Picking the different solutions to the time independent Schrodinger eqaution

Question

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?

Answer 1

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 $p$.

Answer 2

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 $m=-1,0,1$.

Answer 3

... 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.)

Answer 4

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.