Working with X and P operators in QM

[razidan]

Homework Statement

Consider the eigenstates of a particle in an infinite well with walls at $x=\pm a$.
without explicitly evaluating any integrals, what is the expectation value of the following operator
$$\hat{x}^2\hat{p_x}^3+3\hat{x}\hat{p_x}^3\hat{x}+\hat{p_x}^3\hat{x}^2$$

Homework Equations

$1) \langle\hat{A}\rangle=\langle \varphi|\hat{A}|\varphi\rangle$
$2) [x,p_x]=i\hbar$ (at least, I think it's relevant).

The Attempt at a Solution

I finished my QM course almost 4 years ago, and I'm now trying to get back at it, so I pretty much forgot everything, and I'm totally lost.
I tried plugging the operator into equation 1 but i think i missing some math identities... maybe properties of hermitian operators? not sure at all.
I would love some direction...

Thanks
R.

 

[TSny]

Do the energy-eigenstate wavefunctions have any symmetry about x = 0?

 

[razidan]

Do the energy-eigenstate wavefunctions have any symmetry about x = 0?

The wave functions are the symmetric and antisymmetric solutions, if this is what you mean... but I am not sure how this helps me.
I thought the solution to the question is somewhat general, and that I don't need to look into the specifics of the wavefunction.

 

[TSny]

Suppose you have an operator $\hat A$ such that when it acts on an even function it converts it into an odd function; and when it acts on an odd function it converts it into an even function. For a state described by an even or odd function, what can you say about $\langle\hat{A}\rangle$?

 

[razidan]

Suppose you have an operator $\hat A$ such that when it acts on an even function it converts it into an odd function; and when it acts on an odd function it converts it into an even function. For a state described by an even or odd function, what can you say about $\langle\hat{A}\rangle$?

it will be zero.
So i see that $\langle\hat{x}^n\rangle$ or $\langle\hat{p}^n\rangle$ will be zero, if n is odd. but can I say something general about a combination of x and p?
$\hat{p}\hat{x}|\varphi\rangle$ does not have defined parity anymore...

 

[TSny]

If $\phi(x)$ is an even function, then what is the parity of $\hat x \phi(x)$? What is the parity of $\hat p \hat x \phi(x)$?
Think of $\hat p \hat x \phi(x)$ as $\hat p \left( \hat x \phi (x) \right)$.

 

[razidan]

If $\phi(x)$ is an even function, then what is the parity of $\hat x \phi(x)$? What is the parity of $\hat p \hat x \phi(x)$?
Think of $\hat p \hat x \phi(x)$ as $\hat p \left( \hat x \phi (x) \right)$.

Ok, so after reading about it, I saw that the derivative of an even function is always odd, and vice versa.
But is this the only solution? the play on parity? I see that is one way to prove it, but it seems a little... meh.
I thought the solution was going to be more work with the commutator, and then seeing how terms cancel out... but i guess you can't get everything you want...

Thanks for the help!

 

[TSny]

The parity approach is just the way I happened to think about it. Maybe whoever wrote the problem wants you to do it some other way. But, I think you can check that you would get very different results for the expectation value of the operator for the energy eigenstates of a well that extends from x = 0 to x = a instead of x = -a/2 to x = a/2.

 

[DrClaude]

But is this the only solution? the play on parity? I see that is one way to prove it, but it seems a little... meh.

Don't underestimate the power of using symmetries! There are many results that can be obtained just from symmetry arguments that can save you a lot of time.

 

[razidan]

The parity approach is just the way I happened to think about it. Maybe whoever wrote the problem wants you to do it some other way. But, I think you can check that you would get very different results for the expectation value of the operator for the energy eigenstates of a well that extends from x = 0 to x = a instead of x = -a/2 to x = a/2.

what do you mean? won't it be just that $\langle \hat{x} \rangle = a/2$?
a change of origin is arbitrary...

 

[PeroK]

what do you mean? won't it be just that $\langle \hat{x} \rangle = a/2$?
a change of origin is arbitrary...

Your operator includes the momentum operator, so you need to be careful with that assumption.

 

[razidan]

Your operator includes the momentum operator, so you need to be careful with that assumption.

Oh, of course, i just meant $\langle \hat{x} \rangle$, but thinking about it, i have two questions:
1) Is there a way to see what the average value for this operator without doing the calculation?
2) an average value is an observable, so it must have some connection to the zero value we get when the well is from -a/2 to a/2. doesn't it?
how is it that an observable is dependent on the origin?
(I mean, i can think of the example above, where the change of the origin changed the value of $\langle \hat{x} \rangle$, but that was just shift of a/2, directly because we shifted the origin)
I hope my question is clear.

 

[PeroK]

Oh, of course, i just meant $\langle \hat{x} \rangle$, but thinking about it, i have two questions:
1) Is there a way to see what the average value for this operator without doing the calculation?
2) an average value is an observable, so it must have some connection to the zero value we get when the well is from -a/2 to a/2. doesn't it?
how is it that an observable is dependent on the origin?
(I mean, i can think of the example above, where the change of the origin changed the value of $\langle \hat{x} \rangle$, but that was just shift of a/2, directly because we shifted the origin)
I hope my question is clear.

Take a simple example. Let's say a measurement of position was $0$ and a measurement of momentum was $p$. The product of the two is $0$.

If you change the origin, the measurements become $a/2$ and $p$ and the product is $ap/2$.

Also, note that another fundamental problem with your argument is that you are not taking dimensions (mass, length, time) into account.

 

[razidan]

Take a simple example. Let's say a measurement of position was $0$ and a measurement of momentum was $p$. The product of the two is $0$.

If you change the origin, the measurements become $a/2$ and $p$ and the product is $ap/2$.

Also, note that another fundamental problem with your argument is that you are not taking dimensions (mass, length, time) into account.

I understand the example, but my question is more fundamental, WHY does a change of origin affect an observable?

 

[PeroK]

I understand the example, but my question is more fundamental, WHY does a change of origin affect an observable?

It doesn't, except that you get different numerical values.
In fact, this whole question ought to be turned round. You should be justifying your assertion, rather than make an unsubstantiated assertion and ask why it isn't true.