Eigenfunctions of $C\psi(x) = \psi^*(x)$

Question

I am considering the complex conjugation operator $C\psi(x) = \psi^*(x)$.

As $C^2$ is the identity operator, I've deduced that the possible eigenvalues of the $C$ are $\pm 1.$ To determine the eigenfunctions of $C$, I've observed that, if $\lambda = 1$, then an eigenfunction $\psi_{1}$ satisfying $C\psi_{1}(x) = \psi_{1}(x)$ must be purely real. Likewise, if $\lambda = -1$, the eigenfunction $\psi_{-1}$ satisfying $C\psi_{-1}(x) = -\psi_{-1}(x)$ must be purely imaginary.

Am I right to deduce that the eigenfunctions of the complex conjugation operator therefore consist of the set of purely real and imaginary wavefunctions? I feel that something is quite off with this deduction. If it is correct, however, how would I begin to prove that such a set should or should not be complete and/or orthogonal?

Answer 1

Let us denote by $C$ the standard complex conjugation of functions on a Hilbert space $H:=L^2(X)$.

It is already false that the eigenvalues of $C$ are $\pm 1$.

First of all, from $C\psi = \lambda \psi$ you have $CC\psi = \overline{\lambda} C\psi = |\lambda|^2 \psi$. Since $CC=I$ and $\psi\neq 0$, we conclude that $|\lambda|=1$ and not $\lambda = \pm 1$.

Furthermore, suppose that $\psi \neq 0$ is a real ($L^2$) function and consider $\phi_\alpha:= e^{i\alpha} \psi$ for a given real $\alpha$. In this case $$C\phi_\alpha = e^{-i\alpha} C\psi = e^{-i\alpha} \psi = e^{-2i\alpha} e^{i\alpha}\psi \:.$$ In summary, $$C\phi_\alpha = e^{-2i\alpha} \phi_\alpha\:.$$ So that there are also eigenvectors with arbitrary unit complex eigenvalue and not only $\pm 1$.

Notice also that the complex combination of eigenvectors with a given eigenvalue are not eigenvectors (unless the coefficients of the combination are real) so that the notion of eigenspace is more delicate (it makes sense only referring to real combinations).

It is also false that eigenvectors with different eigenvalues are orthogonal as a trivial consequence of the construction above: $\phi_\alpha$ and $\phi_{\alpha'}$ are not orthogonal if $\alpha\neq \alpha'$, but the eigenvalues are different if $\alpha-\alpha' \not \in 2\pi \mathbb{Z}$.

This failure prevents us from obtaining a complete orthogonal decomposition of vectors in terms of all the eigenvectors of $C$. However, it is possible to have complete expansions in terms of some eigenvectors if the Hilbert space is separable. As a matter of fact there is a Hilbert basis of eigenvectors with eigenvalue $1$.

Let $N:=\{u_n\}_{n\in \mathbb{N}}$ be a Hilbert basis (countable because the Hilbert spece is assumed to be separable).

Then we can always extract another Hilbert basis made of real functions, thus eigenvectors of $C$ with eigenvalue $1$.

To this end, decompose every $u_n$ into real and imaginary part and form a new set of vectors $M:=\{\mathfrak{R}\left(u_n\right), \mathfrak{I}\left(u_n\right)\}_{n\in \mathbb N}$ (some of these vectors may be $0$!).

The $L^2$ scalar products of pairs of vectors in $M$ are real by construction.

Therefore the standard orthogonalisation procedure of Gram-Schmidt yields a new orthonormal set of real vectors $N'$.

Since the finite complex linear combinations of $N$ are the same as those of $N'$, we conclude that the complex span of $N'$ is dense in $H$. In other words $N'$ is a Hilbert basis of $H$, made of eigenvectors of $C$ with eigenvalue $1$. (But it does not mean that $H$ is the eigenspace of $C$ with eigenvalue $1$ as it would happen for bounded linear operators, avoinding the contradiction $C=I$!)

All the discussion can be extended to generic Hilbert spaces $H$ for antilinear operators $C:H\to H$ satisfying $CC=I$.

Answer 2

A simple way to see that this is the case is to use the fact that complex numbers are 2-d vectors. We simply write that \begin{align} \psi(x) &\rightarrow \left[\begin{array}{c} \psi_R(x) \\ \psi_I(x)\end{array}\right]. \end{align} In that representation the complex conjugate is just the matrix \begin{align} C &\rightarrow \left[\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right]. \end{align}

Since $C$ is already diagonal in this representation you can see that its eigenvalues are $\{1,-1\}$, and the eigenvectors are the real and imaginary components of the wave function.

So, why does this fail? Well, the above representation doesn't include the full richness of complex numbers. This is especially important because eigenvalues can be complex numbers. There is a matrix representation of complex numbers, but then the complex conjugation operation becomes a similarity transform instead of just a left multiplication.

With that in mind, let's dive in \begin{align} \psi(x) &\rightarrow \left[\begin{array}{cc} \psi_R(x) & \psi_I(x) \\ -\psi_I(x) & \psi_R(x) \end{array}\right] \\ C\psi(x)& \rightarrow c \left[\begin{array}{cc} \psi_R(x) & \psi_I(x) \\ -\psi_I(x) & \psi_R(x) \end{array}\right] c^{-1} \\ &= \left[\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right] \left[\begin{array}{cc} \psi_R(x) & \psi_I(x) \\ -\psi_I(x) & \psi_R(x) \end{array}\right] \left[\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right] \\ &= \left[\begin{array}{cc} \psi_R(x) & -\psi_I(x) \\ \psi_I(x) & \psi_R(x) \end{array}\right]\\ \lambda \psi(x) &\rightarrow \left[\begin{array}{cc} \lambda_R & \lambda_I \\ -\lambda_R & \lambda_R \end{array}\right]\left[\begin{array}{cc} \psi_R(x) & \psi_I(x) \\ -\psi_I(x) & \psi_R(x) \end{array}\right] \\ &= \left[\begin{array}{cc} \lambda_R\psi_R(x) - \lambda_I\psi_I(x) & \lambda_R\psi_I(x) + \lambda_I\psi_R(x) \\ -\lambda_R\psi_I(x) - \lambda_I\psi_R(x) & \lambda_R\psi_R(x) - \lambda_I\psi_I(x) \end{array}\right] \end{align} We can thus get two equations for the real and imaginary parts of our eigenvalues: \begin{align} (\lambda_R - 1)\psi_R(x) &= \lambda_I\psi_I(x) \text{ and}\\ (\lambda_R + 1)\psi_I(x) &= -\lambda_I\psi_R(x). \end{align} That's two equations and four unknowns. Even accounting for the usual ability to ignore the scale of the eigenvector, that still leaves us needing one more equation to fix the eigenvalues. In fact, combining the above equations to reduce it to one equation for the eigenvalues reduces it to $\lambda_R^2 + \lambda_I^2 = 1$.

So, let's say that $\lambda=e^{i\theta}$ then we get that \begin{align} \frac{\psi_R(x)}{\psi_I(x)} &= \frac{\sin\theta}{\cos\theta - 1} \text{ or} \\ \frac{\psi_I(x)}{\psi_R(x)} &= \frac{\cos\theta - 1}{\sin\theta}. \end{align} As a check, if $\psi_R(x) = 0$ then $\theta$ is $0$ or $\pi$, and the $1-\cos\theta$ factor requires $\theta=\pi$. If $\psi_I=0$ then $\theta=0$, consistent with the first attempt. These functions have a range of $(-\infty,\infty)$, so any ratio of real and imaginary parts for the functions can be accommodated, but that ratio cannot depend on $x$.

Thus, every complex function of the form $\psi(x) = R(x)e^{i\theta/2}$ is an eigenfunction with eigenvalue $e^{-i\theta}$, as can be verified by inspection.