Selection of an interval affects the energy eigenstates - impossible

[71GA]

I haven't found a comparison like this in any book that I have been reading so let me explain.

I decided that I will derive an equation for an energy levels of a particle in an infinite potential well in two ways. 1st I tried to derive it using the interval $0<x<d$ where $d$ is a width of a well. What I got was an equation which matches perfectly with any of the books out there:

\begin{align}
&\boxed{E=\frac{N^2\pi^2\hbar^2}{2md^2}}\longrightarrow N=1,2,3\dots\\
&E_1=\pi^2\hbar^2/2md^2\\
&E_2=2^2E_1\\
&E_3=3^2E_1\\
&\dots
\end{align}

So far so good. Now I decided to use the interval $-d/2<x<d/2$ and what I got was a different equation for energy levels. This is impossible because energies can't depend on our selection of an interval. I included my derivation:

\begin{align}
\frac{d^2\psi}{dx^2} &= - \left(\frac{2mE}{\hbar^2}\right)\psi \longleftarrow \substack{\text{Diferential eq. (DE) I get if I}\\\text{set $V=0$ in a Schrödinger. eq. }}\\
&\downarrow\\
\psi &= A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}}x\right)+B \cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}x\right)\longleftarrow \substack{\text{A general solution}\\\text{to the above DE}}
\end{align}

Let us not forget $\psi$ is in fact $\psi(x)$. At this point I used the border equations $\psi(-d/2)=0$ and $\psi(d/2)=0$ and what I got was a system of two equations:

\begin{align}
0 &= A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}} \left(-\tfrac{d}{2}\right)\right)+B \cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}\left(-\tfrac{d}{2}\right)\right)\\
\substack{\text{What I get if I insert $x=-\frac{d}{2}$}}\longrightarrow 0 &= -A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}} \tfrac{d}{2}\right)+B \cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}\tfrac{d}{2}\right)\\
\phantom{d}\\
\substack{\text{What I get if I insert $x=\frac{d}{2}$}}\longrightarrow 0 &=A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}} \tfrac{d}{2}\right)+B \cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}\tfrac{d}{2}\right)\\
\end{align}

I sum these two equations and I get:

\begin{align}
0&=2B\cos\left(\sqrt{\tfrac{2mE}{\hbar^2}}\tfrac{d}{2}\right)
\end{align}

And the possible solutions of this eq. are $B=0$ (this will help me find $\psi$) but if $B\neq 0$ than even $d\neq 0$ or else we would get $0=2B$. So if $d\neq 0$ than it must be true that:

\begin{align}
\sqrt{ \tfrac{2mE}{\hbar^2}} \tfrac{d}{2} &= (2N-1) \tfrac{\pi}{2} \longleftarrow N=1,2,3\dots\\
\sqrt{ \tfrac{2mE}{\hbar^2}} d &= (2N-1) \pi\\
&\!\!\!\!\!\!\boxed{E=\tfrac{(2N-1)^2\pi^2\hbar^2}{2md^2}}\\
E_1&=\pi^2\hbar^2/2md^2\\
E_2&=3^2E_1\\
E_3&=5^2E_1\\
&\dots
\end{align}

Now if I check two boxed equations I see that they are different! Even $N$ is defined the same for those two. Why did I get different energy equation by simply choosing a different interval (only the ground state $E_1$ is the same in both cases while $E_2, E_3, E_4\dots$ differ)? This means that my selection of an interval changes the particle? Please explain. I couldn't spot any mistake in my derivation. Can you?

 

[Bill_K]

Doing it this way (assuming B ≠ 0) you've gotten half the solutions. To get the other half, subtract one equation from the other and assume A ≠ 0.

 

[71GA]

Thank you. If i substract these two equations i get:

\begin{align}
0&=-2A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}}\tfrac{d}{2}\right)
\end{align}

I can solve this with $A=0$ or $d=0$ but the later means that width of the well is 0 and can't be true which means $A=0$ is an only possible solution. Is this assumption ok? If i continue in this maner i get better values for the wavefunctions which must be

\begin{align}
\psi = B\cos\left(\sqrt{\tfrac{2mE}{\hbar^2}} x\right)
\end{align}

If i try to sum up all this in my head now. By summing the equations i got a $B=0$ solution, which now must not be used because if it is, both $A=0$ and $B=0$ which means $\psi=0$ and the particle wasn't supposed to be in the well. But it is! So i am left with only 3 possible equations which are $A=0$ the $\psi = B\cos\left(\sqrt{\frac{2mE}{\hbar^2}} x\right)$ and the one for energies - i mean the second boxed equation.

What now? Am i ready to normalize the wavefunction and Find $B$?

PS: I can't forget the energy equation which is still not the same as for interval $0<x<d$.

 

[Bill_K]

Thank you. If i substract these two equations i get:

\begin{align}
0&=-2A\sin\left(\sqrt{\tfrac{2mE}{\hbar^2}}\tfrac{d}{2}\right)
\end{align}

I can solve this with $A=0$ or $d=0$ but the later means that width of the well is 0 and can't be true which means $A=0$ is an only possible solution. Is this assumption ok?

The solutions are √(2mE/ħ²) d/2 = Nπ.

 

[71GA]

The solutions are √(2mE/ħ²) d/2 = Nπ.

Oh right :) my solution was a bit trivial.

So if i use this i get another equation for energies which looks like:

$$\boxed{W=\frac{2N^2 \pi^2 \hbar^2}{md^2}}$$

I have now 2 equations for the energy levels on an interval $-d/2<x<d/2$ will the sum of those two return the same energy equation as i got for interval $0<x<d$?

 

[jtbell]

[By the way, it would make things easier if each equation were in a separate block of LaTeX, enclosed in its own pair of tags. Then people could easily quote one equation at a time to make a point. I can't figure out how to do that, with your huge blocks of LaTeX.]

Here's another approach. Consider the general solution to the DE, with the coefficients A and B. Apply one of the boundary conditions, say psi(+d/2) = 0. The resulting equation (which you have a little further down the page) can be true if either

Case 1: A = 0 and cos(...) = 0. This gives you solutions of the form psi = B cos(kx), with the values of k (and E) determined by the condition cos(...) = 0.

or

Case 2: B = 0 and sin(...) = 0. This gives you solutions of the form psi = A cos (kx), with the values of k (and E) determined by the condition sin(...) = 0.

If you now start over again with the the general solution, and apply the other boundary condition (psi(-d/2) = 0), you get the same final result. The two boundary conditions are redundant because of the symmetry of the potential function.

Taken together, the allowed values of E for cases 1 and 2 are the same ones as for the other method (interval from 0 to d). The psi functions of course have somewhat different forms because of the shifted origin.

 

[jtbell]

There are also the following cases which don't enter into the final solution:

Case 3: A = 0 and B = 0, which is not useful, for obvious reasons!

Case 4: cos(...) = 0 and sin(...) = 0, which is impossible.

 

[71GA]

I don't think you understood what i wanted to ask. I derived these two equations for an interval $-d/2>x>d/2$:

$$W=\frac{2N^2 \pi^2 \hbar^2}{md^2}$$
$$W=\frac{(2N-1)^2 \pi^2 \hbar^2}{2md^2}$$

and this one using interval $0>x>d$:

$$W=\frac{N^2 \pi^2 \hbar^2}{2md^2}$$

Energy levels should be equal but i still can't see the equality. What should i do with the first two to see that they are in fact the same as the later one?

 

[Bill_K]

Write the first line as W = (2N)²π²ħ²/2md². Then in the third line, N is either odd or even. If it's odd, that's the second line. If it's even, that's the first line.

 

[jtbell]

Re-write the first equation as Bill indicated. Then write out a list of the values of 2N², and of (2N-1)², for N = 1, 2, 3...

See how they "fit together" with each other, and with the third equation?

 

[71GA]

Thanks to both of you. I understand now.

 

[71GA]

One more question. If i now want to derive wavefunctions for an interval $-d/2 >x>d/2$ i am left with two options. Either i take $\psi = A\sin \left(\sqrt{\tfrac{2mW}{\hbar^2}} x\right)$, insert $W=\frac{(2N-1)^2\pi^2\hbar^2}{2md^2}$ and then normalise it OR i take $\psi = B\cos \left(\sqrt{\tfrac{2mW}{\hbar^2}} x\right)$, and insert $W=\frac{(2N)^2\pi^2\hbar^2}{2md^2}$ and then normalise it.

I believe that in both cases i should get the same wavefunction? Please correct me if i am wrong.