Finding Energies For 1D Potential

[flyusx]

Homework Statement

A particle is moving in the potential well $$V(x)=\begin{cases}0&-a\leq x\leq-b\\V_{0}&-b\leq x\leq b\\0&b\leq x\leq a\\+\infty&\text{elsewhere}\end{cases}$$ where $V_{0}$ is positive. In this problem, consider $E<V_{0}$. Let $\psi_{1}(x)$ and $\psi_{2}(x)$ represent the two lowest energy solutions of the Schrodinger equation; call their energies $E_{1}$ and $E_{2}$ respectively.

(a) Calculate $E_{1}$ and $E_{2}$ in eV for the case where $mc^{2}=1$ GeV, $a=10^{-14}$ m and $b=0.4*10^{-14}$ m.
(b) A particular solution of the Schrodinger equation can be constructed by superposing $\psi_{1}(x)\exp\left(\frac{iE_{1}t}{\hbar}\right)$ and $\psi_{2}(x)\exp\left(\frac{iE_{2}t}{\hbar}\right)$. Construct a wave packet $\psi$ which at $t=0$ is (almost) entirely to the left-hand side of the well and describe its motion in time; find the period of oscillations between the two terms of $\psi$.

Relevant Equations

$-\frac{\hbar^{2}}{2m}\frac{\text{d}^{2}\psi}{\text{d}x^{2}}+V\psi(x)=E\psi(x)$

I have been stuck on part (a) for a few days and would appreciate a nudge in the right direction. This is Zettili Exercise 4.4.

I know this is a symmetric potential so that the (non-degenerate) bound states will have definite parity. I have tried to take advantage of this symmetry by focusing on wave functions for $x\geq0$ so that the potential of interest is $$V(x)=\begin{cases}V_{0}&0\leq x\leq b\\0&b\leq x\leq a\\+\infty&x>a\end{cases}$$ and I denote $\psi_{1}(x)$ to be the wave function before $x=b$ and $\psi_{2}(x)$ to be the wave function between $x=b$ and $x=a$.

I first write the Schrodinger equations $$\frac{\text{d}^{2}\psi_{1}}{\text{d}x^{2}}-k_{1}^{2}\psi_{1}(x)=0$$$$\frac{\text{d}^{2}\psi_{2}}{\text{d}x^{2}}+k_{2}^{2}\psi_{2}(x)=0$$ where $$k_{1}=\frac{2m(V_{0}-E)}{\hbar}$$$$k_{2}=\frac{2mE}{\hbar}$$

I found general solution to be $$\psi(x)=\begin{cases}\psi_{1}(x)=A\exp(k_{1}x)+B\exp(-k_{1}x)\\\psi_{2}(x)=C\exp(ik_{2}x)+D\exp(-ik_{2}x)\end{cases}$$

I now impose parity on the wave functions. For the even states $\psi_{+}(x)$, I found $$\psi_{1+}(x)=\psi_{1+}(-x)\rightarrow A\exp(k_{1}x)+B\exp(-k_{1}x)=A\exp(-k_{1}x)+B\exp(-k_{2}x)\rightarrow A=B$$$$\psi_{2+}(x)=\psi_{2+}(-x)\rightarrow C\exp(ik_{2}x)+D\exp(-ik_{2}x)=C\exp(-ik_{2}x)+D\exp(ik_{2}x)\rightarrow C=D$$ I use a few trigonometric-exponential identities and I'll redefine the constants used to be $A$ and $B$ such that the even wave function is $$\psi_{+}(x)=\begin{cases}\psi_{1+}(x)=A\cosh(k_{1}x)\\\psi_{2+}(x)=B\cos(k_{2}x)\end{cases}$$

I did the same for the odd states so that $$\psi_{1-}(x)=-\psi_{1-}(-x)\rightarrow A\exp(k_{1}x)+B\exp(-k_{1}x)=-A\exp(-k_{1}x)-B\exp(k_{1}x)\rightarrow A=-B$$$$\psi_{2-}(x)=-\psi_{2-}(-x)\rightarrow C\exp(ik_{2}x)+ D\exp(-ik_{2}x)=-C\exp(-ik_{2}x)-D\exp(ik_{2}x)\rightarrow C=-D$$ I'll redefine the constants to be $C$ and $D$ so that the odd wave function is $$\psi_{-}(x)=\begin{cases}\psi_{1-}(x)=C\sinh(k_{1}x)\\\psi_{2-}(x)=D\sin(k_{2}x)\end{cases}$$

Here is where I have trouble. When I write down the boundary conditions for the states (I'll take the even state here) $$\psi_{1+}(x)=\psi_{2+}(x)\rightarrow A\cosh(k_{1}b)=B\cosh(k_{2}b)$$$$\left.\frac{\text{d}\psi_{1+}}{\text{d}x}\right\vert_{x=b}=\left.\frac{\text{d}\psi_{2+}}{\text{d}x}\right\vert_{x=b}\rightarrow k_{1}A\sinh(k_{1}b)=-k_{2}B\sin(k_{2}b)$$$$\psi_{2+}(a)=B\cos(k_{2}a)=0$$
Now I see two ways to find the lowest even state energy. I can take the third relation and say $$\cos(k_{2}a)=0\rightarrow\frac{\sqrt{2mE}}{\hbar}a=\frac{\pi}{2}$$
and plugging in values gives $E\approx480.4$ keV. On the other hand, I can take divide the second equation by the first and get $$k_{1}\tanh(k_{1}b)=-k_{2}\tan(k_{2}b)$$ which by doing a change of variables $\gamma=k_{1}b$ and $y=k_{2}b$ gives $$\frac{\gamma}{b}\tanh(\gamma)=-\frac{y}{b}\tan(y)$$ or $$\gamma\tanh(\gamma)=-y\tan(y)$$ When I plot the functions $x\tanh(x)$ and $-x\tan(x)$, I get an intersection point in the first quadrant at $\left(2.36502,\ 2.32364\right)$. Now I take $$y=2.36502=\frac{\sqrt{2mE}}{\hbar}b$$ and I get $E\approx6.8$ MeV.

So I get two energies that are widely in contrast with each other. I'm not sure what I'm doing wrong.

 

[kuruman]

I can understand why you might consider half the region to the right of the origin because of symmetry.
I cannot understand why the wavefunctions, even or odd do not vanish at $\cancel{x=b}$ $x=a$ where you there is a hard wall.

Edited for typo.

 

[TSny]

I found general solution to be $$\psi(x)=\begin{cases}\psi_{1}(x)=A\exp(k_{1}x)+B\exp(-k_{1}x)\\\psi_{2}(x)=C\exp(ik_{2}x)+D\exp(-ik_{2}x)\end{cases}$$

I now impose parity on the wave functions. For the even states $\psi_{+}(x)$, I found $$\psi_{1+}(x)=\psi_{1+}(-x)\rightarrow A\exp(k_{1}x)+B\exp(-k_{1}x)=A\exp(-k_{1}x)+B\exp(-k_{2}x)\rightarrow A=B$$$$\psi_{2+}(x)=\psi_{2+}(-x)\rightarrow C\exp(ik_{2}x)+D\exp(-ik_{2}x)=C\exp(-ik_{2}x)+D\exp(ik_{2}x)\rightarrow C=D$$ I use a few trigonometric-exponential identities and I'll redefine the constants used to be $A$ and $B$ such that the even wave function is $$\psi_{+}(x)=\begin{cases}\psi_{1+}(x)=A\cosh(k_{1}x)\\\psi_{2+}(x)=B\cos(k_{2}x)\end{cases}$$

You should not have excluded the possibility of a term of the form $C\sin(k_2 x)$ in the region $b < x <a$. You can still force the wave function to be even by writing $$\psi(x)= B\cos(k_{2}x) + C\sin(k_{2}|x|)$$ for the regions $b<x<a$ and $-a<x<-b$.

If you don't include the sine function, you will not have enough undetermined constants to match all the boundary conditions.

You might wonder why we don't also include a $\sinh(k_1|x|)$ term. This is excluded because the even wave function must have zero slope at x = 0.

 

[TSny]

I can understand why you might consider half the region to the right of the origin because of symmetry.
I cannot understand why the wavefunctions, even or odd do not vanish at $x=b$ where you there is a hard wall.

The hard walls are at $x = \pm a$. Note $0< b < a$.

 

[kuruman]

The hard walls are at $x = \pm a$. Note $0< b < a$.

Yes of course. Good catch. I meant to say that the wavefunction must vanish at $x=\pm a$.

 

[vela]

You might find it helpful to write the wave function for $b < x < a$ as $\psi_{2+} = C \sin[k_2(x-a)]$ which will automatically satisfy the boundary condition at $x=a$.

 

[flyusx]

You might find it helpful to write the wave function for $b < x < a$ as $\psi_{2+} = C \sin[k_2(x-a)]$ which will automatically satisfy the boundary condition at $x=a$.

So it is generally better to have the wave function vanish at the necessary 'vanishing points' than impose the conditions after writing them like I did. In hindsight, I suppose it would have been better to write the arbitrary wave function of form $A\sin(k_{2}x)+B\cos(k_{2}x)$ instead of with exponentials, force it to be even by writing $A\sin(k_{2}\vert x\vert)+B\cos(k_{2}x)$ and then force it to vanish at $x=\pm a$ by writing $$A\sin(k_{2}(\vert x\vert-a))+B\cos\left(k_{2}\left(\vert{x}\vert-a+\frac{\pi}{2k_{2}}\right)\right)$$
But the cosine term is just a version of the sine term since $\cos\left(x+\frac{\pi}{2}\right)=-\sin(x)$. I'll continue working on this later tonight.

 

[kuruman]

So it is generally better to have the wave function vanish at the necessary 'vanishing points' than impose the conditions after writing them like I did. In hindsight, I suppose it would have been better to write the arbitrary wave function of form $A\sin(k_{2}x)+B\cos(k_{2}x)$ instead of with exponentials, force it to be even by writing $A\sin(k_{2}\vert x\vert)+B\cos(k_{2}x)$ and then force it to vanish at $x=\pm a$ by writing $$A\sin(k_{2}(\vert x\vert-a))+B\cos\left(k_{2}\left(\vert{x}\vert-a+\frac{\pi}{2k_{2}}\right)\right)$$
But the cosine term is just a version of the sine term since $\cos\left(x+\frac{\pi}{2}\right)=-\sin(x)$. I'll continue working on this later tonight.

I think it is more straightforward to observe that

• $\cos(k_2x)$ is an even function and $\cos(k_2a)=0$ implies $k_2=(2n+1)\dfrac{\pi}{2}$ where $n$ is an integer.
• $\sin(k_2x)$ is an odd function and $\sin(k_2a)=0$ implies $k_2=(2n)\dfrac{\pi}{2}$ where $n$ is an integer.

 

[TSny]

So it is generally better to have the wave function vanish at the necessary 'vanishing points' than impose the conditions after writing them like I did. In hindsight, I suppose it would have been better to write the arbitrary wave function of form $A\sin(k_{2}x)+B\cos(k_{2}x)$ instead of with exponentials, force it to be even by writing $A\sin(k_{2}\vert x\vert)+B\cos(k_{2}x)$ and then force it to vanish at $x=\pm a$

Forcing $A\sin(k_2 x) + B\cos(k_2 x)$ to vanish at $x = a$ gives a relation among $A$, $B$,and $k_2$ which is useful. However, @vela 's suggestion is a much better approach. It reduces the amount of algebra. Instead of two constants $A$ and $B$, you have just the one constant $C$ in $C \sin[ k_2 (x-a)]$.

Was the numerical value of $V_0$ given in the problem statement?

 

[kuruman]

However, @vela 's suggestion is a much better approach. It reduces the amount of algebra.

But how does it provide quantized energy levels?

 

[TSny]

But how does it provide quantized energy levels?

The boundary conditions at $x = b$ will restrict the energy to certain discrete values.

 

[vela]

I now impose parity on the wave functions. For the even states $\psi_{+}(x)$, I found $$\psi_{1+}(x)=\psi_{1+}(-x)\rightarrow A\exp(k_{1}x)+B\exp(-k_{1}x)=A\exp(-k_{1}x)+B\exp(-k_{2}x)\rightarrow A=B$$$$\psi_{2+}(x)=\psi_{2+}(-x)\rightarrow C\exp(ik_{2}x)+D\exp(-ik_{2}x)=C\exp(-ik_{2}x)+D\exp(ik_{2}x)\rightarrow C=D$$ I use a few trigonometric-exponential identities and I'll redefine the constants used to be $A$ and $B$ such that the even wave function is $$\psi_{+}(x)=\begin{cases}\psi_{1+}(x)=A\cosh(k_{1}x)\\\psi_{2+}(x)=B\cos(k_{2}x)\end{cases}$$

The problem is that you've assumed that the coefficients of $e^{\pm ik_2 x}$ are the same for $\psi_{2+}$ and $\psi_{2-}$, so imposing the parity condition just picks off the even part of $\psi_{2+}$.

You need to allow for the possibility that the coefficients are different for the two pieces of the solution. So let
\begin{align*}
\psi_{2+} &= A \cos k_2 x + B \sin k_2 x \\
\psi_{2-} &= C \cos k_2 x + D \sin k_2 x
\end{align*} You would get an even solution when $A=C$ and $B = -D$.

 

[flyusx]

Sorry, I'm still a bit confused. It seems I have more confusion with solving the Schrodinger equation than I thought.

You might wonder why we don't also include a $\sinh(k_1|x|)$ term. This is excluded because the even wave function must have zero slope at x = 0.

I can accept that even functions have zero slope at $x=0$, but if I plot a function like $\sin(k_{2}\vert x\vert)+\cos(k_{2}x)$ (where I arbitrarily set $A=B=1$), I get an even function that has an undefined derivative (a cusp) at $x=0$. So why is this allowed?

Forcing $A\sin(k_2 x) + B\cos(k_2 x)$ to vanish at $x = a$ gives a relation among $A$, $B$,and $k_2$ which is useful. However, @vela 's suggestion is a much better approach. It reduces the amount of algebra. Instead of two constants $A$ and $B$, you have just the one constant $C$ in $C \sin[ k_2 (x-a)]$.

Was the numerical value of $V_0$ given in the problem statement?

$V_{0}$ has no numerical value. I see the rather strong advantage of writing as vela suggested by reducing the the continuity of the function, its first derivative and the vanishing condition down to just the two continuity conditions (since I take it the vanishing condition is implicit in the wave function). I assume the proper form is $C\sin(k_{2}(\vert x\vert-a))$ to be an even function?

The problem is that you've assumed that the coefficients of $e^{\pm ik_2 x}$ are the same for $\psi_{2+}$ and $\psi_{2-}$, so imposing the parity condition just picks off the even part of $\psi_{2+}$.

You need to allow for the possibility that the coefficients are different for the two pieces of the solution. So let
\begin{align*}
\psi_{2+} &= A \cos k_2 x + B \sin k_2 x \\
\psi_{2-} &= C \cos k_2 x + D \sin k_2 x
\end{align*} You would get an even solution when $A=C$ and $B = -D$.

I'm not sure I've understand this correctly. I've taken this to mean that a general solution of a differential equation with constant coefficients of the form $\psi''+k_{2}\psi=0$ is $A\sin(k_{2}x)+B\cos(k_{2}x)$ (or an equivalent formalism with complex exponentials). When I considered the two parity cases, I should have written $A\sin(k_{2}x)+B\cos(k_{2}x)$ for an even function and $C\sin(k_{2}x)+D\cos(k_{2})$ for an odd function, and then impose other conditions to make the wave function satisfy the necessary criteria.

It will feel really nice when I understand fully how to properly solve the Schrodinger equation to yield the correct wave functions for this problem. Until then, thanks for the input.

 

[TSny]

I can accept that even functions have zero slope at $x=0$, but if I plot a function like $\sin(k_{2}\vert x\vert)+\cos(k_{2}x)$ (where I arbitrarily set $A=B=1$), I get an even function that has an undefined derivative (a cusp) at $x=0$. So why is this allowed?

The function $\psi(x) =B \sin(k_2 \vert x \vert) + C \cos(k_2 x)$ is to be used only in the regions $b\leq x \leq a$ and $-a \leq x \leq -b$.

For $b \leq x \leq a$, $x$ is positive and so $\psi(x) =B \sin(k_2 x ) + C \cos(k_2 x)$. This function has no cusps for any choice of $B$ and $C$.

In the region $-a \leq x \leq -b$, $x$ is negative. So, in this region $\vert x \vert = -x$. Then, $\psi(x) = B \sin(k_2 (-x )) + C \cos(k_2 x) = -B \sin(k_2 x) + C \cos(k_2 x)$. Again, this function has no cusps.

$\vert x \vert$ has a cusp at $x = 0$, but $x = 0$ is not in the regions where we are using $\psi(x) =B \sin(k_2 \vert x \vert) + C \cos(k_2 x)$.

$V_{0}$ has no numerical value.

Without a value for $V_0$, you cannot determine numerical values for the two lowest energy levels. So, I don't understand why they bothered to give numerical values for the other parameters in the problem statement.

 

[vela]

I'm not sure I've understand this correctly.

Sorry, I misinterpreted your notation, so it's no surprise you found what I wrote confusing. I meant what you wanted to do is write the solution as
$$\psi(x) = \begin{cases}
A \cos k_2 x + B \sin k_2 x, & -a <x < -b \\
F \cosh k_1 x + G \sinh k_1 x, & -b < x < b \\
C \cos k_2 x + D \sin k_2 x, & b < x < a \\
0, & |x| > a
\end{cases}$$ or its equivalent in terms of exponentials. The important thing is to have different constants for the solution for $-a<x<-b$ and $b<x<a$.

Assume $x>b$. Then an even function would satisfy
$$ A \cos k_2 x - B \sin k_2 x = C \cos k_2 x + D \sin k_2 x$$ which implies $A=C$ and $B=-D$. In your work, because you assumed $\psi(x)$ had the same form for positive and negative $x$, you effectively imposed the condition
$$ A \cos k_2 x - B \sin k_2 x = A \cos k_2 x + B \sin k_2 x$$ which led to the result $B=0$.

I see the rather strong advantage of writing as vela suggested by reducing the the continuity of the function, its first derivative and the vanishing condition down to just the two continuity conditions (since I take it the vanishing condition is implicit in the wave function).

I just wanted to point out that you're solving the differential equation the usual way except instead choosing the two independent solutions $\cos k_2 x$ and $\sin k_2 x$, you're using the two solutions $\cos [k_2(x-a)]$ and $\sin [k_2(x-a)]$. The requirement that the wave function vanish at $x=a$ then makes you toss the cosine solution.

 

[TSny]

It will feel really nice when I understand fully how to properly solve the Schrodinger equation to yield the correct wave functions for this problem. Until then, thanks for the input.

When you know that $\psi(x)$ is even or odd about $x = 0$, you only need to worry about solving the problem for $x \ge 0$. Once you have that, the solution for $x < 0$ follows from the even or odd condition.

To solve for $x \ge 0$ you need to deduce the general form of $\psi(x)$ in each region of positive $x$ and you need to know the boundary conditions at $x = 0$, $x = b$, and $x = a$. The even or odd requirement determines the boundary condition at $x = 0$. The boundary conditions at $b$ and $a$ will then determine the allowed energies and the values of the unknown constants $A, B$, etc (except for an overall constant that can be determined later by normalizing the wave function). Once you have found $\psi(x)$ for positive $x$, you automatically know $\psi(x)$ for negative $x$ from the even or odd nature of $\psi(x)$.

 

[flyusx]

I just wanted to point out that you're solving the differential equation the usual way except instead choosing the two independent solutions $\cos k_2 x$ and $\sin k_2 x$, you're using the two solutions $\cos [k_2(x-a)]$ and $\sin [k_2(x-a)]$. The requirement that the wave function vanish at $x=a$ then makes you toss the cosine solution.

So if I look at the entire setup (instead of looking at half of it) and write $$\begin{cases}\frac{\text{d}^{2}\psi_{1}}{\text{d}x^{2}}+k^{2}\psi_{1}(x)=0&-a<x<-b\\\frac{\text{d}^{2}\psi_{2}}{\text{d}x^{2}}-\tilde{k}\psi_{2}(x)=0&-b\leq x\leq b\\\frac{\text{d}^{2}\psi_{3}}{\text{d}x^{2}}+k^{2}\psi_{3}(x)=0&b<x<a\end{cases}$$ where $$k=\frac{\sqrt{2mE}}{\hbar}$$ and $$\tilde{k}=\frac{\sqrt{2m(V_{0}-E)}}{\hbar}$$ (please forgive my redefining of the $k$'s with subscripts to be those without). Would a solution take the form $$\psi(x)=\begin{cases}\psi_{1}(x)=A\sin(k(x+a))+B\cos(k(x+a))\\\psi_{2}(x)=C\sinh(\tilde{k}x)+D\cosh(\tilde{k}x)\\\psi_{3}(x)=E\sin(k(x-a))+F\cos(k(x-a))\end{cases}$$
Where I would then impose parity and vanshing at $x=\pm a$?
$$\psi_{+}(x)=\begin{cases}\psi_{1+}(x)=A\sin(k(x+a))\\\psi_{2+}(x)=B\cosh(\tilde{k}x)\\\psi_{3+}(x)=-A\sin(k(x-a))\end{cases}$$
$$\psi_{-}(x)=\begin{cases}\psi_{1-}(x)=A\sin(k(x+a))\\\psi_{2-}(x)=B\sinh(\tilde{k}x)\\\psi_{3-}(x)=A\sin(k(x-a))\end{cases}$$
So at an arbitrary point $x=c$, then $\psi_{1+}(-c)=\psi_{3+}(c)$ and $\psi_{1-}(-c)=-\psi_{3-}(c)$?

 

[vela]

That looks fine. I'd avoid using $E$ as one of the arbitrary constants, though you did eventually get rid of it here.

 

[flyusx]

That looks fine. I'd avoid using $E$ as one of the arbitrary constants, though you did eventually get rid of it here.

Good catch, I wouldn't like to confuse it for energy.

 

[flyusx]

I was able to derive the energy eigenvalue equations $$\tan(k(a-b))\tanh(\tilde{k}b)=-\frac{k}{\tilde{k}}\text{, even wave function}$$ and $$\tan(k(a-b))\coth(\tilde{k}b)=-\frac{k}{\tilde{k}}\text{, odd wave function}$$ As TSny said, it doesn't seem like I can find numerical energies without these cases because I would have to consider the intersection of these equations with $k^{2}+\tilde{k}^{2}=\frac{2mV_{0}}{\hbar^{2}}$. Regardless, I appreciate the help you all have provided.

 

[TSny]

I was able to derive the energy eigenvalue equations $$\tan(k(a-b))\tanh(\tilde{k}b)=-\frac{k}{\tilde{k}}\text{, even wave function}$$ and $$\tan(k(a-b))\coth(\tilde{k}b)=-\frac{k}{\tilde{k}}\text{, odd wave function}$$

Looks good.