2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ The last two integrals don't seem to have an infinitesimal? $\endgroup$
    – Gert
    Mar 23, 2019 at 0:18

4 Answers 4

6
$\begingroup$
  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 $p$.

$\endgroup$
3
$\begingroup$

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 $m=-1,0,1$.

$\endgroup$
3
$\begingroup$

... 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.)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.