I have a potential which is infinite except between 0 and 2. Between 0 and 1 we have $V=0$, and between 1 and 2 we have $V=1000$.

What’s the ground state energy $E$ of this well? I claim it’s at least 1000.

We can’t have $E$ less than 0, because that’s the lowest value of potential anywhere. (This theorem is Problem 2.2 in Griffiths; it’s true because if $E<V_{min}$ then $\psi$ and its second derivative have the same sign everywhere, which means that the wavefunction can’t be normalized.)

Now let’s consider the possibility of $0<E<1000$. The wavefunction will be oscillating on the left and exponentially decaying on the right, because when potential is constant, the solution to the Schrodinger equation is sinusoidal where $E>V$ and exponential where $E<V$ (eg see these notes).

The right section of the wavefunction can’t just be zero, because the wavefunction has to be twice continuous and a sine wave never has zero value and zero derivative at the same time. So there has to be non-trivial exponential decay on the right. This violates the boundary conditions, because exponential decay can never get to zero.

Therefore, the energy of the wavefunction has to be greater than 1000. This is greater than the ground state energy of the normal infinite square well.

Surely the above argument is wrong--it results in an energy that is higher than the infinite square well of width 1, and I normally wouldn't think that your energy can get arbitrarily large as a result of the potential getting large in just one place. Where is the mistake?


1 Answer 1


I'm not sure I completely follow your argument but for $E<1000$ the wave function on the right is not a single exponential but rather a combination of the form $$ \psi_R(x)\sim A \sinh(\kappa(x-2)) $$ so that it can be made to satisfy the condition $\psi_R(2)=0$.


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