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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

Qmechanic
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user7292119
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2 Answers2

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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: $$\langle T \rangle = \frac{\hbar \omega}{4} = \langle V \rangle.$$

Philip
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  • 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 – user7292119 Jan 27 '21 at 13:02
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    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? – Philip Jan 27 '21 at 13:05
  • I read that all states in a quantum harmonic oscillator are bound: https://physics.stackexchange.com/questions/135456/bound-states-scattering-states-and-infinite-potentials But maybe I am wrong and for classical energies there existes unbound states? – user7292119 Jan 27 '21 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. – Philip Jan 27 '21 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? – user7292119 Jan 27 '21 at 13:36
  • Right, I'm not sure I understand the last bit of your comment, but globally, yes. "Bound" and "scattering" are adjectives used to describe states of definite energy (i.e, eigenfunctions of the Hamiltonian). Eigenfuctions that have a discrete spectrum are called "bound" and those that have a continuous spectrum are called "scattering". The harmonic oscillator only has eigenfunctions that have a discrete energy spectrum, and therefore all states of definite energy are "bound" states. The Virial theorem only holds for such bound states. – Philip Jan 27 '21 at 13:54
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    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. – user7292119 Jan 27 '21 at 14:06
  • That sounds about right, yes! – Philip Jan 27 '21 at 14:07
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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.

                          enter image description here

Philip
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mike stone
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  • The gaussian "breathes" in and out. That's one heck of an image, I'm going to use that phrase from now on. :P – Philip Jan 27 '21 at 13:58
  • I don't think I understand the "breathes in and out" metaphor. Can you expand on that? – user7292119 Jan 27 '21 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.. – mike stone Jan 27 '21 at 15:30
  • Oh, I see, thank you – user7292119 Jan 27 '21 at 16:30
  • @mikestone Hello! I've added a small GIF and a little description illustrating what I think you meant, please feel free to remove it/edit it/rollback the edit, etc. as you see fit! – Philip Jan 29 '21 at 11:51
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    @Phillip That's great! – mike stone Jan 29 '21 at 12:50