I am stuck on a QM homework problem. The setup is this:

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(To be clear, the potential in the left and rightmost regions is $0$ while the potential in the center region is $V_0$, and the wavefunction vanishes when $|x|>b+a/2$.) I'm asked to write the Schrödinger equation for each region, find its solution, set up the BCs, and obtain the transcendental equations for the eigenvalues.

Where I'm at: I understand the infinite potential well easily and I have done a free particle going over a finite barrier before (which I understood less well, but I can deal with it it).

  • The problem asks me to make use of "a symmetry" in the problem, which is a vague hint. Are they trying to get me to make $\psi$ an even function?

  • I am supposed the condition for there to be one and only one bound state for $E<V_0$. How do I go about that?


3 Answers 3


You seem to have trouble to understand the basic approach. Actually there is a systematic way to solve the Schrödinger equation for picewise constant potentials. Maybe this will give you some basic idea how to solve your problem:

Let be the potential given by $$V(z) = \begin{cases} \infty & z < z_1 \\ V_1 & z_1 <= z < z_2 \\ V_2 & z_2 <= z < z_3 \\ ... \end{cases}$$

  • For the above potential the wavefunction for energy eigenvalue $E_n$ is given by $$\Psi_n(z) = \begin{cases} 0 & z < z_1 \\ A_1\exp(-i k_1 z) + B_1\exp(+i k_1 z) & z_1 <= z < z_2 \\ A_2\exp(-i k_2 z) + B_2\exp(+i k_2 z) & z_2 <= z < z_3 \\ ... \end{cases}$$ with $k_i = 2\pi/h \sqrt{2 m e (E_n-V_i)}$ and some (yet to be determined) constants $A_i$ and $B_i$. This is easily verified by plugging in. (In fact each "segment" is the solution to the Schrödinger equation with constant potential). Note that the $k_i$ can be real or imaginary, in which case the wavefunction in the respective segment is either sinusoidal or exponential.

  • As required by physics the wavefunction must be continuous and continuously differentiable everywhere. Hence the constants $A_i$ and $B_i$ must be chosen so that this is fulfilled at each point where this possibly is violated (i.e. the points $z_i$).

  • The above results in a linear equation system for the $A_i$ and $B_i$. This equation system now only contains the energy $E_n$ as remaining unknown. If you do it correctly the equation system contains as many unknowns as equations.
  • Now you compute the determinant of the equation system and set it to zero to find the $E_n$ values for which it is solvable. This is the transcendetal equation for the eigenvalues. This equation has in your case infinitely many discrete solutions $E_n$ (each solution denoted by the running index $n$). For each $E_n$ there are sets of $A_i$ and $B_i$ (which solve the equation system) which give you the wavefunction. In case there is more than one set of linearly independent $A_i$ and $B_i$, you have more than one wavefunction to the same eigenvalue $E_n$. In that case the state is degenerate. (You have degenerate states in your problem!).

Regarding symmetry: The wavefunctions do not need to have the same symmetry as the potential. Of course if you have a solution wavefunction, then the mirrored wavefunction must be a solution as well (if the potential is symmetric as in your case). It needs to belong to the same energy eigenvalue.

Regarding the single bound state: Once you have calculated the $E_n$ you will see that there are conditions where $E_1 < V_0$ and $E_2 > V_0$ ($E_2$ the second largest eigenvalue). This depends on the geometry, i.e. width of your barrier and well. Generally speaking the energy states have higher spacing, if the well is smaller. So probably the single bound state condition will display itself as range specification for $a$ and $b$.


The parity operator commutes with the Hamiltonian because of the symmetry in your potential. This says that all eigenstates of the Hamiltonian are eigenstates of the parity operator. Therefore, the only possible eigenstate solutions to the system are ones with even or odd parity. This fact will allow you to simplify the process of applying the boundary conditions mentioned by Andreas, as you can immediately conclude several things regarding the unknown coefficients.


This problem is somewhat similar to the ammonia inversion.

In that problem the probability densities $\psi_n(x)^2$ are similar for $n =1$ and $2$, for $n = 3$ and $4$ etc. As a result the Hamiltonian levels $E_1$ and $E_2$ are close together, as are $E_3$ and $E_4$, etcetera.

See for example here and here.

See also Energy levels for $NH_3$.


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