An infinite square well problem

quasi-static
Messages
2
Reaction score
0

Homework Statement



Particle in well:

V(x)=0 for |x|<\frac{L}{2}
V(x)=∞ for |x|>\frac{L}{2}

initial wave function \Psi(x,0)=\frac{1}{√L}[cos\frac{\pi*x}{L}+ i*sin\frac{2*\pi*x}{L}]

a) calc P(p,t) (momentum prob density)

Homework Equations



Anything from Griffiths QM

The Attempt at a Solution



I'm getting tripped out from the initial wave function. It is perfectly clear to me the process in which to solve for ψ(x,t) , given the initial wave function ψ(x,0); however, I'm not sure what to do in this case.

I know that, given ψ(x,0), we must do the following:

1)normalize \Psi(x,0)
2)compute the expansion coefficients (aka, c_{n})
3)compute E_{n} and plug into the time dependent solution
4)plug in c_{n}, Normalized "A" value, and E_{n} into ψ(x,t)
 
Last edited:
Physics news on Phys.org
If the initial wave function had been

$$ \Psi(x,0) = \sqrt{\frac{2}{L}}sin(\pi x/L) $$

do you know what \Psi(x,t) would be?
 
king vitamin said:
If the initial wave function had been

$$ \Psi(x,0) = \sqrt{\frac{2}{L}}sin(\pi x/L) $$

do you know what \Psi(x,t) would be?

If you're trying to see that I know how to do the most basic infinite square well problem , yes, and I've done it for various situations (shifted wells and whatnot)

\Psi(x,t)=c_{n}\frac{√2}{√L}sin(\frac{n\pi*x}{L})(e^{\frac{-in^2\pi^{2}ht}{2mL^{2}}})

where, in ψ(x,t), h = h-bar (didn't know how to find h-bar in latex)

was the plan to aid in clarification of the problem at hand once i prove that I'm not just looking for "answers" ?
 
quasi-static said:
(didn't know how to find h-bar in latex)

It's \hbar. :biggrin:
 
I didn't have a "plan," I'm just trying to help based on info given to me. Given that you have no problem attaining \Psi(x,t), do you know the equation you use to find the momentum probability density? You mention expansion coefficients in your OP, but I'm not sure what expansion coefficients you're referring to (are they for momentum?).
 
quasi-static said:

Homework Statement



Particle in well:

V(x)=0 for |x|<\frac{L}{2}
V(x)=∞ for |x|>\frac{L}{2}

initial wave function \Psi(x,0)=\frac{1}{√L}[cos\frac{\pi*x}{L}+ i*sin\frac{2*\pi*x}{L}]

a) calc P(p,t) (momentum prob density)

Homework Equations



Anything from Griffiths QM

The Attempt at a Solution



I'm getting tripped out from the initial wave function. It is perfectly clear to me the process in which to solve for ψ(x,t) , given the initial wave function ψ(x,0); however, I'm not sure what to do in this case.

I know that, given ψ(x,0), we must do the following:

1)normalize \Psi(x,0)
2)compute the expansion coefficients (aka, c_{n})
3)compute E_{n} and plug into the time dependent solution
4)plug in c_{n}, Normalized "A" value, and E_{n} into ψ(x,t)
So what's stopping you from doing these steps? It would help if you were a little more specific than saying you're "getting tripped out."
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top