Trjtdtk

The time independent Schrodinger equation
$$-frachbar^22m fracd^2psidx^2+Vpsi = Epsi$$ 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,frachbarifracddx)psi. $ How do we know for sure that $int a(x)^*Q(x,frachbarifracddx)a(x) $ and $int b(x)^*Q(x,frachbarifracddx)b(x)$ gives the same expectation values?

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 $cpsi$ represent the same state for any $cinmathbbC$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrme^mathrmik$ for some $kin[0,2pi)$.



2. 
   

   If $psi(x) = a(x) + mathrmib(x)$ 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 $psi(x) = mathrme^mathrmipx$ and $psi^ast(x) = mathrme^-mathrmipx$ 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 $P=-ihbarpartial/partial x$.

(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 $x$:
$$psi_1(x),psi_2'(x) - psi_1'(x),psi_2(x) = 0$$
$$psi_1'(x) over psi_1(x) = psi_2'(x) over psi_2(x)$$
$$logpsi_1(x) = logpsi_2(x) + c$$
$$psi_1(x) = psi_2(x),e^c$$
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.