Quantum mechanics, potential well

[fluidistic]

Try multiplying through by sin ka.

Wow. I totally missed it. Thank you very much for all your help. Time for me to sleep. Hopefully I'll solve the problem before tomorrow bed-time.
Have a nice day/night.

 

[fluidistic]

So this gives me:
$\Psi _I (x)=C \cos (ka) e^{\alpha (a+x)}$
$\Psi _{II}(x)=C \cos (kx)$
$\Psi _{III}(x)=C \cos (ka)e^{\alpha (a-x)}$.
Now I know I must normalize but... I think I made some error(s).
The question 3 says I must show that Psi is either odd or even. But the exponentials are a problem to me...
Psi is real valued as I must have showed though, so this work.

 

[vela]

That looks fine. Remember $\alpha a$ term in the exponent merely results in a multiplicative constant because $e^{\alpha(a+x)} = e^{\alpha a} e^{\alpha x}$.

If you wrote the eigenfunction in terms of A instead of C, it becomes even more plain to see:
\begin{align*}
\psi_\mathrm{I}(x) &= Ae^{\alpha x} \\
\psi_\mathrm{II}(x) &= A(e^{-\alpha a}\sec ka)\cos kx \\
\psi_\mathrm{III}(x) &= Ae^{-\alpha x}
\end{align*}

 

[fluidistic]

Ah I see now, even with Psi written in terms of C.
I tried to normalize it. I reached that $\int _{-\infty} ^{-a}|\Psi _I (x)|^2dx=\frac{C^2 \cos ^2 (ka)}{2\alpha}=\int _{a}^{\infty} |\Psi _{III}(x)|^2dx$ (and that's a good sign that they are equal!) while $\int _{-a}^a |\Psi _{II}|dx=C^2 \left [ \frac{\sin (2ak)}{2k}+a \right ]$.
So that $C^2=\left [ \frac{\cos ^2 (ka)}{\alpha}+\frac{\sin(2ak)}{2k}+a \right ] ^{-1}$. Now I suppose I must take the positive square root since the probability must be positive.
Is that okay? This doesn't look very nice on my eyes though.

 

[vela]

Do you need to normalize the wave functions for this problem?

 

[fluidistic]

Do you need to normalize the wave functions for this problem?

I forgot the problem for a moment, I was being enthusiastic.
I think I don't need it.
By the way, I'm going to try to get the other "set" of solutions, when I have to isolate D rather than C. But I don't understand why there are 2 sets of solutions. What does this physically imply?

 

[vela]

Nothing. It's just the way the math works out. The same thing happens with the infinite square well problem when the well is centered at x=0, but if the well is from x=0 to x=a, you don't run into that complication.

 

[fluidistic]

Nothing. It's just the way the math works out. The same thing happens with the infinite square well problem when the well is centered at x=0, but if the well is from x=0 to x=a, you don't run into that complication.

Hmm. Indeed weird to me. By the way thanks for still helping me.
I worked out the other set of solutions by isolating D rather than C as you suggested me. I found out that in that case $\tan (ka)=-\frac{k}{\alpha}$. This gave me, after a few algebra that $C=0$. $A=-G$. And $A=-De^{\alpha a}\sin (ka)$.
All in all, I reach as second set of solutions:
$\Psi _I (x)=-De^{\alpha (a+x)} \sin (ka)$
$\Psi _{II}(x)=D \sin (kx)$
$\Psi _{III}(x)=De^{\alpha (a-x)} \sin (ka)$.
I hope I didn't make any mistake.
What I don't understand is that the solution is either this one or the previous one with cosines. I don't grasp this concept. The Psi function has like 2 choices and must choose 1 and discard the other...

I tried to do 4) but I'm not sure how to proceed. I know that alpha and k depends on E but I don't know why only certain values of E (thus alpha and k) will make Psi mathematically possible. I look at any of the 2 psi functions and don't see why only certain value of alpha and k are possible. I'm missing it.

 

[vela]

There's nothing terribly mysterious going on. It's just like when you solve a quadratic by factoring (x-a)(x-b)=0. You either have x-a=0 or x-b=0, each giving rise to a solution. Same sort of thing here. When you have tan ka = alpha/k, you get one set of solutions. When tan ka = -k/alpha, you get another set of solution.

Pick values for a and V₀, and try plotting these functions for various values of E. Just pick an energy, calculate alpha and k and plot the resulting wave function. And then try it with a value of E where tan ka = alpha/k.

 

[fluidistic]

There's nothing terribly mysterious going on. It's just like when you solve a quadratic by factoring (x-a)(x-b)=0. You either have x-a=0 or x-b=0, each giving rise to a solution. Same sort of thing here. When you have tan ka = alpha/k, you get one set of solutions. When tan ka = -k/alpha, you get another set of solution.

Pick values for a and V₀, and try plotting these functions for various values of E. Just pick an energy, calculate alpha and k and plot the resulting wave function. And then try it with a value of E where tan ka = alpha/k.

Ok thank you.
But say I don't have anything that graph and I want to find algebraically that the energy is quantized, how should I proceed?
My try is to take $\tan (ka)=\frac{\alpha }{k} \Rightarrow \frac{\sqrt {-E} }{\sqrt {V_0 + E}}= \tan \left ( a \sqrt { \frac{2m(V_0+E)}{\hbar ^2} } \right )$.
Now I must show that this is satisfied for only some values of E. Maybe graph the 2 functions and the intersection would show the solutions.
I'll try to graph now... but when I don't have a computer, how do I show what is the ground state? I mean what is the minimum allowed value for the energy?

Edit: I post the graph I obtained with random numbers for a, V_0, etc.

 

[vela]

I misunderstood what you were getting at. Your approach is right. That condition is only satisfied for certain values of E. At other values of E, you can't join the pieces of the wave function together smoothly.

You have to solve for the energy of the ground state numerically, so you're stuck with using a computer or calculator, or doing laborious iterations by hand.

 

[fluidistic]

I misunderstood what you were getting at. Your approach is right. That condition is only satisfied for certain values of E. At other values of E, you can't join the pieces of the wave function together smoothly.

You have to solve for the energy of the ground state numerically, so you're stuck with using a computer or calculator, or doing laborious iterations by hand.

Ah I see! Thank you very much all your help.
I'll try to do 5) alone.
Problem -almost- solved.