Kdjykuj

The time independent Schrodinger equation
$$-frac{hbar^2}{2m} frac{d^2psi}{dx^2}+Vpsi = Epsi$$ 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,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?

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.)

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 $cinmathbb{C}$. 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 $kin[0,2pi)$.



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 $p$.