What Is the Simplified Commutator [A,B] for an Isotropic Harmonic Oscillator?

[Sunshine]

Homework Statement

Simplify the commutator [A,B] and give the expectation value of [A,B] in the ground state for an isotropic harmonic oscillator (mass m) that has the energy $$\hbar \omega /2$$ when
$$A = xp_x$$
$$B = y$$

Homework Equations

$$[AB,C] = A[B,C] + [A,C]B$$
$$[p_i,x_j] = i\hbar\delta_{ij}$$

The Attempt at a Solution

$$[xp_x,y] = x[p_x,y] + [x,y]p_x$$ (first relation)
$$x[p_x,y] = 0$$ (second relation)
Last term with test function f(x)
$$[x,y]p_x f(x) = xy\frac\hbar i \dfrac{\partial}{\partial x}f(x) - yx \frac\hbar i \dfrac{\partial}{\partial x}f(x) = 0 ?$$

I have a feeling that 0 isn't the answer, since I have to find the expectation value as well. If the last equation doesn't become 0 but the middle equation is the most simplified answer, I don't know how to find an expectation value that isn't equal to 0 (because I get that when I put it into the usual integral for expectation value)

 

[nrqed]

Homework Statement

Simplify the commutator [A,B] and give the expectation value of [A,B] in the ground state for an isotropic harmonic oscillator (mass m) that has the energy $$\hbar \omega /2$$ when
$$A = xp_x$$
$$B = y$$

Homework Equations

$$[AB,C] = A[B,C] + [A,C]B$$
$$[p_i,x_j] = i\hbar\delta_{ij}$$

The Attempt at a Solution

$$[xp_x,y] = x[p_x,y] + [x,y]p_x$$ (first relation)
$$x[p_x,y] = 0$$ (second relation)
Last term with test function f(x)
$$[x,y]p_x f(x) = xy\frac\hbar i \dfrac{\partial}{\partial x}f(x) - yx \frac\hbar i \dfrac{\partial}{\partial x}f(x) = 0 ?$$

I have a feeling that 0 isn't the answer, since I have to find the expectation value as well. If the last equation doesn't become 0 but the middle equation is the most simplified answer, I don't know how to find an expectation value that isn't equal to 0 (because I get that when I put it into the usual integral for expectation value)

the commutator of the A and B you gave above is indeed zero which is obvious because y commutes with p_x and with y as well.

Are you sure that A is not x p_y instead of x p_x?? Because then the calculation would be more interesting.

 

[Sunshine]

yes, it is $$xp_x$$.

So the expectation value is 0 as well?

 

[nrqed]

yes, it is $$xp_x$$.

So the expectation value is 0 as well?

Yes, it is zero.


Strange question

 

[Sunshine]

I'm confused because it seems too easy to count expectation value for 0. Especially when question 2 (out of 3 in the assignment) has a trivial expectation value too. Doing it for 3rd time just to make sure I'm not missing something. If it becomes zero again, I'll be almost certain that something is wrong.

 

[nrqed]

I'm confused because it seems too easy to count expectation value for 0. Especially when question 2 (out of 3 in the assignment) has a trivial expectation value too. Doing it for 3rd time just to make sure I'm not missing something. If it becomes zero again, I'll be almost certain that something is wrong.

is that a question from a book or was it typed up by your professor? I would think that he/she may have made a mistake or that there is a typo in the book.
(or do they want the expectation value of AB instead of [A,B]?)

 

[Sunshine]

The professor took the question from an unknown book, typed it into word and emailed it. I don't think it's a typo, and I'd never dare to ask if it was. He wouldn't admit his mistake anyway ;)

ok, second question is like the first one with a few differences:

Now I have
$$A = L_x$$
$$B = z_x^2$$

And it's not a harmonic oscillator, but a hydrogen atom in ground state. My solution:

$$[L_x, zx^2] = [yp_z - zp_y, zx^2]$$
$$=[yp_z,zx^2]-[zp_y,zx^2]$$

First term: $$y(\mathbf{[p_z,z]x^2} + z[p_z,x^2])+([y,z]x^2 + z[y,x^2])p_z$$
Second term:$$z([p_y,z]x^2 + z[p_y,x^2])+([z,z]x^2 + z[z,x^2])p_y$$

Only the term in bold is non-zero, which gives:

$$y[p_z,z]x^2 = -i\hbar y x^2$$

The squared absolute value of wavefunction for hydrogen atom in ground state:
$$|\Psi |^2 = \dfrac{1}{\pi a^3} e^{-2r/a}$$

Hence, the expectation value is (changing to spherical coordinates):
$$-ih<yx^2> = ih<r^2 \sin^3\theta \cos^2\phi \sin\phi > = C \int_0^{2\pi} \int_0^\pi \int_0^a (r^2 \sin^3\theta \cos^2\phi \sin\phi) e^{-2r/a} r^2 \sin \theta dr d\theta d\phi$$

This is 0! (Tried with Mathematica) I'm not sure about the interval for r; if it's from 0 to a or from 0 to infinity. In the last case the integral doesn't converge.

Can you see where I'm going wrong?

 

[nrqed]

The professor took the question from an unknown book, typed it into word and emailed it. I don't think it's a typo, and I'd never dare to ask if it was. He wouldn't admit his mistake anyway ;)

ok, second question is like the first one with a few differences:

Now I have
$$A = L_x$$
$$B = z_x^2$$

And it's not a harmonic oscillator, but a hydrogen atom in ground state. My solution:

$$[L_x, zx^2] = [yp_z - zp_y, zx^2]$$
$$=[yp_z,zx^2]-[zp_y,zx^2]$$

First term: $$y(\mathbf{[p_z,z]x^2} + z[p_z,x^2])+([y,z]x^2 + z[y,x^2])p_z$$
Second term:$$z([p_y,z]x^2 + z[p_y,x^2])+([z,z]x^2 + z[z,x^2])p_y$$

Only the term in bold is non-zero, which gives:

$$y[p_z,z]x^2 = -i\hbar y x^2$$

The squared absolute value of wavefunction for hydrogen atom in ground state:
$$|\Psi |^2 = \dfrac{1}{\pi a^3} e^{-2r/a}$$

Hence, the expectation value is (changing to spherical coordinates):
$$-ih<yx^2> = ih<r^2 \sin^3\theta \cos^2\phi \sin\phi > = C \int_0^{2\pi} \int_0^\pi \int_0^a (r^2 \sin^3\theta \cos^2\phi \sin\phi) e^{-2r/a} r^2 \sin \theta dr d\theta d\phi$$

This is 0! (Tried with Mathematica) I'm not sure about the interval for r; if it's from 0 to a or from 0 to infinity. In the last case the integral doesn't converge.

Can you see where I'm going wrong?

It (almost) all looks right.

Within doing any integral it is clear that the integration will give zero. The ground state wavefunction is spherically symmetric so it's even under $$x,y,z \rightarrow -x,-y,-z$$ but the integrand xy^2 is obviously odd, so the integral must vanish.

The only wrong thing I can see is that you should be able to integrate r from 0 to infinity (the exponential will make any integral with powers of r converge). But that does not change the final conclusion, the integral gives zero.

 

[Sunshine]

You are right, the last integral becomes 0 as well -I accidentally made the assumption that a<0 istead of a>0 in Mathematica last time.

So the conclusion is that the expectation value of both commutators is 0?

 

[nrqed]

You are right, the last integral becomes 0 as well -I accidentally made the assumption that a<0 istead of a>0 in Mathematica last time.

I am not sure what you mean by last integral but if you mean the integral over r, then no, it does not give zero by itself. And I am not sure what "a" is. You mean the Bohr radius? No, the radial integral should be from 0 to infinity, period. But the angular integral will give zero so that the final result is zero.

So the conclusion is that the expectation value of both commutators is 0?

Looks like it to me.