Quantum mechanics, potential well

[fluidistic]

Homework Statement

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.

Homework Equations

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

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!

 

[vela]

Yes, you determine D by normalizing the wave function.

 

[fluidistic]

Yes, you determine D by normalizing the wave function.

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...

 

[vela]

Yes, exactly.

 

[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.

 

[vela]

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

 

[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.

 

[vela]

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.

 

[fluidistic]

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

Hmm ok, good to know.

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

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!

 

[vela]

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.

 

[fluidistic]

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.

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:

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

{{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!

 

[vela]

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.

 

[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.

 

[vela]

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.

 

[fluidistic]

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.

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?

 

[vela]

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.

 

[fluidistic]

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.

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.

 

[vela]

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.)

 

[fluidistic]

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.)

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

 

[fluidistic]

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.

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.

 

[vela]

Again, the first C and D isn't the same as the second C and D. :) They're arbitrary constants, so I just keep reusing the names. I should stop doing that.

After you expand the exponentials, you collect the terms and you'll get
$$\psi_\mathrm{II} = (C+D)\cos kx + i(-C+D)\sin kx$$Since C and D are arbitrary, we can replace the combinations with two new arbitrary constants, so I just renamed them as
\begin{align*}
C+D &\rightarrow C \\
i(-C+D) &\rightarrow D
\end{align*}

 

[fluidistic]

Ah I see, thank you once again.
I think I still reach non sense.
The continuity equations are:
(1) $Ae^{\alpha a}=C \cos (ka)-D \sin (ka)$
(2)$-\alpha Ae^{\alpha a}=Ca \sin (ka)-Da \cos (ka)$
(3) $Ge^{-\alpha a}=C \cos (ka) + D \sin (ka)$
(4) $- \alpha G e^{-\alpha a}=-Ca \sin (ka) + Da \cos (ka)$
From (1) and (2) (divide (2) by - alpha and then equate both equations) I reach that $C=D \frac{[\frac{a}{\alpha}\cos (ka)+sin (ka)]}{[\cos (ka)+ \frac{a}{\alpha}\sin (ka)]}$.
Now $C=D$ + the condition that $a\neq \alpha$ implies that $\tan (ka)=1$ while $C=-D$ + the condition that $a\neq \alpha$ implies that $\tan (ka)=-1$. I'm more than sure that I made at least 1 error since $C= \pm D$ didn't pop up as it should.

 

[vela]

There are a couple of mistakes in your equations. Go back and reread what I wrote in post #12.

If you use sines and cosines, you get either C=0 or D=0, not C=D or C=-D.

 

[fluidistic]

There are a couple of mistakes in your equations. Go back and reread what I wrote in post #12.

If you use sines and cosines, you get either C=0 or D=0, not C=D or C=-D.

Ah I see. I'm digging into this now. I reached your result. I'll try to show that D vanishes.

 

[fluidistic]

I must say I'm getting lost in the algebra. I reached $D=\frac{\alpha}{k} \left [ \frac{Ae^{-\alpha a}}{\cos (ka)}-C \right ]=C \frac{k}{\alpha}-\frac{A e^{-\alpha a}}{\sin (ka)}$.
I isolated C in function of A from this expression. But I couldn't show that D was worth 0 yet.

 

[vela]

Use the fact that $\alpha/k = \tan ka$ (or cot ka, I don't remember which).

 

[fluidistic]

Use the fact that $\alpha/k = \tan ka$ (or cot ka, I don't remember which).

In fact I used it to reach the expression of D I gave in my last post. If I use them once again I go backward one step.
$D= \tan (ka) \left [ \frac{Ae^{-\alpha a}}{\cos (ka)}-C \right ]=C \cot (ka) - \frac{Ae^{-\alpha a}}{\sin (ka)}$.

 

[vela]

What I did was solve for Ae^-αa in two of the equations and then set the results equal to each other:
\begin{align*}
C\cos ka - D \sin ka &= \frac{k}{\alpha}(C\sin ka+D\cos ka) \\
&= (\cot ka)(C\sin ka+D\cos ka)
\end{align*}Can you take it from there?

 

[fluidistic]

What I did was solve for Ae^-αa in two of the equations and then set the results equal to each other:
\begin{align*}
C\cos ka - D \sin ka &= \frac{k}{\alpha}(C\sin ka+D\cos ka) \\
&= (\cot ka)(C\sin ka+D\cos ka)
\end{align*}Can you take it from there?

Oh this helped a lot, thanks for sharing your approach.
I reach $D[\cot (ka) \cos (ka)+\sin (ka)]=0$. It's not very obvious to me that this imply that D=0. But I guess it does?

 

[vela]

Try multiplying through by sin ka.

 

[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?