Why Do Boundary Conditions in a 1D Quantum Box Lead to Different Quantum States?

[itssilva]

Problem: The particle in a 1D box [0, a]

Eqs.: The general solution of the time-independent Schrödinger eq. may be written as ψ(x) = Acos(kx) + Bsin(kx), E = ħ²k²/2m. Imposing the boundary conditions ψ(0) = ψ(a) = 0 , we get immediately A = 0, ka = nπ (for any positive integer n). Using x' = x - a/2 , plus the boundary condition and elementary trigonometry, we also get ψ(x) = Bsin(kx) = Bcos(kx') , same eigenvalue. Everything is nice and dandy up to now, but, if I start from the box [-a/2, a/2] and impose boundary conditions ψ(-a/2) = ψ(a/2) = 0 , I seem to be getting B = 0, ka = nπ (but this time for odd n only)

What's the reason for this inconsistency? Either in the first case we should take only odd n also, and nobody noticed it in QM textbooks (unlikely), or I'm making some silly mistake, but I can't point my finger where; I've checked the sol. to the boundary condition system many times, it seems fine to me.

 

[Orodruin]

What's the reason for this inconsistency?

You are not doing the maths correctly.

 

[itssilva]

You are not doing the maths correctly.

I figured as much; though

ψ(-a/2) = Acos(-ka/2) + Bsin(-ka/2) = Acos(ka/2) - Bsin(ka/2) = 0
ψ(a/2) = Acos(ka/2) + Bsin(ka/2) = 0

=> B = 0 (because symmetry, Fourier series, however you prefer) , cos(ka/2) = 0 <=> ka/2 = (2m+1)π/2

Like I said, it's some silly mistake; but whatever it is, it's in the 3 lines above.

 

[vela]

I figured as much; though

I'm not sure what more of a response you could have gotten considering all you told us was that you got the wrong answer.

ψ(-a/2) = Acos(-ka/2) + Bsin(-ka/2) = Acos(ka/2) - Bsin(ka/2) = 0
ψ(a/2) = Acos(ka/2) + Bsin(ka/2) = 0

=> B = 0 (because symmetry, Fourier series, however you prefer) , cos(ka/2) = 0 <=> ka/2 = (2m+1)π/2

Like I said, it's some silly mistake; but whatever it is, it's in the 3 lines above.

Your mistake is in concluding B=0. If you recast the two equations in matrix notation, you have
$$\begin{bmatrix} \cos \frac{ka}{2} & -\sin \frac{ka}{2} \\ \cos \frac{ka}{2} & \sin \frac{ka}{2} \end{bmatrix}
\begin{bmatrix} A \\ B \end{bmatrix} = 0.$$ To get a non-trivial solution, the determinant of the matrix has to vanish. What does that give you?

 

[Orodruin]

because symmetry

This is wrong. If the potential is symmetric, eigenstates are symmetric or anti-symmetric.

 

[itssilva]

Your mistake is in concluding B=0. If you recast the two equations in matrix notation, you have
$$\begin{bmatrix} \cos \frac{ka}{2} & -\sin \frac{ka}{2} \\ \cos \frac{ka}{2} & \sin \frac{ka}{2} \end{bmatrix}
\begin{bmatrix} A \\ B \end{bmatrix} = 0.$$ To get a non-trivial solution, the determinant of the matrix has to vanish. What does that give you?

Oh, I see; indeed, that was naive of me. Sorry, guys; I'm being way over my head for stuff that's supposed to be simple.

 

[vela]

Let me note that writing the system in matrix form isn't the way most students would work the problem out. The two equations you got were
\begin{align*}
A \cos \frac{ka}{2} &= B \sin\frac{ka}{2} \\
A \cos \frac{ka}{2} &= -B \sin\frac{ka}{2}.
\end{align*} If you add the first equation to the second, you get $2A \cos \frac{ka}{2} = 0$. Now you have two cases to consider: A=0 or $\cos\frac{ka}{2}=0$.