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I need to prove the commutation relation,

$$[p^2,f] = 2 \frac{\hbar}{i}\frac{\partial f}{\partial x} p - \hbar^2 \frac{\partial^2 f}{\partial x^2}$$

where $f \equiv f(\vec{r})$ and $\vec{p} = p_x \vec{i}$

I know

$$[AB,C] = A[B,C] + [A,C]B$$

Applying this, I get

$$[p^2,f] = p[p,f] + [p,f]p$$

where $p = \frac{\hbar}{i}\frac{\partial}{\partial x}$, and $[p,f] \equiv pf - fp$

using a trial function, $g(x)$, I get

$$[p^2,f] = p[p,f]g + [p,f]pg$$

$$= \frac{\hbar}{i}\frac{\partial}{\partial x}\left[\frac{\hbar}{i}\frac{\partial fg}{\partial x} - f\frac{\hbar}{i}\frac{\partial g}{\partial x}\right] + \left[\frac{\hbar}{i}\frac{\partial fg}{\partial x} - f\frac{\hbar}{i}\frac{\partial g}{\partial x}\right] \frac{\hbar}{i}\frac{\partial}{\partial x}$$

using the product rule

$$ = -\hbar^2 \frac{\partial}{\partial x}\left[g \frac{\partial f}{\partial x} + f \frac{\partial g}{\partial x} - f \frac{\partial g}{\partial x} \right] + \left[g \frac{\partial f}{\partial x} + f \frac{\partial g}{\partial x} - f \frac{\partial g}{\partial x} \right] \frac{\hbar}{i}^2 \frac{\partial}{\partial x}$$

cancelling the like terms in the brackets gives

$$= -\hbar^2 \frac{\partial}{\partial x} \left[g \frac{\partial f}{\partial x}\right] - \left[g \frac{\partial f}{\partial x}\right]\hbar^2 \frac{\partial}{\partial x}$$

using the product rule again gives

$$ = -\hbar^2 \left[\frac{\partial g}{\partial x} \frac{\partial f}{\partial x} + g \frac{\partial^2 f}{\partial x^2} \right] - \left[g \frac{\partial f}{\partial x}\right] \hbar^2 \frac{\partial}{\partial x}$$

$\frac{\partial g}{\partial x} = 0$, so

$$ = -\hbar^2 g \frac{\partial^2}{\partial x^2} - \hbar^2 g \frac{\partial f}{\partial x} \frac{\partial}{\partial x}$$

Substituting the momentum operator back in gives

$$ = -\hbar^2 g \frac{\partial^2}{\partial x^2} - \frac{\hbar}{i} g \frac{\partial f}{\partial x} p$$

The trial function, $g$, can now be dropped,

$$ = -\hbar^2 \frac{\partial^2}{\partial x^2} - \frac{\hbar}{i}\frac{\partial f}{\partial x}p$$

But this is not what I was supposed to arrive at. Where did I go wrong?

Art Brown
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user41767
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1 Answers1

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I didn't read your answer, but let's think about just computing the operator $\partial_x^2 f$. First we need to compute the operator $\partial_x f$. Now I am saying "the operator" because we are viewing $\partial_x f$ as a composition of first multiplying by $f$ and then taking the derivative. By the product rule, we know $\partial_x f = (\partial_x f) + f \partial_x$, were by $(\partial_x f)$, I really do just mean multiplication by the derivative of $f$.

Now lets try to compute $\partial_x^2 f$. It is $\partial_x [(\partial_x f) + f \partial_x] = (\partial_x^2 f) + (\partial_x f) \partial_x + (\partial_x f) \partial_x + f \partial_x^2 = (\partial_x^2 f) +2(\partial_x f) \partial_x + f \partial_x^2$.

Then $[\partial_x^2,f] = \partial_x^2f-f \partial_x^2 = (\partial_x^2 f) +2(\partial_x f) \partial_x$. If you understand this then you should get the right answer. You just need to put in the appropriate $i$'s and $\hbar$'s.

Brian Moths
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  • How do you know to treat ∂xf as a composition and apply the product rule to it? – user41767 Mar 07 '14 at 03:21
  • I am not sure what your question is exactly. I probably messed this up the first time I did it too. So I guess the answer is experience, but if you are careful enough to really think about it, you should be able to reason it out. For example, would you expect $\hat{P} \hat{X} |\psi \rangle = -i \hbar |\psi \rangle$? – Brian Moths Mar 07 '14 at 04:16
  • My question was how did you know that ∂f needed to be treated with the product rule? Once you said it was a composition, it made sense, but I didn't see it as a composition when I tried to solve the problem.

    As for

    PX|ψ⟩=−iℏ|ψ⟩

    At first glance, I would expect that. If it is not correct could you explain why?

     – user41767 Mar 07 '14 at 15:38
  • if $\hat{P}\hat{X}|\psi \rangle=−i\hbar |\psi \rangle$. Then the inverse of $\hat{X}$ is basically $\hat{P}$ and so $\hat{X}$ and $\hat{P}$ commute, but quantum mechanics depends very crucially on $\hat{X}$ and $\hat{P}$ not commuting. – Brian Moths Mar 07 '14 at 15:57
  • Ok, well this addresses something else that I am trying to piece together. When do I use commuters? – user41767 Mar 07 '14 at 16:47
  • I was about to say that should be its own question, but I don't think that would go over too well on this site. You can try. Anyway commutators help you prove stuff about eigenstates without explicity finding a basis which diagonalizes the hamiltonian. This is done by defining ladder operators which obey certain commutation relations with the hamiltonian. A key place this is used is to find the spectrum of the Quantum Simple Harmonic Oscillator. – Brian Moths Mar 07 '14 at 16:55
  • The same technique is used when dealing with angular momentum. – Brian Moths Mar 07 '14 at 16:56
  • I understand the use of the commuter in finding the spectrum of the simple harmonic oscillator, but what I don't understand is the reason why the fact that P and X don't commute is the reason that PX|ψ⟩≠−iℏ|ψ⟩. I know that if they did commute, then PX|ψ⟩ = XP|ψ⟩. Maybe I do understand now. Take two different operators A, and B, and say [A,B]=0, and A|ψ⟩=a|ψ⟩ and B|ψ⟩=b|ψ⟩. Then would AB|ψ⟩=ab|ψ⟩? Is that what I did not understand before? – user41767 Mar 07 '14 at 18:01
  • I think what you didn't understand is that the operator $\hat{P}$ can't know what expression it is operating on. Suppose we have a wave function $| k \rangle = e^{ikx}$, and define $|\psi \rangle$ to be $\hat{X} | k \rangle = xe^{ikx}$. Then we must have $\hat{P} \hat{X} | k \rangle = \hat{P} | \psi \rangle$, since $\hat{X} | k \rangle = | \psi \rangle$. The two expressions cannot be different just because one has a $\hat{X}$ to the right of $\hat{P}$. – Brian Moths Mar 07 '14 at 19:07
  • The only way to consistently make sense of these expressions is if you evaluate the operators from right to left so that $\hat{P}$ acts on everything to its right insteand of doing it the way you did it. – Brian Moths Mar 07 '14 at 19:07