Infinite square well – minimum energy

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.3.

We’ve seen that the energy of a system must always be greater than the minimum of the potential function. As a specific example of this we can look at the Schrödinger equation for the square well, between {x=0} and {x=a}:

\displaystyle  \frac{d^{2}\psi}{dx^{2}}=-\frac{2m}{\hbar^{2}}E\psi \ \ \ \ \ (1)

If {E=0}, {\psi"=0}. Integrating gives {\psi=Ax+B.} Attempting to satisfy the boundary conditions, we get {\psi(0)=0} giving {B=0}. Then the condition {\psi(a)=0} gives {A=0}, thus {\psi(x)=0} and cannot be normalized.

If {E<0}, we solve the equation

\displaystyle  \psi"=-\frac{2mE}{\hbar^{2}}\psi\equiv k^{2}\psi \ \ \ \ \ (2)

with {k=\sqrt{-2mE/\hbar^{2}}}. Since {E<0}, {k} is real. The general solution is {\psi(x)=Ae^{kx}+Be^{-kx}}. Applying boundary conditions, we get {\psi(0)=0=A+B}, so {A=-B}. At {x=a}, we have

\displaystyle   \psi(a) \displaystyle  = \displaystyle  0\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  Ae^{ka}+Be^{-ka}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  AC+\frac{B}{C}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  A\left(C-\frac{1}{C}\right) \ \ \ \ \ (6)

where {C\equiv e^{ka}>0}. Since {C} is strictly positive (being an exponential) the only way we can get {C-1/C=0} is for {C=1}, implying {ka=0}. However, neither {k} nor {a} is zero here, so {C\neq1}, so {A=0=B} and {\psi(x)=0} again. Thus if {E\leq0}, the wave function cannot be normalized and satisfy the boundary conditions.

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