Quantum Virial Theorem Derivation

[flyusx]

Homework Statement

Using the time-dependence of an operator equation, show that $\frac{d}{dt}\left\langle\hat{x}\hat{p}\right\rangle=2\langle\widehat{KE}\rangle-\left\langle x\frac{\partial V}{\partial x}\right\rangle$

Relevant Equations

$$\frac{d}{dt}\langle\hat{A}\rangle=\frac{i}{\hbar}\langle[\hat{H},\hat{A}]\rangle+\left\langle\frac{\partial\hat{A}}{\partial t}\right\rangle$$

Using the time derivative of an operator, and expanding out, I got to this:
$$\frac{d}{dt}\langle\hat{x}\hat{p}\rangle=\frac{i}{\hbar}\left\langle\left[\hat{H},\hat{x}\hat{p}\right]\right\rangle+\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle$$
Expanding using $\langle\psi\vert\text{stuff}\vert\psi\rangle$ and noting that the time derivative of $\hat{x}\hat{p}=0$ yields the following integral through all space:
$$\frac{i}{\hbar}\int\psi^{*}\left[\hat{H},\hat{x}\hat{p}\right]\psi\;dx$$
$$\frac{i}{\hbar}\int\psi^{*}\left(\hat{H}\hat{x}\hat{p}-\hat{x}\hat{p}\hat{H}\right)\psi\;dx$$
$$\int\psi^{*}\left(\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\right)x\frac{\partial}{\partial x}-x\frac{\partial}{\partial x}\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\right)\right)\psi\;dx$$

I have tried to simplify the stuff in between $\psi^{*}$ and $\psi$ (ie calculate the commutator between the Hamiltonian and xp) but I get the following, which is wrong:
$$2\left\langle\widehat{KE}\right\rangle-\left\langle x\frac{\partial V}{\partial x}\right\rangle+\left\langle Vx\frac{\partial}{\partial x}\right\rangle$$
The first two parts are correct, but the very last term is just...strange.
However, when I operate the derivatives on $\psi$, simplify and then re-pull out the $\psi$, I get the right answer. It appears that the extraneous term at the very right (with a random partial position derivative) gets cancelled out.

Could someone explain the difference? Is this just me not realising I have to operate on something first? Thanks.

 

[kuruman]

First of all the momentum operator is $\hat p=\dfrac{\hbar}{i} \dfrac{\partial}{\partial x}~$ not what you have.

Secondly, it looks like you messed up the algebra and dropped a $\psi$ somewhere. You cannot end up with a dangling operator $\frac{\partial}{\partial x}$ when you start from the integral of function.

Here is a generous hint
First show that $$[\hat x,\hat H]=-\frac{\hbar}{i}\frac{\hat p}{m}~;~~[\hat p,\hat H]=+\frac{\hbar}{i}\frac{\partial V}{\partial x}$$then use it.

Also, remember that $[AB,C]=A[B,C]+[A,C]B.$

 

[TSny]

$$\int\psi^{*}\left(\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\right)x\frac{\partial}{\partial x}-x\frac{\partial}{\partial x}\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\right)\right)\psi\;dx$$

The terms involving $V$ are $$\int \left[\psi^{*}Vx\frac{\partial \psi} {\partial x}-x\frac{\partial}{\partial x}\left(V \psi \right) \right]\;dx$$ Note the second term where the derivative acts on the product $V \psi$.

The integral will reduce to $\left\langle -x\frac{\partial V}{\partial x}\right\rangle$. So, your approach will yield the desired result.

However, it is very worthwhile to work it using @kuruman's approach.

 

[TSny]

First of all the momentum operator is $\hat p=\dfrac{\hbar}{i} \dfrac{\partial}{\partial x}~$ not what you have.

I think @flyusx did use $\hat p=\dfrac{\hbar}{i} \dfrac{\partial}{\partial x}~$, but the $\dfrac{\hbar}{i}$ was cancelled out by the factor $\dfrac i \hbar$ in front of the integral.

 

[kuruman]

I think @flyusx did use $\hat p=\dfrac{\hbar}{i} \dfrac{\partial}{\partial x}~$, but the $\dfrac{\hbar}{i}$ was cancelled out by the factor $\dfrac i \hbar$ in front of the integral.

Yes, I don't know how I missed that.

 

[flyusx]

Thanks all, I got the correct answer using kuruman's method.
In the future, when commutators are present inside of an integral, is it simply standard to operate them on the ψ instead of directly simplifying them?

 

[kuruman]

Thanks all, I got the correct answer using kuruman's method.
In the future, when commutators are present inside of an integral, is it simply standard to operate them on the ψ instead of directly simplifying them?

Can you show an example of "directly simplifying"? I don't know what it means.

 

[flyusx]

Can you show an example of "directly simplifying"? I don't know what it means.

Yeah.

$$\frac{i}{\hbar}\int\psi^{*}\left[\hat{H},\hat{x}\hat{p}\right]\psi\;dx$$
$$\frac{i}{\hbar}\int\psi^{*}\left(\hat{H}\hat{x}\hat{p}-\hat{x}\hat{p}\hat{H}\right)\psi\;dx$$
$$\int\psi^{*}\left(\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\right)x\frac{\partial}{\partial x}-x\frac{\partial}{\partial x}\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\right)\right)\psi\;dx$$

Originally, I tried to simplify the things in the brackets while ignoring ψ* and ψ, which yielded the incorrect answer:
$$\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\right)x\frac{\partial}{\partial x}-x\frac{\partial}{\partial x}\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\right)$$
$$-\frac{\hbar^{2}}{2m}\frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}\left(x\frac{\partial}{\partial x}\right)\right)+Vx\frac{\partial}{\partial x}+\frac{\hbar^{2}}{2m}x\frac{\partial^{3}}{\partial x^{3}}-x\frac{\partial V}{\partial x}$$
$$-\frac{\hbar^{2}}{2m}\frac{\partial}{\partial x}\left(x\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial}{\partial x}\right)+Vx\frac{\partial}{\partial x}+\frac{\hbar^{2}}{2m}x\frac{\partial^{3}}{\partial x^{3}}-x\frac{\partial V}{\partial x}$$
$$-\frac{\hbar^{2}}{2m}\left(x\frac{\partial^{3}}{\partial x^{3}}+2\frac{\partial^{2}}{\partial x^{2}}\right)+Vx\frac{\partial}{\partial x}+\frac{\hbar^{2}}{2m}x\frac{\partial^{3}}{\partial x^{3}}-x\frac{\partial V}{\partial x}$$
$$2\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\right)+Vx\frac{\partial}{\partial x}-x\frac{\partial V}{\partial x}$$
Where there is the incorrect and extraneous term. Using your method or acting the 'central' stuff on ψ yielded the correct answer. I'm just confused as to why calculating the commutation between the Hamiltonian and the position/momentum and then using the commutation identity works while an issue arises when using the method above.

 

[TSny]

When simplifying a product of operators it's a good idea to consider how the product behaves when acting on an arbitrary state function $\psi(x)$.

For operators $A$ and $B$, $$(A \cdot B) \psi(x) \equiv A[B(\psi(x))].$$ Thus, $$\left( \frac {\partial}{\partial x} \cdot V(x) \right) \psi(x) = \frac {\partial}{\partial x} \left[ V(x) \psi(x) \right] =\frac {\partial V(x)}{\partial x} \psi(x) + V(x) \frac {\partial \psi(x)}{\partial x} .$$ This may be written as $$\left( \frac {\partial}{\partial x} \cdot V(x) \right) \psi(x) = \left[ \frac {\partial V(x)}{\partial x} + V(x) \frac {\partial }{\partial x} \right] \psi(x)$$ Since this is to hold for arbitrary $\psi(x)$, we have the operator identity $$\frac {\partial}{\partial x} \cdot V(x) = \frac {\partial V(x)}{\partial x} + V(x) \frac {\partial }{\partial x}$$

As a little exercise, verify the important commutator $[\hat x, \hat p] = i \hbar$ where $\hat x = x$ and $\hat p = - i \hbar\frac {\partial}{\partial x}$.

 

[kuruman]

You can't do that. You start with an expectation value, which is a constant, and you propose to end up with an operator. It's like mixing birthday cakes and diesel engines.

 

[flyusx]

Thanks. I've noticed that whenever I operate with respect to ψ, I get the correct answer.