Continuity Equation for a Dimensionless Harmonic Oscillator

[flyusx]

Homework Statement

Consider that the wave function of a dimensionless harmonic oscillator, whose Hamiltonian is ##\hat{H}=\frac{1}{2}\hat{p}^{2}+\frac{1}{2}\hat{x}^{2}## is given at time ##t=0## by $$\psi(x,0)=\frac{1}{\sqrt{8\pi}}\phi_{0}(x)+\frac{1}{\sqrt{18\pi}}\phi_{2}(x)=\frac{1}{\sqrt{8\pi}}\exp(-\frac{x^{2}}{2})+\frac{1}{\sqrt{18\pi}}(1-2x^{2})\exp(-\frac{x^{2}}{2})$$ (this may be too big to fit so I've attached a PDF file with the full question). Find the expression of the oscillator's wave function at any later time ##t##. Calculate the probability density ρ(x,t) and the current density J(x,t) (both at a later time ##t##). Verify that probability is conserved; that is, show that ##\frac{\partial\rho}{\partial t}+\frac{\partial J}{\partial x}=0##.

Relevant Equations

$$\rho=\psi^{*}\psi$$
$$J=\frac{i\hbar}{2m}\left(\psi\frac{\partial\psi^{*}}{\partial x}-\psi^{*}\frac{\partial\psi}{\partial x}\right)$$

I've tried to solve this problem (Zettili, Exercise 3.5) four times at this point. I believe my equation for the wave function at a later time ##t## is correct. The problem is my continuity equation is not satisfied; it does not equal zero. It's close but I'm off by a factor of ##m## and ##\hbar##.
I think the problem may be the phrase 'dimensionless harmonic oscillator'. In Exercise 3.4, Zettili stated that the momentum operator for this dimensionless system is given by $$\hat{p}=-i\frac{d}{dx}$$

Without the ##\hbar##. Zettili covers harmonic oscillators in the next chapter (Ch4) alongside other simple systems (potential step/well, for instance), so I don't have experience regarding that formalism. I'm quite certain this problem is just meant to get me comfortable with calculating ρ, J and using the continuity equation.

I've linked a PDF file of my work. In that document, I use ##\psi## for the wave function at ##t=0## and ##\Psi## for the wave function at a later time. Maybe I shouldn't be multiplying by ##\frac{i\hbar}{2m}##?

 

[kuruman]

Homework Statement: Consider that the wave function of a dimensionless harmonic oscillator, whose Hamiltonian is ##\hat{H}=\frac{1}{2}\hat{p}^{2}+\frac{1}{2}\hat{x}^{2}## is given at time ##t=0## by $$\psi(x,0)=\frac{1}{\sqrt{8\pi}}\phi_{0}(x)+\frac{1}{\sqrt{18\pi}}\phi_{2}(x)=\frac{1}{\sqrt{8\pi}}\exp(-\frac{x^{2}}{2})+\frac{1}{\sqrt{18\pi}}(1-2x^{2})\exp(-\frac{x^{2}}{2})$$ (this may be too big to fit so I've attached a PDF file with the full question). Find the expression of the oscillator's wave function at any later time ##t##. Calculate the probability density ρ(x,t) and the current density J(x,t) (both at a later time ##t##). Verify that probability is conserved; that is, show that ##\frac{\partial\rho}{\partial t}+\frac{\partial J}{\partial x}=0##.
Relevant Equations: $$\rho=\psi^{*}\psi$$
$$J=\frac{i\hbar}{2m}\left(\psi\frac{\partial\psi^{*}}{\partial x}-\psi^{*}\frac{\partial\psi}{\partial x}\right)$$

I've tried to solve this problem (Zettili, Exercise 3.5) four times at this point. I believe my equation for the wave function at a later time ##t## is correct. The problem is my continuity equation is not satisfied; it does not equal zero. It's close but I'm off by a factor of ##m## and ##\hbar##.
I think the problem may be the phrase 'dimensionless harmonic oscillator'. In Exercise 3.4, Zettili stated that the momentum operator for this dimensionless system is given by $$\hat{p}=-i\frac{d}{dx}$$

Without the ##\hbar##. Zettili covers harmonic oscillators in the next chapter (Ch4) alongside other simple systems (potential step/well, for instance), so I don't have experience regarding that formalism. I'm quite certain this problem is just meant to get me comfortable with calculating ρ, J and using the continuity equation.

I've linked a PDF file of my work. In that document, I use ##\psi## for the wave function at ##t=0## and ##\Psi## for the wave function at a later time. Maybe I shouldn't be multiplying by ##\frac{i\hbar}{2m}##?

I don't think that setting ##\frac{\hbar}{2m}## equal to 1 matters. Setting ##i=1## would be problematic. I suspect that you did it correctly because, as you say, you're off "by a factor of ##m## and ##\hbar## which in Zettilli's world are set equal to 1. Of course. one would not be able to tell for sure unless you post your work.

 

[flyusx]

Of course. one would not be able to tell for sure unless you post your work.

If I set ##\hbar=m=1##, my answer obeys the continuity equation perfectly. Is this what 'dimensionless' implies (though t is not set to 1)?
I uploaded a PDF of my work typed in Latex. Is my document not showing/do you prefer I type it up here?

 

[kuruman]

Is my PDF of my Latex document not showing?

It's showing fine. When you said "I've attached a PDF file with the full question" I thought you meant the question as it appears in Zettelli, not your attempt to answer it, so I didn't look at it at first.

After looking at it, what leapt out immediately is the last factor your last equation

##\dfrac{E_2-E_0}{\hbar}## has dimensions of time ##[T]##
##\dfrac{2\hbar}{m}## has dimensions of ##[L^2T^{-1}]## which makes the factor dimensionally incorrect.

There is probably a mistake in applying the differentiation chain rule somewhere in there. I think that you could have simplified things a bit by using ##E_0=\frac{1}{2}\hbar \omega## and ##E_0=\frac{5}{2}\hbar \omega##. Then ##\dfrac{(E_2-E_0)t}{\hbar}=2\omega t## which is easier to handle.

Also, I would note that ##\Psi^*\dfrac{\partial \Psi}{\partial t}-\Psi\dfrac{\partial \Psi^*}{\partial t}## is the difference of a function ##f(x,t)=\Psi^*\dfrac{\partial \Psi}{\partial t}## and its complex conjugate and is therefore, equal to ##2i~Im(f)##.

Perhaps if you redid this slightly differently from the 4 times that you did it already, the problem will disappear.

 

[flyusx]

Thanks. I find that if I set ##\hbar=m=1##, I get the right answer.

 

[kuruman]

Thanks. I find that if I set ##\hbar=m=1##, I get the right answer.

So you're saying that you should not have done this?

If so, it seems to me that elsewhere in your solution you implicitly assumed that ##\hbar=m=1## but I could not find where. Anyway, I think you're good.

 

[Rosenthal]

I assume you mean that you are off by a "term", not by a "factor". You can't be off from zero by a factor.
The wave functions are written with a dimensionless normalization, meaning that you are working with a dimensionless displacement. The current operator, however, is written assuming the wave function has
dimension L^(-1/2)
Write the fully dimensioned Hamiltonian and divide both sides by hbar*omega to get something dimensionless. The momentum term is now p^2/(2*m*hbar*omega). Define a new dimensionless momentum by pi = p/sqrt(m*hbar*omega) and similarly a dimensionless displacement by q = x*sqrt(m*omega/hbar). The Hamiltonian is then H/(hbar*omega)=pi^2/2+q^2/2 and these satisfy the commutation relation [q,pi]=i.
Rewrite J by replacing x with q and everything works out.

 

[vela]

Thanks. I find that if I set ##\hbar=m=1##, I get the right answer.

You might get the answer you're looking for, but that's by coincidence. It's not the correct way to solve your problem.

 

[flyusx]

Write the fully dimensioned Hamiltonian and divide both sides by hbar*omega to get something dimensionless. The momentum term is now p^2/(2*m*hbar*omega). Define a new dimensionless momentum by pi = p/sqrt(m*hbar*omega) and similarly a dimensionless displacement by q = x*sqrt(m*omega/hbar). The Hamiltonian is then H/(hbar*omega)=pi^2/2+q^2/2 and these satisfy the commutation relation [q,pi]=i.
Rewrite J by replacing x with q and everything works out.

I understand the point about getting dimensionless operators but I don't quite see how this would help. In this exercise, Zettili explicitly states the Hamiltonian is $$\hat{H}=\frac{1}{2}\hat{p}+\frac{1}{2}\hat{x}$$ and also explicitly states the momentum operator is $$\hat{p}=-i\frac{d}{dx}$$ which conflicts with your definition of a dimensionless momentum.

 

[flyusx]

You might get the answer you're looking for, but that's by coincidence. It's not the correct way to solve your problem.

So ##\hbar=m=1## isn't valid? According to kurkman, ##\hbar=m=1## is what's done in Zettili's world.

 

[Orodruin]

First of all, I strongly discourage posting your work in a pdf for several reasons:
- It is an extra step for people trying to help you to open it.
- People in general (me for example) may be hesitant to download and open unknown files.
- The text is not directly quotable here uaing the quote functionality

Instead, please type out the attempt in your post. If you already have the LaTeX it is a simple question of changing the math delimiters.

I understand the point about getting dimensionless operators but I don't quite see how this would help. In this exercise, Zettili explicitly states the Hamiltonian is $$\hat{H}=\frac{1}{2}\hat{p}+\frac{1}{2}\hat{x}$$ and also explicitly states the momentum operator is $$\hat{p}=-i\frac{d}{dx}$$ which conflicts with your definition of a dimensionless momentum.

It does not. For the Hamiltonian to make sense dimensionally both x and p must be dimensionless here.

You might get the answer you're looking for, but that's by coincidence. It's not the correct way to solve your problem.

It is because clearly those are the units the problem statement is working with.

 

[vela]

I understand the point about getting dimensionless operators but I don't quite see how this would help. In this exercise, Zettili explicitly states the Hamiltonian is $$\hat{H}=\frac{1}{2}\hat{p}+\frac{1}{2}\hat{x}$$ and also explicitly states the momentum operator is $$\hat{p}=-i\frac{d}{dx}$$ which conflicts with your definition of a dimensionless momentum.

Zettili is unfortunately using ##\hat X## and ##\hat P## to represent both the regular position and momentum operators in some contexts and the dimensionless position and momentum operators in others, which is what's causing your confusion.

The natural length scale for the harmonic oscillator is ##\alpha = \sqrt{\hbar/m\omega}##. If you change variables from ##x## to the dimensionless variable ##u=x/\alpha##, the Schrödinger equation becomes
$$\hbar\omega\left[-\frac 12 \frac{\partial^2 \Psi}{\partial u^2} + \frac 12 u^2 \Psi\right] = i\hbar \frac{\partial \Psi}{\partial t}.$$ Dividing through by ##\hbar \omega## gives you
$$-\frac 12 \frac{\partial^2 \Psi}{\partial u^2} + \frac 12 u^2 \Psi = i \frac{\partial \Psi}{\partial \tau},$$ where ##\tau = \omega t## is the dimensionless time variable. From this, you can derive the probability current ##j(u,\tau)## and show that ##\partial \rho/\partial \tau + \partial j/\partial u = 0##.

 

[flyusx]

It is because clearly those are the units the problem statement is working with.

I'm sort of confused. Are you saying ##\hbar=m=1## is coincidentally the right answer and you agree with vela in that:

From this, you can derive the probability current ##j(u,\tau)## and show that ##\partial \rho/\partial \tau + \partial j/\partial u = 0##.

I must derive ##J## and ##\rho## from the new Schrodinger equation with respect to ##\tau## and ##u##, and use that in regards to my quantum state that is dependent on ##x##?

Zettili doesn't get to the specifics of quantum harmonic oscillators until Chapter 4 section 8 which is an entire chapter ahead of where I am at right now. Flipping there, I see where he introduces the dimensionless momentum and position operators. I'm just confused as to how this could be the right approach when Zettili hasn't introduced this method yet.

 

[vela]

I'm sort of confused. Are you saying ##\hbar=m=1## is coincidentally the right answer and you agree with vela in that:

No, he's saying taking ##\hbar=m=1## is effectively the same as what I explained above about using the natural length scale for the harmonic oscillator, though I think you also need ##\omega = 1##.

I must derive ##J## and ##\rho## from the new Schrodinger equation with respect to ##\tau## and ##u##, and use that in regards to my quantum state that is dependent on ##x##?

Your (dimensionless) ##x## is the same as my ##u## but different than the ##x## in the expression for ##J## in the original post. I just used ##u## to make it explicit that it's not the same as the ##x## Zettili used in the rest of the chapter.

 

[flyusx]

No, he's saying taking ##\hbar=m=1## is effectively the same as what I explained above about using the natural length scale for the harmonic oscillator, though I think you also need ##\omega = 1##.

Your (dimensionless) ##x## is the same as my ##u## but different than the ##x## in the expression for ##J## in the original post. I just used ##u## to make it explicit that it's not the same as the ##x## Zettili used in the rest of the chapter.

So ##\hbar=m=1## gets the right answer but the proper way to solve the problem is to define non-dimensional momentum and position operators.

I'll try deriving the probability current.
Probability density is given by $$\rho=\Psi^{*}\Psi$$
Using the product rule for derivatives gives $$\frac{\partial\rho}{\partial \tau}=\Psi^{*}\frac{\partial\Psi}{\partial\tau}+\Psi\frac{\partial\Psi^{*}}{\partial\tau}$$

From the non-dimensional Schrodinger equation, $$\frac{\partial\Psi}{\partial\tau}=\frac{i}{2}\frac{\partial^{2}\Psi}{\partial u^{2}}-\frac{i}{2}u^{2}\Psi$$ and $$\frac{\partial\Psi^{*}}{\partial\tau}=-\frac{i}{2}\frac{\partial^{2}\Psi^{*}}{\partial u^{2}}+\frac{i}{2}u^{2}\Psi^{*}$$

Substituting this into ##\partial\rho/\partial\tau## gives $$\frac{\partial\rho}{\partial\tau}=\Psi^{*}\left(\frac{i}{2}\frac{\partial^{2}\Psi}{\partial u^{2}}-\frac{i}{2}u^{2}\Psi\right)+\Psi\left(-\frac{i}{2}\frac{\partial^{2}\Psi^{*}}{\partial u^{2}}+\frac{i}{2}u^{2}\Psi^{*}\right)$$

The terms without the derivatives cancel out which gives $$\frac{\partial\rho}{\partial\tau}=\frac{i}{2}\left(\Psi^{*}\frac{\partial^{2}\Psi}{\partial u^{2}}-\Psi\frac{\partial^{2}\Psi^{*}}{\partial u^{2}}\right)$$

This is ##\frac{\partial J}{\partial u}## and leads to $$J(u,\tau)=\frac{i}{2}\left(\Psi^{*}\frac{\partial\Psi}{\partial u}-\Psi\frac{\partial\Psi^{*}}{\partial u}\right)$$

I hope I didn't screw up my derivation.

 

[Orodruin]

So ℏ=m=1 gets the right answer but the proper way to solve the problem is to define non-dimensional momentum and position operators.

These are the same things. The non-dimensional operators are defined in such a way that it is the same as putting the constants to one.

 

[flyusx]

These are the same things. The non-dimensional operators are defined in such a way that it is the same as putting the constants to one.

In this particular case, since I wasn't introduced to the concept of dimensionless systems, would it be okay to use the ##\hbar=m=ω=1## method?
In the future, when I see 'dimensionless', can I set ##\omega=\hbar=m=1## or would this be a really bad habit?
Edit: There's an appendix in the back where Zettili gives a code for solving the Schrodinger equation numerically. He temporarily uses ##\hbar=m=ω=1##. I doubt it means anything though.