# Particle-in-a-box eigenstates

• itssilva

#### 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.

What's the reason for this inconsistency?
You are not doing the maths correctly.

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.

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?

• itssilva
because symmetry
This is wrong. If the potential is symmetric, eigenstates are symmetric or anti-symmetric.

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.

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##.