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Quantum Harmonic Oscillator

  • Thread starter chris_avfc
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Homework Statement



Particle of mass m undergoes simple harmonic motion along the x axis

Normalised eigenfunctions of the particle correspond to the energy levels

[tex] E_n = (n+ 1/2)\hbar\omega\ \ \ \ (n=0,1,2,3...) [/tex]
For the two lowest energy levels the eigenfunctions expressed in natural units are:

[tex] u_0 = C_0 \exp^{-q^2 /2} [/tex]
[tex] u_1 = C_1 q \exp^{-q^2 /2} [/tex]

At time [itex] t = 0 [/itex] the wave function of the particle is given by an equal superposition of the two eigenstates represented by the two eigenfunctions [itex] u_0 [/itex] and [itex] u_1 [/itex].

Assume [itex] \psi [/itex] is normalised.

Calculate the expectation values of the momentum operator [itex] \hat{P} [/itex] and position operator [itex] \hat{X} [/itex] at time [itex] t [/itex].


Homework Equations



Given two standard integrals:
[tex] \int e^{-ax^2}\,dx = 1/2 \sqrt{\pi/a} [/tex][tex] \int x^2 e^{-ax^2}\,dx = 1/4 \sqrt{\pi/a^2}[/tex]

The Attempt at a Solution



I've calculated the two normalisation constants, but then I am seriously stuck, I don't have a clue what to do, could somebody point me in the right direction?

Thanks,
Chris
 
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Answers and Replies

  • #2
Mute
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Do you know how to calculate the expectation value of an operator? Hint: the formula is an integral involving the wave function.

Do you know what the time-dependent wave function is in this problem?
 
  • #3
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Do you know how to calculate the expectation value of an operator? Hint: the formula is an integral involving the wave function.

Do you know what the time-dependent wave function is in this problem?
Wow, thanks for the quick reply!

Is the formula:

[tex] \langle X \rangle = \int \psi^* x \psi dx [/tex]

Not given a time dependent wave function in this question.
 
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  • #4
Mute
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Wow, thanks for the quick reply!

Is the formula:

[tex] <X> = \int \psi* x \psi dx [\tex]

Not given a time dependent wave function in this question.
Tip: [strike]The forum doesn't use the usual latex tags, rather[/strike] For simplicity, itex tags can be replaced by double # signs and tex can be replaced by the double $ signs.

Anywho:

You do have a time-dependent wave function in this problem. The problem specifically asks about the time-dependence of the averages you need to calculate. You are told what superposition of states the system has been prepared in at time t=0; for later times the overall wave-function will vary in time. Do you know what the time-dependent wave function will look like?

Once you have figured that out, you can calculate the averages. You are correct that

$$\langle x \rangle = \int dx~\psi^\ast(x,t)~x~\psi(x,t),$$
where I have noted the time-dependence of ##\psi(x,t)##. The formula for ##\langle p \rangle## is similar; however, as you have represented ##\psi## in the position basis (i.e., as a function of space), the momentum operator will be a true operator (unlike the position operator which just becomes a factor of x in the integral).
 
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  • #5
vela
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Tip: The forum doesn't use the usual latex tags, rather itex is replaced by double # signs and tex is replaced by the double $ signs.
The tex and itex tags will work as well, but the closing tags have a forward slash in them, not a backslash. I fixed the tags and some of the markup in the earlier posts.
 
  • #6
Mute
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The tex and itex tags will work as well, but the closing tags have a forward slash in them, not a backslash. I fixed the tags and some of the markup in the earlier posts.
Ah, right! I've been using the # and $ signs for a long enough while now that I got confused and thought it was the itex and tex tags that didn't work as opposed to the way the syntax works in other latex clients. :tongue:
 
  • #7
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Tip: The forum doesn't use the usual latex tags, rather itex is replaced by double # signs and tex is replaced by the double $ signs.

Anywho:

You do have a time-dependent wave function in this problem. The problem specifically asks about the time-dependence of the averages you need to calculate. You are told what superposition of states the system has been prepared in at time t=0; for later times the overall wave-function will vary in time. Do you know what the time-dependent wave function will look like?

Once you have figured that out, you can calculate the averages. You are correct that

$$\langle x \rangle = \int dx~\psi^\ast(x,t)~x~\psi(x,t),$$
where I have noted the time-dependence of ##\psi(x,t)##. The formula for ##\langle p \rangle## is similar; however, as you have represented ##\psi## in the position basis (i.e., as a function of space), the momentum operator will be a true operator (unlike the position operator which just becomes a factor of x in the integral).
Right okay, so I have the u1 and u0 eigenfunctions, but I'm confused how I get wavefunction from that, I thought the eigenfunctions represents the wavefunctions, in most cases?

So once I found how to get the wavefunction ##\psi## (currently googling how to, but I am unsure about the superposition of the states, I remember it in lectures from last semester so I will go over them), I can substitute it in to the formula including the momentum and position operators and get the expectation values?

The tex and itex tags will work as well, but the closing tags have a forward slash in them, not a backslash. I fixed the tags and some of the markup in the earlier posts.
Oops, thanks for that!
 
  • #8
Mute
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Right okay, so I have the u1 and u0 eigenfunctions, but I'm confused how I get wavefunction from that, I thought the eigenfunctions represents the wavefunctions, in most cases?
The eigenfunctions correspond to the wave functions of stationary states - quantum states with fixed energy values. If you prepare a system in one of these states, it will stay in that state. However, if you prepare a system in a superposition of states then the superposition will change in time as amount of the superposition in each state oscillates between them.

Remember that there is a time-dependent and a time-independent Schrodinger equation. The eigenstates come from solving the time-independent equation. Do you remember or have any idea of how to put the time dependence back in?

Hint: Strictly speaking, even the wave function of a state prepared in a single energy state will have some time-dependence, but it's trivial - it's just a phase factor ##\exp(iE_nt/\hbar)## that doesn't affect the overall probability. From this, can you guess what the time-dependent wavefunction of a superposition of states might be?
 
  • #9
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The eigenfunctions correspond to the wave functions of stationary states - quantum states with fixed energy values. If you prepare a system in one of these states, it will stay in that state.
Am I right in thinking that is until you try to measure something, and then it changes its state, because you act on the sytem?

However, if you prepare a system in a superposition of states then the superposition will change in time as amount of the superposition in each state oscillates between them.

Remember that there is a time-dependent and a time-independent Schrodinger equation. The eigenstates come from solving the time-independent equation. Do you remember or have any idea of how to put the time dependence back in?
The superposition and the oscillation rings a bell actually, I'm sure its in the notes somewhere, will look it up.

Hint: Strictly speaking, even the wave function of a state prepared in a single energy state will have some time-dependence, but it's trivial - it's just a phase factor ##\exp(iE_nt/\hbar)## that doesn't affect the overall probability. From this, can you guess what the time-dependent wavefunction of a superposition of states might be?
Ah okay I definitely remember that factor.
Looking at the notes I'm thinking if I have ## u(x) ## that if I multiply by the probability amplitude and the phase factor, and sum over all the eigenstates I get the ## \psi ##?

Is that sounding along the right lines?
 
  • #10
Mute
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Am I right in thinking that is until you try to measure something, and then it changes its state, because you act on the sytem?
Yes, once you make a measurement you will get a definite result (the system will 'collapse' into a single eigenstate).



The superposition and the oscillation rings a bell actually, I'm sure its in the notes somewhere, will look it up.



Ah okay I definitely remember that factor.
Looking at the notes I'm thinking if I have ## u(x) ## that if I multiply by the probability amplitude and the phase factor, and sum over all the eigenstates I get the ## \psi ##?

Is that sounding along the right lines?
It's close - I think you may have it, it may just be that your description wasn't clear enough. Let me give you a couple more hints just to make sure you've got it. You don't just have a ##u(x)##, you have a set of them, ##u_n(x)## for n = 0, 1, 2, 3...; i.e., these are your eigenfunctions. Each of them has its own energy, as ##\mathcal H u_n(x) = E_n u_n(x)##, right? So, that means each one has its own phase-factor.

So, if your initial state is ##\psi(x,t=0) = c_0 u_0(x) + c_1 u_1(x)## (where c0 and c1 are constants you have to chose according to what the initial probabilities of each state are supposed to be. That info is given to you in your problem statement), then given what you just told me plus the information I gave you above, what do you think the time-dependent wave function is?
 
  • #11
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Yes, once you make a measurement you will get a definite result (the system will 'collapse' into a single eigenstate).





It's close - I think you may have it, it may just be that your description wasn't clear enough. Let me give you a couple more hints just to make sure you've got it. You don't just have a ##u(x)##, you have a set of them, ##u_n(x)## for n = 0, 1, 2, 3...; i.e., these are your eigenfunctions. Each of them has its own energy, as ##\mathcal H u_n(x) = E_n u_n(x)##, right? So, that means each one has its own phase-factor.

So, if your initial state is ##\psi(x,t=0) = c_0 u_0(x) + c_1 u_1(x)## (where c0 and c1 are constants you have to chose according to what the initial probabilities of each state are supposed to be. That info is given to you in your problem statement), then given what you just told me plus the information I gave you above, what do you think the time-dependent wave function is?
Hmm, so is that ##c_0## and ##c_1## different from the normalisation constants? Are they to do with the fact that it is an equal superposition?

I get that they have their own phase factor now, but I don't know how they would be different?

I am so confused.
 
  • #12
Mute
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Hmm, so is that ##c_0## and ##c_1## different from the normalisation constants? Are they to do with the fact that it is an equal superposition?

I get that they have their own phase factor now, but I don't know how they would be different?

I am so confused.
The c0 and c1 are normalization constants; i.e., |c0|2 is the probability of finding the initial superposition in state 0 and |c1|2 is the probability of finding the initial superposition in state 1. You're told that at t=0 the system has equal probability of being in either state, so what should the c's be?

So, if each state has its own phase factor that means:

If the initial system were prepared only in the 0 state, the time-dependent wave function would be ##u_0(x)\exp(iE_0 t/\hbar)##.

If the initial system were prepared only in the 1 state, the time-dependent wave function would be ##u_1(x)\exp(iE_1 t/\hbar)##.

So what should the time-dependent wave function look like if the system is formed from a superposition of those two states?
 
  • #13
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The c0 and c1 are normalization constants; i.e., |c0|2 is the probability of finding the initial superposition in state 0 and |c1|2 is the probability of finding the initial superposition in state 1. You're told that at t=0 the system has equal probability of being in either state, so what should the c's be?

So, if each state has its own phase factor that means:

If the initial system were prepared only in the 0 state, the wave function would be ##u_0(x)\exp(iE_0 t/\hbar)##.

If the initial system were prepared only in the 1 state, the wave function would be ##u_1(x)\exp(iE_1 t/\hbar)##.

So what should the wave function look like if the system is formed from a superposition of those two states?
Ohhhhh, okay, that makes sense, I think.

Am I right in thinking that ## E ## is

$$ E = n^2\pi^2\hbar^2 /2ml^2 $$

So I'm thinking the wave function would be

$$ \psi = [c_0 exp(-q^2/2) exp( iE_0 t/\hbar)] + [c_1 q exp(-q^2/2) exp( iE_1 t/\hbar)] $$

Then use that in the integral?
 
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  • #14
Mute
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Ohhhhh, okay, that makes sense, I think.

Am I right in thinking that ## E ## is

$$ E = n^2\pi^2\hbar^2 /2ml^2 $$

So I'm thinking the wave function would be

$$ \psi = [c_0 exp(-q^2/2) exp( iE_0 t/\hbar)] + [c_1 q exp(-q^2/2) exp( iE_1 t/\hbar)] $$

Then use that in the integral?
Hey, sorry I hadn't replied yet. I've been out of town with barely functional hotel wifi.

You're almost correct. The energy you wrote down in this post is the energy of a particle in a box. You want the energy of a harmonic oscillator (which you wrote down in your first post). Also, keep in mind you need to include the normalization factors for ##u_1(x)## and ##u_0(x)## - the big C0 and C1 in your first post, which are not equal to the little c0 and c1. I didn't notice when I wrote ##\psi(x,t=0) = c_0 u_0(x) + c_1 u_1(x)## in a previous post that you had already used c's for some constants; I should have chosen another letter, like d perhaps. That is, your time dependent function should be

$$\psi(x,t) = d_0 C_0 \exp(-q^2/2) \exp(iE_0 t/\hbar) + d_1 C_1 q \exp(-q^2/2) \exp(iE_1 t/\hbar),$$
where the uppercase C's are the normalization factors you mentioned in your first post and the lowercase d's are the probability factors from the fact that you have a superposition of states.

Does that make sense?
 
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