmonicOscDiagonalize:100}
\begin{aligned}
\ddt{x^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{\frac{p^2}{2m}} U \\
&= \inv{2 m i \Hbar} U^\dagger \lr{ i \Hbar \PD{p}{p^2} } U \\
&= \inv{m } U^\dagger p U \\
&= \inv{m } p^{(H)},
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:harmonicOscDiagonalize:120}
\begin{aligned}
\ddt{p^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{\inv{2} m \omega^2 x^2 } U \\
&= \frac{m \omega^2}{2 i \Hbar} U^\dagger \lr{ -i \Hbar \PD{x}{x^2} } U \\
&= -m \omega^2 U^\dagger x U \\
&= -m \omega^2 x^{(H)}.
\end{aligned}
\end{equation}

In the Heisenberg picture the equations of motion are precisely those of classical Hamiltonian mechanics, except that we are dealing with operators instead of scalars

\begin{equation}\label{eqn:harmonicOscDiagonalize:140}
\begin{aligned}
\ddt{p^{(H)}} &= -m \omega^2 x^{(H)} \\
\ddt{x^{(H)}} &= \inv{m } p^{(H)}.
\end{aligned}
\end{equation}

In the text the ladder operators are used to simplify the solution of these coupled equations, since they can decouple them. That’s not really required since we can solve them directly in matrix form with little work

\begin{equation}\label{eqn:harmonicOscDiagonalize:160}
\ddt{}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix}
=
\begin{bmatrix}
0 & -m \omega^2 \\
\inv{m} & 0
\end{bmatrix}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix},
\end{equation}

or, with length scaled variables

\begin{equation}\label{eqn:harmonicOscDiagonalize:180}
\begin{aligned}
\ddt{}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}
&=
\begin{bmatrix}
0 & -\omega \\
\omega & 0
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\sigma_y
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}.
\end{aligned}
\end{equation}

Writing \( y = \begin{bmatrix} \frac{p^{(H)}}{m \omega} \\ x^{(H)} \end{bmatrix} \), the solution can then be written immediately as

\begin{equation}\label{eqn:harmonicOscDiagonalize:200}
\begin{aligned}
y(t)
&=
\exp\lr{ -i \omega \sigma_y t } y(0) \\
&=
\lr{ \cos \lr{ \omega t } I – i \sigma_y \sin\lr{ \omega t } } y(0) \\
&=
\begin{bmatrix}
\cos\lr{ \omega t } & \sin\lr{ \omega t } \\
-\sin\lr{ \omega t } & \cos\lr{ \omega t }
\end{bmatrix}
y(0),
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:harmonicOscDiagonalize:220}
\begin{aligned}
\frac{p^{(H)}(t)}{m \omega} &= \cos\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \sin\lr{ \omega t } x^{(H)}(0) \\
x^{(H)}(t) &= -\sin\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \cos\lr{ \omega t } x^{(H)}(0).
\end{aligned}
\end{equation}

This solution depends on being lucky enough to recognize that the matrix has a Pauli matrix as a factor (which squares to unity, and allows the exponential to be evaluated easily.)

If we hadn’t been that observant, then the first tool we’d have used instead would have been to diagonalize the matrix. For such diagonalization, it’s natural to work in completely dimensionless variables. Such a non-dimensionalisation can be had by defining

\begin{equation}\label{eqn:harmonicOscDiagonalize:240}
x_0 = \sqrt{\frac{\Hbar}{m \omega}},
\end{equation}

and dividing the working (operator) variables through by those values. Let \( z = \inv{x_0} y \), and \( \tau = \omega t \) so that the equations of motion are

\begin{equation}\label{eqn:harmonicOscDiagonalize:260}
\frac{dz}{d\tau}
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
z.
\end{equation}

This matrix can be diagonalized as

\begin{equation}\label{eqn:harmonicOscDiagonalize:280}
A
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
=
V
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
V^{-1},
\end{equation}

where

\begin{equation}\label{eqn:harmonicOscDiagonalize:300}
V =
\inv{\sqrt{2}}
\begin{bmatrix}
i & -i \\
1 & 1
\end{bmatrix}.
\end{equation}

The equations of motion can now be written

\begin{equation}\label{eqn:harmonicOscDiagonalize:320}
\frac{d}{d\tau} \lr{ V^{-1} z } =
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
\lr{ V^{-1} z }.
\end{equation}

This final change of variables \( V^{-1} z \) decouples the system as desired. Expanding that gives

\begin{equation}\label{eqn:harmonicOscDiagonalize:340}
\begin{aligned}
V^{-1} z
&=
\inv{\sqrt{2}}
\begin{bmatrix}
-i & 1 \\
i & 1
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i \frac{p^{(H)}}{m \omega} + x^{(H)} \\
i \frac{p^{(H)}}{m \omega} + x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
a^\dagger \\
a
\end{bmatrix},
\end{aligned}
\end{equation}

where
\begin{equation}\label{eqn:harmonicOscDiagonalize:n}
\begin{aligned}
a^\dagger &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ -i \frac{p^{(H)}}{m \omega} + x^{(H)} } \\
a &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ i \frac{p^{(H)}}{m \omega} + x^{(H)} }.
\end{aligned}
\end{equation}

Lo and behold, we have the standard form of the raising and lowering operators, and can write the system equations as

\begin{equation}\label{eqn:harmonicOscDiagonalize:360}
\begin{aligned}
\ddt{a^\dagger} &= i \omega a^\dagger \\
\ddt{a} &= -i \omega a.
\end{aligned}
\end{equation}

It is actually a bit fluky that this matched exactly, since we could have chosen eigenvectors that differ by constant phase factors, like

\begin{equation}\label{eqn:harmonicOscDiagonalize:380}
V = \inv{\sqrt{2}}
\begin{bmatrix}
i e^{i\phi} & -i e^{i \psi} \\
1 e^{i\phi} & e^{i \psi}
\end{bmatrix},
\end{equation}

so

\begin{equation}\label{eqn:harmonicOscDiagonalize:341}
\begin{aligned}
V^{-1} z
&=
\frac{e^{-i(\phi + \psi)}}{\sqrt{2}}
\begin{bmatrix}
-i e^{i\psi} & e^{i \psi} \\
i e^{i\phi} & e^{i \phi}
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i e^{i\phi} \frac{p^{(H)}}{m \omega} + e^{i\phi} x^{(H)} \\
i e^{i\psi} \frac{p^{(H)}}{m \omega} + e^{i\psi} x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{i\phi} a^\dagger \\
e^{i\psi} a
\end{bmatrix}.
\end{aligned}
\end{equation}

To make the resulting pairs of operators Hermitian conjugates, we’d want to constrain those constant phase factors by setting \( \phi = -\psi \). If we were only interested in solving the time evolution problem no such additional constraints are required.

The raising and lowering operators are seen to naturally occur when seeking the solution of the Heisenberg equations of motion. This is found using the standard technique of non-dimensionalisation and then seeking a change of basis that diagonalizes the system matrix. Because the Heisenberg equations of motion are identical to the classical Hamiltonian equations of motion in this case, what we call the raising and lowering operators in quantum mechanics could also be utilized in the classical simple harmonic oscillator problem. However, in a classical context we wouldn’t have a justification to call this more than a change of basis.

References

[1] Eli Lansey. The Quantum Harmonic Oscillator Ladder Operators, 2009. URL http://behindtheguesses.blogspot.ca/2009/03/quantum-harmonic-oscillator-ladder.html. [Online; accessed 18-August-2015].

[2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics, chapter {Time Development of the Oscillator}. Pearson Higher Ed, 2014.

Heisenberg picture position commutator

August 14, 2015 phy1520 No comments , , , ,

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Question: Heisenberg picture position commutator ([1] pr. 2.5)

Evaluate

\begin{equation}\label{eqn:positionCommutator:20}
\antisymmetric{x(t)}{x(0)},
\end{equation}

for a Heisenberg picture operator \( x(t) \) for a free particle.

Answer

The free particle Hamiltonian is

\begin{equation}\label{eqn:positionCommutator:40}
H = \frac{p^2}{2m},
\end{equation}

so the time evolution operator is

\begin{equation}\label{eqn:positionCommutator:60}
U(t) = e^{-i p^2 t/(2 m \Hbar)}.
\end{equation}

The Heisenberg picture position operator is

\begin{equation}\label{eqn:positionCommutator:80}
\begin{aligned}
x^\textrm{H}
&= U^\dagger x U \\
&= e^{i p^2 t/(2 m \Hbar)} x e^{-i p^2 t/(2 m \Hbar)} \\
&= \sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i p^2 t}{2 m \Hbar} }^k
x
e^{-i p^2 t/(2 m \Hbar)} \\
&= \sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k p^{2k} x
e^{-i p^2 t/(2 m \Hbar)} \\
&=
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ \antisymmetric{p^{2k}}{x} + x p^{2k} }
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \antisymmetric{p^{2k}}{x}
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ -i \Hbar \PD{p}{p^{2k}} }
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ -i \Hbar 2 k p^{2 k -1} }
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
-2 i \Hbar p \frac{i t}{2 m \Hbar} \sum_{k = 1}^\infty \inv{(k-1)!} \lr{ \frac{i t}{2 m \Hbar} }^{k-1} p^{2(k – 1)}
e^{-i p^2 t/(2 m \Hbar)} \\
&= x + t \frac{p}{m}.
\end{aligned}
\end{equation}

This has the structure of a classical free particle \( x(t) = x + v t \), but in this case \( x,p \) are operators.

The evolved position commutator is
\begin{equation}\label{eqn:positionCommutator:100}
\begin{aligned}
\antisymmetric{x(t)}{x(0)}
&=
\antisymmetric{x + t p/m}{x} \\
&=
\frac{t}{m} \antisymmetric{p}{x} \\
&=
-i \Hbar \frac{t}{m}.
\end{aligned}
\end{equation}

Compare this to the classical Poisson bracket
\begin{equation}\label{eqn:positionCommutator:120}
\antisymmetric{x(t)}{x(0)}_{\textrm{classical}}
=
\PD{x}{}\lr{x + p t/m} \PD{p}{x} – \PD{p}{}\lr{x + p t/m} \PD{x}{x}
=
– \frac{t}{m}.
\end{equation}

This has the expected relation \( \antisymmetric{x(t)}{x(0)} = i \Hbar \antisymmetric{x(t)}{x(0)}_{\textrm{classical}} \).

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Heisenberg picture spin precession

August 13, 2015 phy1520 No comments , , , , ,

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Question: Heisenberg picture spin precession ([1] pr. 2.1)

For the spin Hamiltonian

\begin{equation}\label{eqn:heisenbergSpinPrecession:20}
H = -\frac{e B}{m c} S_z = \omega S_z,
\end{equation}

express and solve the Heisenberg equations of motion for \( S_x(t), S_y(t) \), and \( S_z(t) \).

Answer

The equations of motion are of the form

\begin{equation}\label{eqn:heisenbergSpinPrecession:40}
\begin{aligned}
\frac{dS_i^\textrm{H}}{dt}
&= \inv{i \Hbar} \antisymmetric{S_i^\textrm{H}}{H} \\
&= \inv{i \Hbar} \antisymmetric{U^\dagger S_i U}{H} \\
&= \inv{i \Hbar} \lr{U^\dagger S_i U H – H U^\dagger S_i U } \\
&= \inv{i \Hbar} U^\dagger \lr{ S_i H – H S_i } U \\
&= \frac{\omega}{i \Hbar} U^\dagger \antisymmetric{ S_i}{S_z } U.
\end{aligned}
\end{equation}

These are

\begin{equation}\label{eqn:heisenbergSpinPrecession:60}
\begin{aligned}
\frac{dS_x^\textrm{H}}{dt} &= -\omega U^\dagger S_y U \\
\frac{dS_y^\textrm{H}}{dt} &= \omega U^\dagger S_x U \\
\frac{dS_z^\textrm{H}}{dt} &= 0.
\end{aligned}
\end{equation}

To completely specify these equations, we need to expand \( U(t) \), which is

\begin{equation}\label{eqn:heisenbergSpinPrecession:80}
\begin{aligned}
U(t)
&= e^{-i H t /\Hbar} \\
&= e^{-i \omega S_z t /\Hbar} \\
&= e^{-i \omega \sigma_z t /2} \\
&= \cos\lr{ \omega t/2 } -i \sigma_z \sin\lr{ \omega t/2 } \\
&=
\begin{bmatrix}
\cos\lr{ \omega t/2 } -i \sin\lr{ \omega t/2 } & 0 \\
0 & \cos\lr{ \omega t/2 } + i \sin\lr{ \omega t/2 }
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix}.
\end{aligned}
\end{equation}

The equations of motion can now be written out in full. To do so seems a bit silly since we also know that \( S_x^\textrm{H} = U^\dagger S_x U, S_y^\textrm{H} U^\dagger S_x U \). However, if that is temporarily forgotten, we can show that the Heisenberg equations of motion can be solved for these too.

\begin{equation}\label{eqn:heisenbergSpinPrecession:100}
\begin{aligned}
U^\dagger S_x U
&=
\frac{\Hbar}{2}
\begin{bmatrix}
e^{i\omega t/2} & 0 \\
0 & e^{-i\omega t/2}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t/2} \\
e^{-i\omega t/2} & 0
\end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix},
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:heisenbergSpinPrecession:120}
\begin{aligned}
U^\dagger S_y U
&=
\frac{\Hbar}{2}
\begin{bmatrix}
e^{i\omega t/2} & 0 \\
0 & e^{-i\omega t/2}
\end{bmatrix}
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{i\Hbar}{2}
\begin{bmatrix}
0 & -e^{i\omega t/2} \\
e^{-i\omega t/2} & 0
\end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{i \Hbar}{2}
\begin{bmatrix}
0 & -e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix}.
\end{aligned}
\end{equation}

The equations of motion are now fully specified

\begin{equation}\label{eqn:heisenbergSpinPrecession:140}
\begin{aligned}
\frac{dS_x^\textrm{H}}{dt} &=
-\frac{i \Hbar \omega}{2}
\begin{bmatrix}
0 & -e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
\frac{dS_y^\textrm{H}}{dt} &=
\frac{\Hbar \omega}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
\frac{dS_z^\textrm{H}}{dt} &= 0.
\end{aligned}
\end{equation}

Integration gives

\begin{equation}\label{eqn:heisenbergSpinPrecession:160}
\begin{aligned}
S_x^\textrm{H} &=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} + C \\
S_y^\textrm{H} &=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & -i e^{i\omega t} \\
i e^{-i\omega t} & 0
\end{bmatrix} + C \\
S_z^\textrm{H} &= C.
\end{aligned}
\end{equation}

The integration constants are fixed by the boundary condition \( S_i^\textrm{H}(0) = S_i \), so

\begin{equation}\label{eqn:heisenbergSpinPrecession:180}
\begin{aligned}
S_x^\textrm{H} &=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
S_y^\textrm{H} &=
\frac{i \Hbar}{2}
\begin{bmatrix}
0 & – e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
S_z^\textrm{H} &= S_z.
\end{aligned}
\end{equation}

Observe that these integrated values \( S_x^\textrm{H}, S_y^\textrm{H} \) match \ref{eqn:heisenbergSpinPrecession:100}, and \ref{eqn:heisenbergSpinPrecession:120} as expected.

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Dynamics of non-Hermitian Hamiltonian

August 13, 2015 phy1520 No comments , , , , ,

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Question: Dynamics of non-Hermitian Hamiltonian ([1] pr. 2.2)

Revisiting an earlier Hamiltonian, but assuming it was entered incorrectly as

\begin{equation}\label{eqn:dynamicsNonHermitian:20}
H = H_{11} \ket{1}\bra{1}
+ H_{22} \ket{2}\bra{2}
+ H_{12} \ket{1}\bra{2}.
\end{equation}

What principle is now violated? Illustrate your point explicitly by attempting to solve the most generaqtl time-dependent problem using an illegal Hamiltonian of this kind. You may assume that \( H_{11} = H_{22} \) for simplicity.

Answer

In matrix form this Hamiltonian is

\begin{equation}\label{eqn:dynamicsNonHermitian:40}
\begin{aligned}
H
&=
\begin{bmatrix}
\bra{1} H \ket{1} & \bra{1} H \ket{2} \\
\bra{2} H \ket{1} & \bra{2} H \ket{2} \\
\end{bmatrix} \\
&=
\begin{bmatrix}
H_{11} & H_{12} \\
0 & H_{22} \\
\end{bmatrix}.
\end{aligned}
\end{equation}

This is not a Hermitian operator. What is the physical implication of this non-Hermicity? Consider the simpler case where \( H_{11} = H_{22} \). Such a Hamiltonian has the form

\begin{equation}\label{eqn:dynamicsNonHermitian:60}
H =
\begin{bmatrix}
a & b \\
0 & a
\end{bmatrix}.
\end{equation}

This has only one unique eigenvector ( \( (1,0) \), but we can still solve the time evolution equation

\begin{equation}\label{eqn:dynamicsNonHermitian:80}
i \Hbar \PD{t}{U} = H U,
\end{equation}

since for constant \( H \), we have

\begin{equation}\label{eqn:dynamicsNonHermitian:100}
U = e^{-i H t/\Hbar}.
\end{equation}

To exponentiate, note that we have

\begin{equation}\label{eqn:dynamicsNonHermitian:120}
{\begin{bmatrix}
a & b \\
0 & a
\end{bmatrix}}^n
=
\begin{bmatrix}
a^n & n a^{n-1} b \\
0 & a^n
\end{bmatrix}.
\end{equation}

To prove the induction, the \( n = 2 \) case follows easily

\begin{equation}\label{eqn:dynamicsNonHermitian:140}
\begin{bmatrix}
a & b \\
0 & a
\end{bmatrix}
\begin{bmatrix}
a & b \\
0 & a
\end{bmatrix}
=
\begin{bmatrix}
a^2 & 2 a b \\
0 & a^2
\end{bmatrix},
\end{equation}

as does the general case

\begin{equation}\label{eqn:dynamicsNonHermitian:160}
\begin{bmatrix}
a^n & n a^{n-1} b \\
0 & a^n
\end{bmatrix}
\begin{bmatrix}
a & b \\
0 & a
\end{bmatrix}
=
\begin{bmatrix}
a^{n+1} & (n +1 ) a^{n} b \\
0 & a^{n+1}
\end{bmatrix}.
\end{equation}

The exponential sum is thus
\begin{equation}\label{eqn:dynamicsNonHermitian:180}
e^{H \tau}
=
\begin{bmatrix}
e^{a \tau} & 0 + \frac{b \tau}{1!} + \frac{2 a b \tau^2}{2!} + \frac{3 a^2 b \tau^3}{3!} + \cdots \\
0 & e^{a \tau}
\end{bmatrix}.
\end{equation}

That sum simplifies to

\begin{equation}\label{eqn:dynamicsNonHermitian:200}
\frac{b \tau}{0!} + \frac{a b \tau^2}{1!} + \frac{a^2 b \tau^3}{2!} + \cdots \\
=
b \tau \lr{ 1 + \frac{a \tau}{1!} + \frac{(a \tau)^2}{2!} + \cdots }
=
b \tau e^{a \tau}.
\end{equation}

The exponential is thus
\begin{equation}\label{eqn:dynamicsNonHermitian:220}
e^{H \tau}
=
\begin{bmatrix}
e^{a\tau} & b \tau e^{a\tau} \\
0 & e^{a\tau}
\end{bmatrix}
=
\begin{bmatrix}
1 & b \tau \\
0 & 1
\end{bmatrix}
e^{a\tau}.
\end{equation}

In particular

\begin{equation}\label{eqn:dynamicsNonHermitian:240}
U = e^{-i H t/\Hbar} =
\begin{bmatrix}
1 & -i b t/\Hbar \\
0 & 1
\end{bmatrix}
e^{-i a t /\Hbar }.
\end{equation}

We can verify that this is a solution to \ref{eqn:dynamicsNonHermitian:80}. The left hand side is

\begin{equation}\label{eqn:dynamicsNonHermitian:260}
\begin{aligned}
i \Hbar \PD{t}{U}
&=
i \Hbar
\begin{bmatrix}
-i a/\Hbar & -i b /\Hbar + (-i b t/\Hbar)(-i a/\Hbar) \\
0 & -i a /\Hbar
\end{bmatrix}
e^{-i a t /\Hbar } \\
&=
\begin{bmatrix}
a & b – i a b t/\Hbar \\
0 & a
\end{bmatrix}
e^{-i a t /\Hbar },
\end{aligned}
\end{equation}

and for the right hand side
\begin{equation}\label{eqn:dynamicsNonHermitian:280}
\begin{aligned}
H U
&=
\begin{bmatrix}
a & b \\
0 & a
\end{bmatrix}
\begin{bmatrix}
1 & -i b t/\Hbar \\
0 & 1
\end{bmatrix}
e^{-i a t /\Hbar } \\
&=
\begin{bmatrix}
a & b – i a b t/\Hbar \\
0 & a
\end{bmatrix}
e^{-i a t /\Hbar } \\
&=
i \Hbar \PD{t}{U}.
\end{aligned}
\end{equation}

While the Schr\”{o}dinger is satisfied, we don’t have the unitary invertion physical property that is desired for the time evolution operator \( U \). Namely

\begin{equation}\label{eqn:dynamicsNonHermitian:300}
\begin{aligned}
U^\dagger U
&=
\begin{bmatrix}
1 & 0 \\
i b t/\Hbar & 1
\end{bmatrix}
e^{i a t /\Hbar }
\begin{bmatrix}
1 & -i b t/\Hbar \\
0 & 1
\end{bmatrix}
e^{-i a t /\Hbar } \\
&=
\begin{bmatrix}
1 & -i b t/\Hbar \\
i b t/\Hbar & (b t)^2/\Hbar^2
\end{bmatrix} \\
&\ne I.
\end{aligned}
\end{equation}

We required \( U^\dagger U = I \) for the time evolution operator, but don’t have that property for this non-Hermitian Hamiltonian.

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Translation operator problems

August 7, 2015 phy1520 No comments , , , , , , ,

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Question: One dimensional translation operator. ([1] pr. 1.28)

(a)

Evaluate the classical Poisson bracket

\begin{equation}\label{eqn:translation:420}
\antisymmetric{x}{F(p)}_{\textrm{classical}}
\end{equation}

(b)

Evaluate the commutator

\begin{equation}\label{eqn:translation:440}
\antisymmetric{x}{e^{i p a/\Hbar}}
\end{equation}

(c)

Using the result in \ref{problem:translation:28:b}, prove that
\begin{equation}\label{eqn:translation:460}
e^{i p a/\Hbar} \ket{x’},
\end{equation}

is an eigenstate of the coordinate operator \( x \).

Answer

(a)

\begin{equation}\label{eqn:translation:480}
\begin{aligned}
\antisymmetric{x}{F(p)}_{\textrm{classical}}
&=
\PD{x}{x} \PD{p}{F(p)} – \PD{p}{x} \PD{x}{F(p)} \\
&=
\PD{p}{F(p)}.
\end{aligned}
\end{equation}

(b)

Having worked backwards through these problems, the answer for this one dimensional problem can be obtained from \ref{eqn:translation:140} and is

\begin{equation}\label{eqn:translation:500}
\antisymmetric{x}{e^{i p a/\Hbar}} = a e^{i p a/\Hbar}.
\end{equation}

(c)

\begin{equation}\label{eqn:translation:520}
\begin{aligned}
x e^{i p a/\Hbar} \ket{x’}
&=
\lr{
\antisymmetric{x}{e^{i p a/\Hbar}}
e^{i p a/\Hbar} x
+
}
\ket{x’} \\
&=
\lr{ a e^{i p a/\Hbar} + e^{i p a/\Hbar} x ‘ } \ket{x’} \\
&= \lr{ a + x’ } \ket{x’}.
\end{aligned}
\end{equation}

This demonstrates that \( e^{i p a/\Hbar} \ket{x’} \) is an eigenstate of \( x \) with eigenvalue \( a + x’ \).

Question: Polynomial commutators. ([1] pr. 1.29)

(a)

For power series \( F, G \), verify

\begin{equation}\label{eqn:translation:180}
\antisymmetric{x_k}{G(\Bp)} = i \Hbar \PD{p_k}{G}, \qquad
\antisymmetric{p_k}{F(\Bx)} = -i \Hbar \PD{x_k}{F}.
\end{equation}

(b)

Evaluate \( \antisymmetric{x^2}{p^2} \), and compare to the classical Poisson bracket \( \antisymmetric{x^2}{p^2}_{\textrm{classical}} \).

Answer

(a)

Let

\begin{equation}\label{eqn:translation:200}
\begin{aligned}
G(\Bp) &= \sum_{k l m} a_{k l m} p_1^k p_2^l p_3^m \\
F(\Bx) &= \sum_{k l m} b_{k l m} x_1^k x_2^l x_3^m.
\end{aligned}
\end{equation}

It is simpler to work with a specific \( x_k \), say \( x_k = y \). The validity of the general result will still be clear doing so. Expanding the commutator gives

\begin{equation}\label{eqn:translation:220}
\begin{aligned}
\antisymmetric{y}{G(\Bp)}
&=
\sum_{k l m} a_{k l m} \antisymmetric{y}{p_1^k p_2^l p_3^m } \\
&=
\sum_{k l m} a_{k l m} \lr{
y p_1^k p_2^l p_3^m – p_1^k p_2^l p_3^m y
} \\
&=
\sum_{k l m} a_{k l m} \lr{
p_1^k y p_2^l p_3^m – p_1^k y p_2^l p_3^m
} \\
&=
\sum_{k l m} a_{k l m}
p_1^k
\antisymmetric{y}{p_2^l}
p_3^m.
\end{aligned}
\end{equation}

From \ref{eqn:translation:100}, we have \( \antisymmetric{y}{p_2^l} = l i \Hbar p_2^{l-1} \), so

\begin{equation}\label{eqn:translation:240}
\begin{aligned}
\antisymmetric{y}{G(\Bp)}
&=
\sum_{k l m} a_{k l m}
p_1^k
\antisymmetric{y}{p_2^l}
\lr{ l
i \Hbar p_2^{l-1}
}
p_3^m \\
&=
i \Hbar \PD{y}{G(\Bp)}.
\end{aligned}
\end{equation}

It is straightforward to show that
\( \antisymmetric{p}{x^l} = -l i \Hbar x^{l-1} \), allowing for a similar computation of the momentum commutator

\begin{equation}\label{eqn:translation:260}
\begin{aligned}
\antisymmetric{p_y}{F(\Bx)}
&=
\sum_{k l m} b_{k l m} \antisymmetric{p_y}{x_1^k x_2^l x_3^m } \\
&=
\sum_{k l m} b_{k l m} \lr{
p_y x_1^k x_2^l x_3^m – x_1^k x_2^l x_3^m p_y
} \\
&=
\sum_{k l m} b_{k l m} \lr{
x_1^k p_y x_2^l x_3^m – x_1^k p_y x_2^l x_3^m
} \\
&=
\sum_{k l m} b_{k l m}
x_1^k
\antisymmetric{p_y}{x_2^l}
x_3^m \\
&=
\sum_{k l m} b_{k l m}
x_1^k
\lr{ -l i \Hbar x_2^{l-1}}
x_3^m \\
&=
-i \Hbar \PD{p_y}{F(\Bx)}.
\end{aligned}
\end{equation}

(b)

It isn’t clear to me how the results above can be used directly to compute \( \antisymmetric{x^2}{p^2} \). However, when the first term of such a commutator is a monomomial, it can be expanded in terms of an \( x \) commutator

\begin{equation}\label{eqn:translation:280}
\begin{aligned}
\antisymmetric{x^2}{G(\Bp)}
&= x^2 G – G x^2 \\
&= x \lr{ x G } – G x^2 \\
&= x \lr{ \antisymmetric{ x }{ G } + G x } – G x^2 \\
&= x \antisymmetric{ x }{ G } + \lr{ x G } x – G x^2 \\
&= x \antisymmetric{ x }{ G } + \lr{ \antisymmetric{ x }{ G } + G x } x – G x^2 \\
&= x \antisymmetric{ x }{ G } + \antisymmetric{ x }{ G } x.
\end{aligned}
\end{equation}

Similarily,

\begin{equation}\label{eqn:translation:300}
\antisymmetric{x^3}{G(\Bp)} = x^2 \antisymmetric{ x }{ G } + x \antisymmetric{ x }{ G } x + \antisymmetric{ x }{ G } x^2.
\end{equation}

An induction hypothesis can be formed

\begin{equation}\label{eqn:translation:320}
\antisymmetric{x^k}{G(\Bp)} = \sum_{j = 0}^{k-1} x^{k-1-j} \antisymmetric{ x }{ G } x^j,
\end{equation}

and demonstrated

\begin{equation}\label{eqn:translation:340}
\begin{aligned}
\antisymmetric{x^{k+1}}{G(\Bp)}
&=
x^{k+1} G – G x^{k+1} \\
&=
x \lr{ x^{k} G } – G x^{k+1} \\
&=
x \lr{ \antisymmetric{x^{k}}{G} + G x^k } – G x^{k+1} \\
&=
x \antisymmetric{x^{k}}{G} + \lr{ x G } x^k – G x^{k+1} \\
&=
x \antisymmetric{x^{k}}{G} + \lr{ \antisymmetric{x}{G} + G x } x^k – G x^{k+1} \\
&=
x \antisymmetric{x^{k}}{G} + \antisymmetric{x}{G} x^k \\
&=
x \sum_{j = 0}^{k-1} x^{k-1-j} \antisymmetric{ x }{ G } x^j + \antisymmetric{x}{G} x^k \\
&=
\sum_{j = 0}^{k-1} x^{(k+1)-1-j} \antisymmetric{ x }{ G } x^j + \antisymmetric{x}{G} x^k \\
&=
\sum_{j = 0}^{k} x^{(k+1)-1-j} \antisymmetric{ x }{ G } x^j.
\end{aligned}
\end{equation}

That was a bit overkill for this problem, but may be useful later. Application of this to the problem gives

\begin{equation}\label{eqn:translation:360}
\begin{aligned}
\antisymmetric{x^2}{p^2}
&=
x \antisymmetric{x}{p^2}
+ \antisymmetric{x}{p^2} x \\
&=
x i \Hbar \PD{x}{p^2}
+ i \Hbar \PD{x}{p^2} x \\
&=
x 2 i \Hbar p
+ 2 i \Hbar p x \\
&= i \Hbar \lr{ 2 x p + 2 p x }.
\end{aligned}
\end{equation}

The classical commutator is
\begin{equation}\label{eqn:translation:380}
\begin{aligned}
\antisymmetric{x^2}{p^2}_{\textrm{classical}}
&=
\PD{x}{x^2} \PD{p}{p^2} – \PD{p}{x^2} \PD{x}{p^2} \\
&=
2 x 2 p \\
&= 2 x p + 2 p x.
\end{aligned}
\end{equation}

This demonstrates the expected relation between the classical and quantum commutators

\begin{equation}\label{eqn:translation:400}
\antisymmetric{x^2}{p^2} = i \Hbar \antisymmetric{x^2}{p^2}_{\textrm{classical}}.
\end{equation}

Question: Translation operator and position expectation. ([1] pr. 1.30)

The translation operator for a finite spatial displacement is given by

\begin{equation}\label{eqn:translation:20}
J(\Bl) = \exp\lr{ -i \Bp \cdot \Bl/\Hbar },
\end{equation}

where \( \Bp \) is the momentum operator.

(a)

Evaluate

\begin{equation}\label{eqn:translation:40}
\antisymmetric{x_i}{J(\Bl)}.
\end{equation}

(b)

Demonstrate how the expectation value \( \expectation{\Bx} \) changes under translation.

Answer

(a)

For clarity, let’s set \( x_i = y \). The general result will be clear despite doing so.

\begin{equation}\label{eqn:translation:60}
\antisymmetric{y}{J(\Bl)}
=
\sum_{k= 0} \inv{k!} \lr{\frac{-i}{\Hbar}}
\antisymmetric{y}{
\lr{ \Bp \cdot \Bl }^k
}.
\end{equation}

The commutator expands as

\begin{equation}\label{eqn:translation:80}
\begin{aligned}
\antisymmetric{y}{
\lr{ \Bp \cdot \Bl }^k
}
+ \lr{ \Bp \cdot \Bl }^k y
&=
y \lr{ \Bp \cdot \Bl }^k \\
&=
y \lr{ p_x l_x + p_y l_y + p_z l_z } \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ p_x l_x y + y p_y l_y + p_z l_z y } \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ p_x l_x y + l_y \lr{ p_y y + i \Hbar } + p_z l_z y } \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ \Bp \cdot \Bl } y \lr{ \Bp \cdot \Bl }^{k-1}
+ i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1} \\
&= \cdots \\
&=
\lr{ \Bp \cdot \Bl }^{k-1} y \lr{ \Bp \cdot \Bl }^{k-(k-1)}
+ (k-1) i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ \Bp \cdot \Bl }^{k} y
+ k i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1}.
\end{aligned}
\end{equation}

In the above expansion, the commutation of \( y \) with \( p_x, p_z \) has been used. This gives, for \( k \ne 0 \),

\begin{equation}\label{eqn:translation:100}
\antisymmetric{y}{
\lr{ \Bp \cdot \Bl }^k
}
=
k i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1}.
\end{equation}

Note that this also holds for the \( k = 0 \) case, since \( y \) commutes with the identity operator. Plugging back into the \( J \) commutator, we have

\begin{equation}\label{eqn:translation:120}
\begin{aligned}
\antisymmetric{y}{J(\Bl)}
&=
\sum_{k = 1} \inv{k!} \lr{\frac{-i}{\Hbar}}
k i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
l_y \sum_{k = 1} \inv{(k-1)!} \lr{\frac{-i}{\Hbar}}
\lr{ \Bp \cdot \Bl }^{k-1} \\
&=
l_y J(\Bl).
\end{aligned}
\end{equation}

The same pattern clearly applies with the other \( x_i \) values, providing the desired relation.

\begin{equation}\label{eqn:translation:140}
\antisymmetric{\Bx}{J(\Bl)} = \sum_{m = 1}^3 \Be_m l_m J(\Bl) = \Bl J(\Bl).
\end{equation}

(b)

Suppose that the translated state is defined as \( \ket{\alpha_{\Bl}} = J(\Bl) \ket{\alpha} \). The expectation value with respect to this state is

\begin{equation}\label{eqn:translation:160}
\begin{aligned}
\expectation{\Bx’}
&=
\bra{\alpha_{\Bl}} \Bx \ket{\alpha_{\Bl}} \\
&=
\bra{\alpha} J^\dagger(\Bl) \Bx J(\Bl) \ket{\alpha} \\
&=
\bra{\alpha} J^\dagger(\Bl) \lr{ \Bx J(\Bl) } \ket{\alpha} \\
&=
\bra{\alpha} J^\dagger(\Bl) \lr{ J(\Bl) \Bx + \Bl J(\Bl) } \ket{\alpha} \\
&=
\bra{\alpha} J^\dagger J \Bx + \Bl J^\dagger J \ket{\alpha} \\
&=
\bra{\alpha} \Bx \ket{\alpha} + \Bl \braket{\alpha}{\alpha} \\
&=
\expectation{\Bx} + \Bl.
\end{aligned}
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.