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my class is using Griffiths, and I've been assigned to prove the Virial theorem. Of course, there are innumerable resources devoted to working through solutions to his problems, and I use this gentleman's excellent work to check my answers and clarify things I don't understand: http://www.physicspages.com/2012/10/09/virial-theorem/

In this case, however, I'm not able to follow one of his steps and I wonder if I'm misunderstanding how to compute commutators.

I follow him up until step 4; step 5 simply doesn't seem to follow and I can't figure out what I'm doing wrong.

For step 5 I'm getting : $-\frac{\hbar^3}{2mi}\left[\frac{\partial ^2x}{\partial x^2}\frac{\partial}{\partial x} + x\frac{\partial ^3}{\partial x^3}\right]f + \frac{x\hbar}{i}\left[V\frac{\partial f}{\partial x} - \frac{\partial V}{\partial x} f \right] + \frac{\hbar^3}{2mi}x\frac{\partial ^3 f}{\partial x^3}$.

I feel like $\frac{\partial ^2x}{\partial x^2}\frac{\partial}{\partial x}= 0$ because the second derivative of x is zero, and then the other term in the brackets on the left should cancel with the last term. That only leaves the middle term; I don't think this is correct but I can't for life of me figure out what mistake I'm making.

P.S. I'm actually supposed to prove the 3D version, but I'm working through the 1D version first; I'm a little concerned that if I can't figure out the 1D version the 3D will be a nightmare; though it seems superficially like it ought to be a straightforward generalization.

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  • $\begingroup$ It is not just $\frac{\partial^2{x}}{\partial{x}^2}\frac{\partial}{\partial{x}}$, it is $\frac{\partial^2{x}}{\partial{x}^2}\frac{\partial{f}}{\partial{x}}$, Which is $\frac{\partial}{\partial{x}}\frac{\partial{x}}{\partial{x}}\frac{\partial{f}}{\partial{x}}$, the $\frac{\partial{x}}{\partial{x}}=1$, so it becomes $$\frac{\partial^2{f}}{\partial{x}^2}$$ $\endgroup$ Nov 28 '16 at 2:56
  • $\begingroup$ you can also prove Virial's theorem using Heisenberg's equation of motion also.. Please Refer the following link: Heisenberg's EOM $\endgroup$ Nov 28 '16 at 3:04
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Note that

\begin{eqnarray} \frac{\partial^2}{\partial x^2}\left(x \frac{\partial f}{\partial x}\right) &=& \frac{\partial }{\partial x}\left[\frac{\partial }{\partial x} \left(x \frac{\partial f}{\partial x}\right)\right] = \frac{\partial }{\partial x}\left[x\frac{\partial^2f }{\partial x^2} + \frac{\partial f}{\partial x}\right] \\ &=& x \frac{\partial^3f }{\partial x^3} + \frac{\partial^2f }{\partial x^2} + \frac{\partial^2f }{\partial x^2} = x \frac{\partial^3f }{\partial x^3} + 2\frac{\partial^2f }{\partial x^2} \tag{1} \end{eqnarray}

Now

\begin{eqnarray} [H, xp]f &=& \left(-\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial x^2} + V(x)\right)\left(x + \frac{\hbar}{i} \frac{\partial }{\partial x}\right)f - \left(x + \frac{\hbar}{i} \frac{\partial }{\partial x}\right)\left(-\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial x^2} + V(x)\right)f \\ &=&-\frac{\hbar^3}{2mi}\frac{\partial^2 }{\partial x^2} \left(x \frac{\partial f}{\partial x}\right) + \frac{\hbar}{i}V(x)x\frac{\partial f}{\partial x} + \frac{\hbar^3}{2mi} x \frac{\partial^3f }{\partial x^3} - x\frac{\hbar}{i}\frac{\partial }{\partial x}\left(V(x)f \right) \\ &\stackrel{(1)}{=}& -\frac{\hbar^3}{2mi} \left(x \frac{\partial^3f }{\partial x^3} + 2\frac{\partial^2f }{\partial x^2} \right) + \frac{\hbar}{i}V(x)x\frac{\partial f}{\partial x} + \frac{\hbar^3}{2mi} x \frac{\partial^3f }{\partial x^3} - x\frac{\hbar}{i}\left(\frac{\partial V}{\partial x}f + V\frac{\partial f}{\partial x} \right) \\ &=& -\frac{\hbar^3}{mi} \frac{\partial^2f }{\partial x^2} - \frac{\hbar}{i}x\frac{\partial V}{\partial x}f \tag{2} \end{eqnarray}

We can then conclude that

$$ \frac{i}{\hbar}[H, xp] = -\frac{\hbar^2}{m} \frac{\partial^2 }{\partial x^2} - x\frac{\partial V}{\partial x} = 2\frac{p^2}{2m} - x\frac{\partial V}{\partial x} = 2 T - x\frac{\partial V}{\partial x}\tag{3} $$

Using Ehrenfest Theorem we arrive to

$$ \frac{d}{dt}\langle xp \rangle =2 \langle T \rangle - \left\langle x \frac{\partial V}{\partial x}\right\rangle $$

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  • $\begingroup$ Thanks! I was making a dumb mistake regarding the order in which operators act (right to left) and so I missed to instances in which the product rule was required. Should have known better. $\endgroup$
    – BenL
    Nov 28 '16 at 13:38

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