# Is a non-degenerate wavefunction real or complex?

In this video it is stated that:

It can easily be verified that the wavefunction of a non-degenerate quantum mechanical system will be real.

However the presenter does not explain why this statement is true. How can we prove this? Does the professor assume a real Hamiltonian, i.e. one that includes only kinetic energy and Coulomb interaction terms?

• Possible duplicate: physics.stackexchange.com/q/77894/2451 and links therein. – Qmechanic Nov 3 '15 at 11:52
• A complex ground state $\psi$ can be written as $\psi_r + i\psi_i$, where $\psi_r$ and $\psi_i$ are real. Since $H\psi - E\psi = 0$ substituting for $\psi$ gives us the two equations $H\psi_r - E\psi_r = 0$ and $H\psi_i - E\psi_i = 0$. So $\psi_r$ and $\psi_i$ are both eigenfunctions of $H$ with the same energy $E$ as $\psi$, and therefore the ground state must be degenerate, which contradicts our initial assumption. – John Rennie Nov 4 '15 at 10:11
• It might be better to say that the wavefunction "can be chosen to be real". For instance in John Rennie's example you could also have $\psi_r = \psi_i$, in which case the state is not degenerate, but instead we have $\psi = (1+i)\psi_r$. You can choose to drop the overall phase factor. – Tim Goodman Nov 5 '15 at 17:48
• @JohnRennie 1) That should be an answer, as I'm sure you are already aware. 2) Does one need to additionally stipulate time-reversal invariance for your argument to hold? – Mark Mitchison Nov 5 '15 at 19:46
• @MarkMitchison: I answered in a comment because the question had been closed. Now it's been reopened I'll convert my comment to an answer. – John Rennie Nov 5 '15 at 20:43

We prove this by a reductio ad absurdum. We start by assuming that the wavefunction of a non-degenerate ground state is complex, then show this means the wavefunction must be degenerate.

Suppose we have a complex ground state. Then we can write it as a sum of real and imagniary parts:

$$\psi = \psi_r + i\psi_i \tag{1}$$

The ground state obeys Schrodinger's equation:

$$H\psi - E\psi = 0$$

and if we use equation (1) to substitute for $\psi$ we get:

$$H\psi_r + iH\psi_i - E\psi_r - Ei\psi_i = 0$$

For a complex number to be zero both its real and imaginary parts must be zero, so we get the two equations:

\begin{align} H\psi_r - E\psi_r &= 0 \\ H\psi_i - E\psi_i &= 0 \end{align}

But this means that $\psi_r$ and $\psi_i$ are also eigenstates with the same energy $E$ as $\psi$. That means $\psi$ is degenerate, and that contradicts our initial assumption.

• How do you define "complex", "real" and "imaginary" part here? If someone hands you a generic Hilbert space, there is no unique notion of "real" and "imaginary" on it. Also, you need to handle the case $\psi_r = \psi_i$. – ACuriousMind Nov 5 '15 at 20:50
• Indeed, the statement is valid for Hamiltonians of the form $-\frac{\hbar^2}{2m}\Delta + V(x)$ with $V$ real and referring to the Hilbert space $L^2(\mathbb R^n,dx)$, using the usual notion of real and complex valued function...It can be extended to more general cases when there is an antilinear operator $C: \cal H \to \cal H$, surjective, isometric such that $CC=I$, commuting with the Hamiltonian operator... – Valter Moretti Nov 5 '15 at 21:43
• @ValterMoretti Are there nontrivial examples of such $C$s which do not reduce to complex conjugation in a particular basis? It's a nice formalization (and it covers cases such as $H=\tfrac12 \hat p^2 +\hat p+\tfrac12\hat x^2$) but it feels like it could be little more than just a formalization. Or maybe there are more interesting examples out there, and if there are then I'm really curious. – Emilio Pisanty Dec 13 '15 at 19:29
• @Emilio Pisanty Actually I do not know. Let me think about... – Valter Moretti Dec 14 '15 at 8:06
• @Valter Fair enough. I'm glad I piqued your interest, though =). – Emilio Pisanty Dec 14 '15 at 8:10