Infinite square well doubled with time

[Bananen]

A particle is in its ground state of an infinite square well of width a <xl i>=√2/a*sin(πx/a) and since it's an eigenstate of the Hamiltonian it will evolve as <xlα(t)>=√2/a*sin(πx/a)e^(-iE₁t/ħ) where E=π²ħ²/2ma²

If the well now suddenly expands to witdh 2a
If the well suddenly expands to 2a the particle's state remains the same and the wavefunction is given by

<xli>=2/a*sin(πx/a) if 0 < x< a and 0 for x>a

which is not the ground state of the new Hamiltonian i.e. √1/a*sin(πx/2a)

compute <xlα(t)> for this new system. Is there a time-dependende and/or conservation laws?Okay so this is a really hard question for me, I know how to solve the infinite square well for the first case but I don't really have any intuition when it comes to the second part of this problem. My initial thought was to calculate <xlα(t)> using the new Hamiltonian stated above and the time dependence e^(-iE₁t/ħ) but this I feel like is way too easy. Then I read somewhere that the wavefunction is the sum over all states and so I thought that maybe I should sum over the two states (the old and new Hamiltonian) times the time-dependence but it feels wrong aswell.. Could anyone give me a push in the right direction with this problem?

I'm thankful for any answers.

 

[vela]

First suggestion: Learn LaTeX and proofread your posts.

My initial thought was to calculate <xlα(t)> using the new Hamiltonian stated above and the time dependence e^(-iE₁t/ħ) but this I feel like is way too easy.

Not sure what this means. It's kind of vague.

Then I read somewhere that the wavefunction is the sum over all states and so I thought that maybe I should sum over the two states (the old and new Hamiltonian) times the time-dependence but it feels wrong aswell.. Could anyone give me a push in the right direction with this problem?

It sounds like you're confusing the Hamiltonian with states. They're not the same thing.

Your textbook should cover the time evolution of a state (not just eigenstates). I suggest you read up on that, make sure you understand all the reasoning there, and then come back to this problem.

 

[stevendaryl]

Could you clarify what you mean by <xli> and <xlα(t)>? Do you mean the wave function?

If you mean the wave function, then I think the question boils down to this:

You have an initial wave function $\psi_i(x) \propto sin(\pi x/a)$. You have a new set of basis states:

$\phi_n(x, t) \propto sin(n \pi x/(2a)) e^{-i \omega_n t}$

The first problem is to write $\psi_i(x)$ in terms of the new basis. Then include the time dependence to get an expression for $\psi_i(x,t)$.

 

[Bananen]

First suggestion: Learn LaTeX and proofread your posts.Not sure what this means. It's kind of vague.

It sounds like you're confusing the Hamiltonian with states. They're not the same thing.

Your textbook should cover the time evolution of a state (not just eigenstates). I suggest you read up on that, make sure you understand all the reasoning there, and then come back to this problem.

I'm sorry for being vague and not knowing how to use LaTex, to me it's perfectly readable. However I didn't come here to be talked down to, I wanted some kind of help to get going with my assignment. This assignment is something my teacher has come up with so there's not really anything relatable in the textbook to this particular question.

Could you clarify what you mean by <xli> and <xlα(t)>? Do you mean the wave function?

I guess <xli> and <xlα(t)> are wavefunctions it's not stated in the problem.

If you mean the wave function, then I think the question boils down to this:

You have an initial wave function ψi(x)∝sin(πx/a)ψi(x)∝sin(πx/a)\psi_i(x) \propto sin(\pi x/a). You have a new set of basis states:

ϕn(x,t)∝sin(nπx/(2a))e−iωntϕn(x,t)∝sin(nπx/(2a))e−iωnt\phi_n(x, t) \propto sin(n \pi x/(2a)) e^{-i \omega_n t}

The first problem is to write ψi(x)ψi(x)\psi_i(x) in terms of the new basis. Then include the time dependence to get an expression for ψi(x,t)ψi(x,t)\psi_i(x,t).

Thank you I will try that.

 

[vela]

I'm sorry for being vague and not knowing how to use LaTex, to me it's perfectly readable.

That's not really what's important, though, is it? You're asking others to help so you want to make your post readable to others, not yourself.

Note if you had written $\langle x \vert i \rangle$, stevendaryl wouldn't have had to ask you what <xli> meant.

However I didn't come here to be talked down to, I wanted some kind of help to get going with my assignment. This assignment is something my teacher has come up with so there's not really anything relatable in the textbook to this particular question.

I doubt that your textbook doesn't go over this topic. It's a basic topic that every quantum mechanics book goes over.

Anyway, my intent wasn't to talk down to you, but your original post suggests you have kind of a vague idea of what's going on as evidenced by the fact that you seem to use the words Hamiltonian and state as synonyms, not understanding how to time evolve a general state, etc. Now that you have a specific problem you're trying to solve, you'll probably get more out of rereading the textbook and help you clarify some of these concepts in your mind.

 

[Bananen]

That's not really what's important, though, is it? You're asking others to help so you want to make your post readable to others, not yourself.

Note if you had written $\langle x \vert i \rangle$, stevendaryl wouldn't have had to ask you what <xli> meant.I doubt that your textbook doesn't go over this topic. It's a basic topic that every quantum mechanics book goes over.

Anyway, my intent wasn't to talk down to you, but your original post suggests you have kind of a vague idea of what's going on as evidenced by the fact that you seem to use the words Hamiltonian and state as synonyms, not understanding how to time evolve a general state, etc. Now that you have a specific problem you're trying to solve, you'll probably get more out of rereading the textbook and help you clarify some of these concepts in your mind.

Yes okay I know where to never ask for help again thank you very much