# Picking the different solutions to the time independent Schrodinger eqaution

The time independent Schrodinger equation $$-\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2}+V\psi = E\psi$$ can have many different solutions of $$\psi$$ for a particular value of $$E$$.

For example, if we found a complex solution $$\psi(x)$$ for a particular value of $$E$$, say $$E_0$$, we can write $$\psi(x)=a(x)+ib(x)$$. Then $$a(x)$$ and $$b(x)$$ will also be solutions to the T.I.S.E with $$E=E_0$$. Furthermore $$c_1a(x)+c_2b(x)$$ will also be solutions with $$c_1$$ and $$c_2$$ being arbitrary constants.

I read that one can always choose any of these solutions as the solution for the stationary state with energy $$E_0$$. But does that mean all these different solutions represent the same physical state of a particle?

The expectation value of any dynamical variable $$Q(x,p)$$ is given by $$\int \psi^*Q(x,\frac{\hbar}{i}\frac{d}{dx})\psi.$$ How do we know for sure that $$\int a(x)^*Q(x,\frac{\hbar}{i}\frac{d}{dx})a(x)$$ and $$\int b(x)^*Q(x,\frac{\hbar}{i}\frac{d}{dx})b(x)$$ gives the same expectation values?

• The last two integrals don't seem to have an infinitesimal? – Gert Mar 23 '19 at 0:18

1. In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $$\psi$$ and $$c\psi$$ represent the same state for any $$c\in\mathbb{C}$$. If you insist on wavefunctions being normalized, then $$c$$ is restricted to complex numbers of absolute value 1, i.e. number of the form $$\mathrm{e}^{\mathrm{i}k}$$ for some $$k\in[0,2\pi)$$.

2. If $$\psi(x) = a(x) + \mathrm{i}b(x)$$ is a solution of the Schrödinger equation, then it is not automatically true that $$a(x)$$ and $$b(x)$$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $$\psi^\ast(x)$$ is a solution if $$\psi(x)$$ is, and $$a(x)$$ and $$b(x)$$ can be obtained by linear combinations of $$\psi(x)$$ and $$\psi^\ast(x)$$.

There are now two cases: If $$\psi$$ and $$\psi^\ast$$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.

Note that already the free particle with $$V=0$$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $$\psi(x) = \mathrm{e}^{\mathrm{i}px}$$ and $$\psi^\ast(x) = \mathrm{e}^{-\mathrm{i}px}$$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $$p$$.

Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $$z$$ component.

For example the hydrogenic wave functions with $$n=2$$: $$2s$$ and $$2p$$. And then there are three different $$2p$$ wave functions that can be written as linear combinations of $$2p_x,\ 2p_y,\ 2p_z$$ or of the spherical harmonics $$Y_{\ell,m}$$ with $$\ell=1$$ and $$m=-1,0,1$$.

... one can always choose any of these solutions as the solution for the stationary state with energy $$E_0$$.

That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.

A given observable will typically have different expectation values in two different states with the same energy. For example, when $$V=0$$, the functions $$\exp(\pm ipx/\hbar)$$ both have energy $$E=p^2/2m$$, but they have different eigenvalues ($$\pm p$$) of the momentum operator $$P=-i\hbar\partial/\partial x$$.

(I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)

For a onedimensional equation like the one you propose something more certain may be said.

Schrödinger equation is a second-order, linear and homogeneous ODE. Then a unique solution is determined giving $$\psi(x_0)$$ and $$\psi'(x_0)$$. As a consequence, there can't be more than two independent solutions (for a given $$E$$).

The homogeneous boundary value problem ($$\psi(a)=\psi(b)=0$$ for given $$a$$, $$b$$, even infinite) is still more restricted: for each eigenvalue there is only one independent solution. In other words, eigenvalues are never degenerate.

The proof makes use of the Wronskian. Let $$\psi_1$$, $$\psi_2$$ two solutions (for the same $$E$$ and the same boundary conditions). Define $$W(x) = \psi_1(x)\,\psi_2'(x) - \psi_1'(x)\,\psi_2(x).$$ It can be shown that $$W(x)$$ is a constant, the same for all $$x$$. If the boundary conditions are homogeneous, clearly $$W(x)=0$$ for all $$x$$: $$\psi_1(x)\,\psi_2'(x) - \psi_1'(x)\,\psi_2(x) = 0$$ $${\psi_1'(x) \over \psi_1(x)} = {\psi_2'(x) \over \psi_2(x)}$$ $$\log\psi_1(x) = \log\psi_2(x) + c$$ $$\psi_1(x) = \psi_2(x)\,e^c$$ q.e.d.

The case of a free particle, with two solutions, is no counterexample, since in that case there are no homogeneous boundary conditions. Anyhow you get two independent solutions and no more.