# Quantum mechanics, potential well

1. Oct 14, 2011

### fluidistic

1. The problem statement, all variables and given/known data
A particle of mass m is find to be inside a uni-dimensional potential well of the form: $V(x)=0$ for $x \leq -a$ and $a\leq x$ and $V(x)=-V_0$ for $-a <x<a$.
1)Write down the corresponding Schrödinger's equation.
2)Consider the case $-V_0<E<0$. Determine the contour conditions and conditions of continuity that an eigenfunction must satisfy inside this potential well.
3)Show that the eigenfunctions have a definite parity and that they are real for the "linked states". I'm not really sure about the word "linked state", this is my own translation of the original problem.
4)Show that the energy of the linked states only take discrete values. (Maybe linked states means eigenstates.)
5)Consider now that E>0. Find and graph the wave function of the particle. Calculate the transmission coefficient when the particle moves along the potential well.
2. Relevant equations

$-\frac{\hbar ^2 }{2m} \frac{d ^2 \Psi (x)}{d x^2}+[V(x)-E]\Psi (x)=0$.

3. The attempt at a solution
Part 1) is solved if you take the equation above and replace V(0) by its values, for each region of the potential well.
I won't put all my work for 2), since it's extremely long.
I reached that in the first region $(x \leq -a)$, $\Psi _I (x)=De^{\alpha x}[e^{x(\alpha+\beta )}-e^{x(\alpha - \beta)}]$.
For the region of the well, region II, $\Psi _{II} (x)=D(e^{-\beta x}-e^{\beta x})$.
For the third region, when $x \geq a$, $\Psi _{III}(x)=De^{-\alpha x} [e^{x(\alpha - \beta)}-e^{x(\alpha + \beta)}]$.
Where $\alpha = \frac{\sqrt {-2mE}}{\hbar}$ and $\beta = \frac{\sqrt{2m(-V_0-E)}}{\hbar}$.
I never met any complex exponential since despite its negative look, all the arguments I met under the square roots were positive.
For the contour condition, I assumed that Psi should not diverge when x tends to negative and positive infinity. From this, I could "drop" 2 constants, or more precisely, equal them to 0.
For the continuity, I assumed that the function Psi and its first derivative must be continuous over the whole 3 regions. This is how I could reduce the number of unknown constant to 1 (I started with A, B, C and D).
My question is now... how do I obtain D? Should I normalize something?
I don't think I'm done for part 2).

For the first part of part 3), it's easy. In previous exercise I showed that if V(x)=V(-x) then Psi is either odd or even. And this is the case in this exercise so by the previous exercise, the eigenfunctions have a definite parity. I'll have to show that they are real and take discrete values.
But first, I want to deal with D... but have no idea how to.
Any help will be appreciated! Thanks!

2. Oct 14, 2011

### vela

Staff Emeritus
Yes, you determine D by normalizing the wave function.

3. Oct 14, 2011

### fluidistic

Ah ok thanks! So I must do $\int _{-\infty} ^{\infty} |\Psi (x)|^2dx=1$, right? Of course I'll have to split the integral according to the 3 regions...

4. Oct 14, 2011

### vela

Staff Emeritus
Yes, exactly.

5. Oct 15, 2011

### fluidistic

I'm depressed. I made at least an error somewhere. If you look at my Psi II and Psi III, you can see that they are equal and this shouldn't be so... I realized this when trying to normalize the Psi function.

6. Oct 15, 2011

### vela

Staff Emeritus
Yeah, I was wondering why V0 makes an appearance in your solutions for regions I and III.

7. Oct 15, 2011

### fluidistic

Ok let's retake this.
Do you buy that the solution to S. equation for the region I is of the form $\Psi _I (x)=Ae^{\alpha x}+Be^{-\alpha x}$ (but B is worth 0 from the contour conditions), thus $\Psi ' _I (x)=\alpha A e^{\alpha x}$?
For the other 2 regions I get $\Psi _{II}(x)= C e^{\beta x}+De^{- \beta x}$, thus $\Psi ' _{II}(x)= \beta C e^{\beta x}-\beta D e^{- \beta x}$.
And $\Psi _{III}(x)=Fe^{\alpha x }+Ge^{-\alpha x}$ (G=0 for the later mentioned reason), thus $\Psi ' _{III}(x)=-\alpha G e^{-\alpha x}$.
Now I say that F and B are worth 0 as conditions of no divergence of Psi when x tends to infinities.
I get the continuity equations:
(1) $Ae^{- \alpha a}=Ce^{-\beta a}+D e^{\beta a}$
(2) $Ce^{\beta a}+D e^{-\beta a}=Ge^{-\alpha a}$
(3) $\alpha A e^{-\alpha a}=\beta C e^{-\beta a}-\beta D e^{\beta a}$
(4) $\beta C e^{\beta a}-\beta D e^{-\beta a}=G e^{-\alpha a}$.
Is what I've done so far okay?
After this, what I've done of my lengthy draft is playing with algebra to reach that, assuming C=-D, G must equal -A. I also isolated A to be worth $D[e^{a (\alpha + \beta)}-e^{a(\alpha - \beta)}]$.
So that I have A, G, and C in terms of D... the remaining unknown constant to be determined with the normalization of the Psi function.

8. Oct 15, 2011

### vela

Staff Emeritus
You should get sinusoidal solutions inside the well when E > -V0

Probably just a typo, but you're missing an alpha on the righthand side of (4). Your approach is fine.

Last edited: Oct 15, 2011
9. Oct 15, 2011

### fluidistic

Hmm ok, good to know.
A typo that costs me a lot! Originally a typo on my draft. I wrote $\Psi ' _{II}(a)= \Psi _{III}(a)$ instead of $\Psi ' _{II}(a)= \Psi' _{III}(a)$. I must redo the algebra.
By the way, is it ok to assume C=-D?
Thank you very much for your help!

10. Oct 15, 2011

### vela

Staff Emeritus
From the symmetry of the potential, you should expect to get even and odd solutions, which will correspond to C=D and C=-D or, respectively, cosines and sines. Those two conditions should come out of the math somehow.

11. Oct 15, 2011

### fluidistic

Ah I see. Well I'll try to reach this mathematically. I've tried to deal with matrices but after about 1 hour of algebra, I reach non sense so I made at least 1 error as usual. I even tried wolfram alpha (failed to calculate the coefficient A, C, D and G.)
Even mathematica 7:
Code (Text):
Solve[{A*e^(-k*a) - C*e^(-b*a) - D*e^(b*a) == 0,
C*e^(b*a) + D*e^(-b*a) - G*e^(-a*k) == 0,
k*A*e^(-a*k) - C*b*e^(-b*a) + b*D*e^(b*a) == 0,
b*C*e^(b*a) - b*D*e^(-b*a) - k*G*e^(-k*a) == 0}, {A, C, D, G}]
returns
Code (Text):
{{G -> 0, A -> 0, C -> 0, D -> 0}}
...
It's almost 2 am, time to sleep for me. I hope I'll wake up and attack this when I just wake up!

12. Oct 16, 2011

### vela

Staff Emeritus
Being lazy, I tried the same thing in Mathematica and got the same result. The problem is if you have four independent equations, the only solution is the trivial solution. You have to find the appropriate values for $\alpha$ and $\beta$ so the equations are no longer independent. These values happen to correspond to the allowed energies of the bound state.

I'm going to suggest you switch to sines and cosines for region II, just to simplify the algebra. Your four equations become:
\begin{align*}
C\cos k_2 a - D \sin k_2 a &= A e^{-k_1 a} \\
C \sin k_2 a + D \cos k_2 a &= \frac{k_1}{k_2} A e^{-k_1 a} \\
C\cos k_2 a + D \sin k_2 a &= G e^{-k_1 a} \\
C \sin k_2 a - D \cos k_2 a &= \frac{k_1}{k_2} G e^{-k_1 a}
\end{align*}where I replaced your $\alpha$ and $\beta$ by k1 and k2 because they're shorter to type.

If you add the first and third equations together, you get
$$2C\cos k_2 a = (A+G)e^{-k_1 a}$$Similarly, if you add the second and fourth equations together, you get
$$2C\sin k_2 a = \frac{k_1}{k_2}(A+G)e^{-k_1 a}$$If you divide these two equations, a bunch of factors cancel, and you get
$$\tan k_2 a = \frac{k_1}{k_2}$$This is one of the conditions you are looking for. You should be able to show that when this condition is satisfied, D has to vanish. Then it follows that A=G, and you have your wave function up to the normalization constant. If you do the same thing except this time isolating the terms with D in it, you'll get another condition, which leads to the rest of the solutions.

13. Oct 16, 2011

### fluidistic

Ok thank you very much vela. I do have a few questions though, I don't really get what you get.
From your first equation it seems like $e^{- \beta a}$ transforms into $\cos (\beta a)$ and $e^{\beta a}$ transforms into $-\sin (\beta a)$. However if I transform all equations according to this rule, I get different equations than yours, but for equation 1. So I didn't really understand how you simplified the exponential function for the region II.

14. Oct 16, 2011

### vela

Staff Emeritus
Your C and D aren't the same as my C and D. I should have made that clear. Remember that $\beta=ik_2$ will be complex, so $e^{\beta a} = \cos k_2a + i\sin k_2a$ and $e^{-\beta a} = \cos k_2a - i\sin k_2a$. Then you can collect terms and rename the constants. Alternately, you can write the solution to region II down as $\psi_\mathrm{II}(x) = C \cos k_2 x + D \sin k_2 x$ and then apply the boundary conditions.

15. Oct 16, 2011

### fluidistic

Ok, I'm going to retake all. I've tried to keep exponentials and reached $C=D \frac{\frac{\beta }{\alpha }+1}{1-\frac{\beta}{\alpha}}$ but I think it's wrong because $C \neq \pm D$ in this case I think.
So basically if I restart the whole exercise, I should define $\beta = \frac{\sqrt{2m(-V_0-E)}}{\hbar}= \frac{i \sqrt {2m (V_0+E)}}{\hbar}$ for convenience?

16. Oct 16, 2011

### vela

Staff Emeritus
I would actually pull the i out and define β as a real quantity, so your solution in region II would be
$$\psi(x) = C e^{i\beta x} + D e^{-i\beta x}$$or
$$\psi(x) = C \cos \beta x + D \sin \beta x$$I find it's simpler to work with real-valued variables if you can.

17. Oct 16, 2011

### fluidistic

Hmm I don't really understand. My beta was supposed to be real. It is/was worth $\frac{\sqrt {2m(-V_0-E)}}{\hbar}$ where both $V_0$ and $E$ are negative, so that $-V_0-E>0$. Thus the square root is real and so is beta.
If I factorizes by "i", I get a negative argument in the square root, which is worth a complex number (I can't define the new "beta" to be real in this case, which is what you've done I think. I don't really understand this part). In all cases beta is real valued.

18. Oct 16, 2011

### vela

Staff Emeritus
The way you originally stated the problem, the potential is -V0 at the bottom of the well, which implied V0>0 so that -V0<0. Since E<0 and it needs to be greater than -V0, you have E > -V0 so 0 > -V0-E. (These signs always mess me up too.)

19. Oct 16, 2011

### fluidistic

Oh... I totally missed this! Well thank you very much for pointing this out. I'm going to redo everything with this in mind!

20. Oct 16, 2011

### fluidistic

Okay your beta is worth $\frac{ \sqrt{2m(V_0+E)}}{\hbar}$ which is indeed real.
But I don't understand how you get such values for Psi.
$\Psi _{II}(x)=Ce^{i kx}+De^{-ikx}$ where my k is worth your beta.
Now, this is worth $C[\cos (kx)-i \sin (kx)]+D[\cos (kx)+i \sin (kx)]$. But from here, I don't undestand how you simplified to get $C \cos (kx)+D \sin (kx)$. Could you explain a bit more please?
Thanks for all your help and time.