# Quantum Harmonic Oscillator Virial theorem is not holding

I'm asked to calculate the average Kinetic and Potential Energies for a given state of a quantum harmonic oscillator. The state is: $$\psi(x,0) = \left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{\frac{-2m\omega}{\hbar}x^2}$$ The thing is, calculating $$\langle T\rangle=\int_{-\infty}^{\infty}\psi(x)(-i\hbar)^2\frac{d^2}{dx}\psi dx=\left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{2}\int_{-\infty}^{\infty}e^{\frac{-4m\omega}{h}x^2}dx-\left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{2}\int_{-\infty}^{\infty}x^2e^{\frac{-4m\omega}{h}x^2}dx=\hbar\omega$$

Where I used that the momentum operator is $$p=-i\hbar\frac{d}{dx}$$

$$\langle V\rangle=\dfrac{m\omega^2}{2}\left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{2}\int_{-\infty}^{\infty}x^2e^{\frac{-4m\omega}{h}x^2}dx=\dfrac{\hbar\omega}{16}$$

But then the Virial Theorem is not satisfied. I've read the virial theorem holds for any bound state and all states in a Quantum Harmonic Oscillator are bound. Can someone point out where I am going wrong? Thank you

• This is the ground state of the Hamiltonian $\frac{1}{2m}\nabla^2 + 8m\omega^2 x^2$ - are you sure that's the state they meant? Jan 27 at 13:20
• Yes @jacob1729 it's the state I'm given. It's not a ground state, more an infinite sum of QHO eigenstates Jan 27 at 13:40
• Maybe you could give us more context; what does this state describe etc Jan 27 at 13:49
• It's a previous exam question, it just states that a QHO of mass m and frequency $\omega$ is in said state at t=0 and asks to calculate <V> and <T> Jan 27 at 16:29

The ground state of the harmonic oscillator is (see Wikipedia for example): $$\psi_0(x) = \left(\frac{\alpha}{\pi}\right)^{1/4} e^{-\alpha x^2/2},\quad \quad \text{where }\quad \alpha =\frac{ m \omega}{\hbar}$$

Your math is correct, it's just that the state you have is not a bound state of the harmonic oscillator, the parameters are slightly off. If you use the state provided above, you can indeed show that: $$\langle T \rangle = \frac{\hbar \omega}{4} = \langle V \rangle.$$

• The state they give me is indeed not the ground state, I need to calculate it for the state given, with that factor of 4 Jan 27 at 13:02
• Sure, except that the Virial theorem in Quantum Mechanics is only true for bound states, and the state you've provided isn't a bound state of the Harmonic Oscillator. So it's normal that it doesn't satisfy the theorem. Perhaps there is a misunderstanding as to what a bound state is? Jan 27 at 13:05
• I read that all states in a quantum harmonic oscillator are bound: physics.stackexchange.com/questions/135456/… But maybe I am wrong and for classical energies there existes unbound states? Jan 27 at 13:19
• That is also correct. However, not all functions are bound states! A bound state for a particular Hamiltonian $H$ is a state that satisfies$$H \psi = E\psi,$$ where $E$ is a constant number, which we understand to be the energy. I urge you to plug the state that you have into this differential equation to see if it satisfies it. Jan 27 at 13:21
• That relation is satisfied by eigenfunctions right? That's why they have a definite energy E but my state is an infinte sum of the QHO eigenfunctions with different weight coefficients. So it does not satisfy $H\psi=E\psi$. Thus it is not a bound state so Virial does not hold. Therefore Virial only holds for eigenstates? And Bound states are those of the eigenfunctions? Jan 27 at 13:36

You can have Gaussian fields that are not eigenstates, but then they are not time independent -- and time independence is the essential element of the virial theorem. For example, the harmonic oscillator time-dependent Schrödinger equation $$i\frac{\partial \psi}{\partial t} = -\frac 12 \frac {\partial^2 \psi}{\partial x^2} +\frac 12 \omega^2 x^2 \psi$$ has a time-dependent solution $$\psi(x,t)= \left(\frac{\omega}{\pi}\right)^{1/4}\frac 1{\sqrt{e^{i \omega t} +R e^{-i\omega t}}}\exp\left\{ - \frac \omega 2 \left(\frac{1-R\,e^{-2i\omega t}}{1+R\,e^{-2i\omega t}}\right)x^2\right\},$$ where the parameter $$|R|<1$$. Only if $$R=0$$ are its $$x$$ and $$p$$ distributions time independent. If $$R\ne 0$$ the gaussian "breathes" in and out. Your wavefunction is a snapshot of this one at some particular time.

Below is a visualisation of $$|\psi(x,t)|^2$$ (taking $$\omega=1$$) for different values of $$R$$, showing how the Gaussian "breathes". As you can see, as $$R\to 0$$, the probability distribution tends to not change as much.

• The gaussian "breathes" in and out. That's one heck of an image, I'm going to use that phrase from now on. :P Jan 27 at 13:58
• I don't think I understand the "breathes in and out" metaphor. Can you expand on that? Jan 27 at 15:16
• I mean that if you plot $|\psi(x,t)|^2$ for my solution as a function of time you will see that the gaussian expands and contracts with frequency $2\omega$. Thus $<x^2>$. and $<p^2>$ shrink and grow periodically. Of course $<p^2>$ grows as $<x^2>$ shrinks.. Jan 27 at 15:30
• Oh, I see, thank you Jan 27 at 16:30
• @Phillip That's great! Jan 29 at 12:50