Quantum Mechanics: Translation and Wave Function

[Robben]

Homework Statement

Let $|\psi\rangle \to |\psi'\rangle = \hat{T}(\delta x)|\psi\rangle$ for infinitesimal $\delta x$. Show that $\langle x \rangle = \langle x \rangle + \delta x$ and $\langle p_x \rangle = \langle p_x\rangle.$

Homework Equations

$\hat{T}(\delta x) = e^{-i\hat{p}_x\delta x/\hbar}$

The Attempt at a Solution

I am confused. Why would $\langle x \rangle = \langle x \rangle + \delta x$?
Shouldn't it equal $\langle x \rangle?$
Since, $\langle x\rangle = \langle \psi'|\hat{x}|\psi'\rangle = \langle \psi'|x\hat{T}(\delta x)|\psi\rangle = \langle \psi |\hat{T}^{\dagger}(\delta x)\hat{T}(\delta x)x|\psi\rangle = \langle x\rangle.$

 

[jfizzix]

The translation operator doesn't commute with the position observable.

So what you actually have, is:
\langle x\rangle = \langle \psi'|\hat{x}|\psi'\rangle = \langle \psi'|\hat{x}\hat{T}(\delta x)|\psi\rangle = \langle \psi |\hat{T}^{\dagger}(\delta x)\hat{x}\hat{T}(\delta x)|\psi\rangle

If you evaluate \langle \psi |\hat{T}^{\dagger}(\delta x)\hat{x}\hat{T}(\delta x)|\psi\rangle
while keeping in mind the order of the observables, and using the approximation
\hat{T}(\delta x) = e^{-i\frac{\hat{p}_x\delta x}{\hbar}}\approx I - i\frac{\hat{p}_{x}}{\hbar} \delta x
since \delta x is indeed an infinitesimal translation,
I expect you'll get the result you're looking for.

 

[Robben]

The translation operator doesn't commute with the position observable.

So what you actually have, is:
\langle x\rangle = \langle \psi'|\hat{x}|\psi'\rangle = \langle \psi'|\hat{x}\hat{T}(\delta x)|\psi\rangle = \langle \psi |\hat{T}^{\dagger}(\delta x)\hat{x}\hat{T}(\delta x)|\psi\rangle

If you evaluate \langle \psi |\hat{T}^{\dagger}(\delta x)\hat{x}\hat{T}(\delta x)|\psi\rangle
while keeping in mind the order of the observables, and using the approximation
\hat{T}(\delta x) = e^{-i\frac{\hat{p}_x\delta x}{\hbar}}\approx I - i\frac{\hat{p}_{x}}{\hbar} \delta x
since \delta x is indeed an infinitesimal translation,
I expect you'll get the result you're looking for.

Oh I see, so my method is wrong because I cannot commute the translation operator with position, but I am not sure how $I - i\frac{\hat{p}_{x}}{\hbar} \delta x$ will help me here?

 

[TSny]

In the expression $\hat{T}^{\dagger}(\delta x)\hat{x}\hat{T}(\delta x)$, approximate $\hat{T}^{\dagger}(\delta x)$ as well as $\hat{T}(\delta x)$ and simplify the overall expression.

 

[Robben]

In the expression $\hat{T}^{\dagger}(\delta x)\hat{x}\hat{T}(\delta x)$, approximate $\hat{T}^{\dagger}(\delta x)$ as well as $\hat{T}(\delta x)$ and simplify the overall expression.

So using the suggestion you guys provided, I got: $$\langle x\rangle = \langle \psi'|\hat{x}|\psi'\rangle = \langle \psi'|\hat{x}\hat{T}(\delta x)|\psi\rangle = \langle \psi |\hat{T}^{\dagger}(\delta x)\hat{x}\hat{T}(\delta x)|\psi\rangle$$ $$ = \langle \psi|\hat{T}(\delta x) x + [x,\hat{T}(\delta x)])|\psi\rangle$$ $$ = \langle\psi|\hat{T}^{\dagger}(\delta x)\hat{T}(\delta x)x|\psi\rangle + \langle\psi|\hat{T}(\delta x)|[x,e^{-ip_x\delta_x/\hbar}]|\psi\rangle, $$ but I am not sure how to proceed after replacing $e^{-ip_x\delta_x/\hbar}$ with $1 - i\frac{\hat{p}_{x}}{\hbar} \delta x?$

 

[jfizzix]

Oh I see, so my method is wrong because I cannot commute the translation operator with position, but I am not sure how $I - i\frac{\hat{p}_{x}}{\hbar} \delta x$ will help me here?

If
\bar{T}(\delta x)\approx I - i\frac{\hat{p}_{x}}{\hbar}\delta\hat{x},
then
\langle\psi|\bar{T}^{\dagger}(\delta x)\hat{x}\bar{T}(\delta x)|\psi\rangle \approx \langle\psi|( I +i\frac{\hat{p}_{x}}{\hbar}\delta\hat{x})\hat{x}( I - i\frac{\hat{p}_{x}}{\hbar}\delta\hat{x})\psi\rangle
\approx \langle\psi| \hat{x} +\frac{i}{\hbar}\hat{p}_{x}\delta\hat{x}\hat{x} - \frac{i}{\hbar}\hat{x}\hat{p}_{x}\delta\hat{x}|\psi\rangle
Note that the last term in the expansion was of the order (\delta x)^{2}, and so can be neglected.

Knowing the commutator [\hat{x},\hat{p}_{x}]=i\hbar, the rest is straightforward.

 

[Robben]

If
\bar{T}(\delta x)\approx I - i\frac{\hat{p}_{x}}{\hbar}\delta\hat{x},
then
\langle\psi|\bar{T}^{\dagger}(\delta x)\hat{x}\bar{T}(\delta x)|\psi\rangle \approx \langle\psi|( I +i\frac{\hat{p}_{x}}{\hbar}\delta\hat{x})\hat{x}( I - i\frac{\hat{p}_{x}}{\hbar}\delta\hat{x})\psi\rangle
\approx \langle\psi| \hat{x} +\frac{i}{\hbar}\hat{p}_{x}\delta\hat{x}\hat{x} - \frac{i}{\hbar}\hat{x}\hat{p}_{x}\delta\hat{x}|\psi\rangle
Note that the last term in the expansion was of the order (\delta x)^{2}, and so can be neglected.

Knowing the commutator [\hat{x},\hat{p}_{x}]=i\hbar, the rest is straightforward.

Oh, I see! Thank you very much!

 

[Robben]

I actually ran into a problem that is similar to this. It says if we modify the wave equation by a position dependent phase, i.e. $e^{ip_ox/\hbar}$ then $\langle x\rangle = \langle x \rangle$ and $\langle p_x \rangle = \langle p_x \rangle + p_o$, but why is that?

 

[jfizzix]

I actually ran into a problem that is similar to this. It says if we modify the wave equation by a position dependent phase, i.e. $e^{ip_ox/\hbar}$ then $\langle x\rangle = \langle x \rangle$ and $\langle p_x \rangle = \langle p_x \rangle + p_o$, but why is that?

e^{i\frac{p_{0}}{\hbar}\hat{x}} is a translation operator in momentum space.

You can see this for yourself by expanding this exponential as a power series, and noting that the position operator acts like a derivative in momentum space, giving you a Taylor series expansion of the displaced momentum-space wavefunction.

Translations in momentum space will shift the expectation values of momentum observables,
but they will not alter the expectation values of position observables.
(e.g., changing your reference frame in a single instant doesn't change where things are, only how they are moving relative to you)

 

[Robben]

e^{i\frac{p_{0}}{\hbar}\hat{x}} is a translation operator in momentum space.

You can see this for yourself by expanding this exponential as a power series, and noting that the position operator acts like a derivative in momentum space, giving you a Taylor series expansion of the displaced momentum-space wavefunction.

Translations in momentum space will shift the expectation values of momentum observables,
but they will not alter the expectation values of position observables.
(e.g., changing your reference frame in a single instant doesn't change where things are, only how they are moving relative to you)

Expanding Taylor series for e^{i\frac{p_{0}}{\hbar}\hat{x}} gives $1+\frac{ip_ox}{\hbar}$.
Expanding Taylor series for e^{-i\frac{p_{x}}{\hbar}\delta x} gives $1-\frac{ip_x\delta x}{\hbar}$.

Since they are both translation operators why does one comute with the position observable and the other does not?

 

[DrClaude]

Since they are both translation operators why does one comute with the position observable and the other does not?

One is the translation operator in position space, the other in momentum space. Look at the operators: you have $e^{i\frac{p_{0}}{\hbar}\hat{x}}$ and $e^{-i\frac{\hat{p}_{x}}{\hbar}\delta x}$. The position operator $\hat{x}$ commutes with itself, while $\hat{p}_x$ doesn't commute with $\hat{x}$.

 

[Robben]

One is the translation operator in position space, the other in momentum space. Look at the operators: you have $e^{i\frac{p_{0}}{\hbar}\hat{x}}$ and $e^{-i\frac{\hat{p}_{x}}{\hbar}\delta x}$. The position operator $\hat{x}$ commutes with itself, while $\hat{p}_x$ doesn't commute with $\hat{x}$.

What makes one an operator of momentum space and the other an operator of position space? I tried to search this online but I couldn't find any links that distinguished the two operators. Also, why does $p_o$ commute with x?

 

[DrClaude]

What makes one an operator of momentum space and the other an operator of position space? I tried to search this online but I couldn't find any links that distinguished the two operators.

But that exactly the point of these exercises!

Homework Statement

Let $|\psi\rangle \to |\psi'\rangle = \hat{T}(\delta x)|\psi\rangle$ for infinitesimal $\delta x$. Show that $\langle x \rangle = \langle x \rangle + \delta x$ and $\langle p_x \rangle = \langle p_x\rangle.$

This shows that $\hat{T}(\delta x)$ is an operator that translates a wave function in position space, while leaving the momentum unchanged. Likewise for the problem in post #8.

Also, why does $p_o$ commute with x?

It's a number, not an operator.