Quantum mechanics, harmonic oscillator and wavefunction

[fluidistic]

Homework Statement

A harmonic oscillator is initially in the state \psi (x,0)=Ae^{-\frac{\alpha ^2 x^2}{2}} \alpha x (2\alpha x +i). Where \alpha ^2 =\frac{m \omega}{\hbar}.
1)Find the wavefunction for all t>0.
2)Calculate the probability to measure the values \frac{5\hbar \omega }{2} and \frac{7\omega \hbar }{2} for the energy.
3)Calculate the mean value of the energy of this oscillator.

Homework Equations

\Psi (x,t)= \psi (x)e^{-\frac{iEt}{\hbar}}.
E=(n+1)\frac{\hbar }{2}.
Probability to measure E_n associated to the wavefunction \psi _n is |c_n|^2 in the expression \Psi (x,t)= \sum _i ^{\infty } c_i \psi _i (x,t). I'm only using memory for all of this, so I might be wrong for some things.

The Attempt at a Solution

Now that I think about it... can I consider that the particle is in a stationary state? I.e. that \psi (x,0)=\psi (x,t)? Hmm I think not, that would be too easy to answer to 1).
Now even more confused. Stationary state would imply that \Psi (x,t)=\Psi(x) and so the equation \Psi (x,t)= \psi (x)e^{-\frac{iEt}{\hbar}} looks wrong, which is I think impossible.
I hope someone can shred some light on my understanding of intro to QM.
Thank you very much.

 

[BruceW]

I think you've misunderstood the meaning of stationary state. Maybe you should read through its definition again.

I would personally just use the time-dependent Schrodinger equation. But I think you could answer the questions with the relevant equations you have given.

I should also say - in usual notation, capital Psi means the actual wave function (which is a function of time and space), and if the wave function can be separated into a function of space multiplied by a function of time, then the function of space is often denoted by lower-case psi.

Your notation seems to disobey this convention, which makes it a bit confusing.

 

[fluidistic]

I think you've misunderstood the meaning of stationary state. Maybe you should read through its definition again.

I would personally just use the time-dependent Schrodinger equation. But I think you could answer the questions with the relevant equations you have given.

Ok thank you, heading into this. (I mean the read part)

I should also say - in usual notation, capital Psi means the actual wave function (which is a function of time and space), and if the wave function can be separated into a function of space multiplied by a function of time, then the function of space is often denoted by lower-case psi.

Your notation seems to disobey this convention, which makes it a bit confusing.

Indeed, and thanks for pointing this out. I didn't know about the convention, although I used it on my draft unlike here.

 

[fluidistic]

I just read about stationary state. And I can say that I can't consider the particle to be in a stationary state. A stationary state implies a definite energy (I think with a probability 1). For the harmonic oscillator a stationary state could be for example if the particle is in the ground state (n=0 with probability 1), or the first excited state (n= 1 with probability 1).
So I would have \Psi (x,t) = \psi _n (x) e^{-\frac{i E_n t}{\hbar}} for some n.
I hope I'm right on this. (Please correct me if I said something wrong)
Ok, I'll reconsider the problem now. Will post here if I'm still stuck.

 

[BruceW]

Yep, that's all right.

 

[fluidistic]

Yep, that's all right.

Ok thank you for confirming.
Well I'm having some hard time with the problem.
My idea is to calculate \psi _i (x,t). Because \Psi (x,t)= \sum _{i=0} \psi _i (x,t) with \sum _i |c_i|^2=1.
Thanks to a previous problem, I know that c_n=\int _{-\infty}^{\infty} \psi _i ^* (x) \psi (x,0)dx.
Where \psi _i (x) are worth a somehow "nasty" expressions but which is somehow a real valued function. (Hermite polynomials with constants of normalization).
In other words c_n=\int _{-\infty}^{\infty} \psi _i ^* (x) \psi (x,0)dx=c_n=\int _{-\infty}^{\infty} \psi _i (x) \psi (x,0)dx.
But in a solution to another problem, I've seen that c_n=\int _{-\infty}^{\infty} \psi ^* (x,0)\psi _i (x)dx which would here give a totally different result. So I have a doubt about what c_n I should consider.
There might be a quicker way to solve the problem that I'm missing. My friend told me he solved the problem within 5 minutes and the answer was something like 1/2 and 1/2 for the probabilities or something like that (I'm not sure I remember so well!). I don't think he had used Hermite polynomials though...

 

[BruceW]

I know that c_n=\int _{-\infty}^{\infty} \psi _i ^* (x) \psi (x,0)dx.
Where \psi _i (x) are worth a somehow "nasty" expressions but which is somehow a real valued function. (Hermite polynomials with constants of normalization).
In other words c_n=\int _{-\infty}^{\infty} \psi _i ^* (x) \psi (x,0)dx=c_n=\int _{-\infty}^{\infty} \psi _i (x) \psi (x,0)dx.

This is correct. (Where i and n are the same). So from this, you could explicitly do the integration to find the coefficients. Its an overly-complicated way of doing it though.

You have \Psi(x,0) and you are given the \psi_n(x) right? So have a look at their form and think what the coefficients would be. You don't need to do integrals to find the answer.

 

[fluidistic]

This is correct. (Where i and n are the same). So from this, you could explicitly do the integration to find the coefficients. Its an overly-complicated way of doing it though.

You have \Psi(x,0) and you are given the \psi_n(x) right? So have a look at their form and think what the coefficients would be. You don't need to do integrals to find the answer.

1)\Psi (x,t)= \sum _{n=0}^{\infty } c_n \psi _n (x,t)= \sum _{n=0}^{\infty } c_n \psi _n (x) e^{-\frac{iEt}{\hbar}}.
So that \Psi (x,0)= \sum _{n=0}^{\infty} c_n \psi _n (x)=2 \alpha ^2 x^2 Ae^{-\frac{\alpha ^2 x^2}{2}}+i \alpha x A e^{-\alpha ^2 \frac{x^2}{2}}.
So my answer for part 1 would be this multiplied by e^{-\frac{iEt}{\hbar}}.
The first 2 Hermite polynomials are 1 and 2x. So I can already smell that the oscillator is in the superposition of the states n=0 and n=1.
For part 2), I must look at \psi _n (y)= \left ( \frac{\beta}{\pi} \right ) ^{1/4} \frac{1}{\sqrt {2^n n!}}H_n (y)e^{-y^2/2} where y=x \sqrt \beta and \beta =\alpha ^2 and identify the coefficients c_n?
Though if I'm correct that the oscillator is only in n=0 and n=1, I could already answer "0" to part 2.

 

[vela]

But in a solution to another problem, I've seen that c_n=\int _{-\infty}^{\infty} \psi ^* (x,0)\psi _i (x)dx which would here give a totally different result. So I have a doubt about what c_n I should consider.

In Dirac notation, you would write
| \psi \rangle = \sum_i | \phi_i \rangle\langle \phi_i | \psi \ranglewhere c_i = \langle \phi_i | \psi \rangle = \int \phi_i^*(x)\psi(x)\,dx. This corresponds to the expression
\psi(x) = \sum_i c_i \phi_i(x)In other words, you definitely want the eigenstate ϕ_i being the one conjugated when expanding a state ψ in terms of them.

 

[vela]

1)\Psi (x,t)= \sum _{n=0}^{\infty } c_n \psi _n (x,t)= \sum _{n=0}^{\infty } c_n \psi _n (x) e^{-\frac{iEt}{\hbar}}.
So that \Psi (x,0)= \sum _{n=0}^{\infty} c_n \psi _n (x)=2 \alpha ^2 x^2 Ae^{-\frac{\alpha ^2 x^2}{2}}+i \alpha x A e^{-\alpha ^2 \frac{x^2}{2}}. (That would be my answer for part 1)
The first 2 Hermite polynomials are 1 and 2x. So I can already smell that the oscillator is in the superposition of the states n=0 and n=1.

How do you get x² without using the n=2 state?

 

[fluidistic]

In Dirac notation, you would write
| \psi \rangle = \sum_i | \phi_i \rangle\langle \phi_i | \psi \ranglewhere c_i = \langle \phi_i | \psi \rangle = \int \phi_i^*(x)\psi(x)\,dx. This corresponds to the expression
\psi(x) = \sum_i c_i \phi_i(x)In other words, you definitely want the eigenstates ϕ_i being the one conjugated and expanding a state ψ in terms of them.

Thanks, I'm going to spend some time understanding this (I didn't learn Dirac's notation yet, but I've sneaked into it).

How do you get x² without using the n=2 state?

My bad, you're right, it's impossible. So there's indeed the state with E_2=\frac{3 \hbar \omega}{2}.
By the way I edited my previous post (for answer to part 1).

 

[fluidistic]

Part 2:
After several tries (and errors!), I have come to the conclusion that the oscillator is in the superposition of the states n=0, n=1 and n=2. (100% sure)
\Psi (x,0)=Ae^{-\frac{\alpha ^2 x^2}{2}} (2\alpha ^2 x^2 +i\alpha x)=c_0 \psi _0 (x)+c_1 \psi _1 (x)+c_2 \psi _2 (x).
Where \psi _0 (\alpha x )= \pi ^{-1/4} \sqrt { \frac{m\omega }{\hbar}} e^{-\frac{\alpha ^2 x^2}{2}}.
\psi _1 (\alpha x)= \pi ^{-1/4} \sqrt { \frac{m \omega}{2 \hbar }} \cdot 2 \alpha x e^{-\frac{\alpha ^2 x^2}{2}}.
\psi _2 (\alpha x)=\pi ^{-1/4} \sqrt { \frac{m \omega }{8\hbar }} (4 \alpha ^2 x^2 -2)e^{-\frac{\alpha ^2 x^2}{2}}.
I skip the arithmetics. I reach that |c_0|^2=\frac{1}{11}, |c_1|^2=\frac{2}{11}, |c_2|^2=\frac{8}{11}.
However it's only for t=0. Did they ask for all t's?

So the answer for t=0, would be 0 probability for E_3=\frac{5\hbar \omega }{2} and E_4=\frac{7\hbar \omega }{2}.

Edit: I made an error in the energy expression (bad memory of mine!). It's E_n= \left ( n+\frac{1}{2} \right ) \hbar \omega.
So I reach that the probability to measure E_2=\frac{5\hbar \omega }{2} is 8/11, or around 72.7% while the probabilty to measure E_=\frac{7\hbar \omega }{2} is 0.

 

[vela]

Your expression for the time evolution needs a slight correction. It should be
\Psi(x,t) = \sum_{n=0}^\infty c_n\psi_n(x)e^{-iE_nt/\hbar}Each component evolves with frequency \omega_n = E_n/\hbar corresponding to its energy. You left the subscript n off of the energy earlier.

By the way, I got a slightly different expansion than you did. I found c_0 = \sqrt{2/7}, c_1 = i/\sqrt{7}, and c_2 = 2/\sqrt{7}.

 

[fluidistic]

Your expression for the time evolution needs a slight correction. It should be
\Psi(x,t) = \sum_{n=0}^\infty c_n\psi_n(x)e^{-iE_nt/\hbar}Each component evolves with frequency \omega_n = E_n/\hbar corresponding to its energy. You left the subscript n off of the energy earlier.

Ah you're right, I shouldn't have dropped the n subscript.

By the way, I got a slightly different expansion than you did. I found c_0 = \sqrt{2/7}, c_1 = i/\sqrt{7}, and c_2 = 2/\sqrt{7}.

Ok I post my work:
(c_0 \psi _0 +c_1 \psi _1+c_2 \psi _2)(\alpha x )=\pi ^{-1/4} e^{-\frac{\alpha ^2 x^2}{2}} \left [ c_0 \sqrt {\frac{m \omega }{\hbar}} - c_1 \sqrt {\frac{m \omega }{ 2 \hbar}} \cdot 2 \alpha x + \sqrt {\frac{m \omega }{ 8 \hbar}} c_2 (4 \alpha ^2 x^2 -2) \right ].
But here I have the condition that the term that does not contains x must vanish, because this expression is worth Ae^{-\frac{\alpha ^2 x^2}{2}}(2 \alpha ^2 x^2 +i \alpha x) and there's no term that doesn't contain x.
Mathematically this implies that c_0 \sqrt {\frac{m \omega }{ \hbar}}-c_1 \sqrt {\frac{m \omega }{ 2 \hbar}} =0 \Rightarrow c_0=\frac{c_1}{\sqrt 2 }.
Now equating the terms with x^2: A(2\alpha ^2 x^2)=\pi ^{-1/4} c_2 \alpha ^2 x^2 \sqrt {\frac{m \omega }{ \hbar}} \Rightarrow A= \pi ^{-1/4} c_2 \sqrt {\frac{m \omega }{ 2 \hbar}}.
On the other hand, by equating the terms with x: Ai \alpha x= \pi ^{-1/4} c_1 \sqrt {\frac{m \omega }{ 2 \hbar}} \cdot 2 \alpha x \Rightarrow A=-2i \pi ^{-1/4}c_1 \sqrt {\frac{m \omega }{ 2 \hbar}}.
These 2 equalities with A gives me c_2 \sqrt {\frac{m \omega }{ 2 \hbar}}=-2i c_1\sqrt {\frac{m \omega }{ 2 \hbar}} \Rightarrow c_2=-2i c_1.
Now I use the fact that |c_0|^2+|c_1|^2+|c_2|^2=1. So that \frac{|c_1|^2}{2}+|c_1|^2+4|c_1|^2=1 \Rightarrow |c_1|^2=\frac{2}{11} \Rightarrow |c_0|^2=\frac{1}{11} and |c_2|^2=\frac{8}{11}.

 

[vela]

When you looked at the constant term, you should be considering c₂, not c₁.

 

[fluidistic]

When you looked at the constant term, you should be considering c₂, not c₁.

Indeed, thank you. I've redone the algebra and reached the same result as you. I'll try to do part 3) now.

 

[fluidistic]

Hmm actually I didn't find the values of c_i's; only their modulus to the 2nd power. And I need them for part 3). How did you find out that for instance c_1 was worth i/\sqrt 7 rather than 1/\sqrt 7?
I'm about to try \langle E \rangle= \int _{-\infty}^{\infty} \Psi ^* (x,t) \hat H \Psi (x,t) dx.

 

[vela]

Remember the wave function is unique only up to a complex phase. You can arbitrarily pick the phase for one of the constants. Your equations will then allow you to determine the relative phases of the other constants.

 

[Redbelly98]

Please note: I have moved this thread to Advanced Physics.

 

[fluidistic]

Remember the wave function is unique only up to a complex phase. You can arbitrarily pick the phase for one of the constants. Your equations will then allow you to determine the relative phases of the other constants.

Ah I didn't know this, good to know.
To calculate \langle E \rangle, can I just say it's equal to |c_0|^2E_0+|c_1|^2E_1+|c_2|^2E_2?

 

[vela]

Yup!

 

[fluidistic]

Yup!

Ah ok good! We can finally say "problem solved". Thank you guys for all!