# 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, 2021 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, 2021 at 13:40
• Maybe you could give us more context; what does this state describe etc Jan 27, 2021 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, 2021 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, 2021 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, 2021 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, 2021 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, 2021 at 13:21
• Thank you very much! Then as a summary, to make sure I have understood: Virial Theorem holds for bound states, that means eigenfunctions with discrete energy. Singe eigenfunctions of the QHO have discrete energy, all eigenstates are bound. So Virial holds for any QHO eigenstate. However, my state does not have a definite enrgy, since it isn't an eigenstate, so it's not bound, so no Virial. Jan 27, 2021 at 14:06

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, 2021 at 13:58
• I don't think I understand the "breathes in and out" metaphor. Can you expand on that? Jan 27, 2021 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, 2021 at 15:30
• Oh, I see, thank you Jan 27, 2021 at 16:30
• @Phillip That's great! Jan 29, 2021 at 12:50