Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

We know $\Psi(x,t)$ is complex, but can $\Psi(x)$ be complex? I have seen particle in a box, well and harmonic oscillator. All have real solutions for time-independent Schrödinger equation. Hence, I am curious to know examples where it is complex. This question says that it is possible, hence my request is for examples and references.

share|improve this question

5 Answers 5

up vote 4 down vote accepted

A charged particle in external magnetic field has the following Hamiltonian:

$$\hat H=\frac1{2m}\left(\hat{\vec p}-q\vec A\right)^2+qV,$$

where $\vec A$ is vector potential, $V$ is scalar potential and $\hat{\vec p}=-i\hbar\nabla$ is momentum operator.

If you set $\vec A\not=0$, you'll get non-trivially complex wavefunction, and there will be no degeneracy due to time inversion symmetry as is the case for usual running wave, because magnetic field breaks time-inversion symmetry.

share|improve this answer
    
Equation 6 in arxiv.org/ftp/arxiv/papers/0712/0712.4201.pdf gives the solution in 2-d polar cordinates. It is complex in the sense there is an $e^{il\theta}$ factor which is a complex sinusoid. –  Rajesh D Dec 26 '13 at 10:56
1  
Right, but this time it's not arbitrary. It's a phase which depends on spatial coordinate. You can't just simply multiply the solution by phase coefficient or take linear combination of some eigenstates to form a real wavefunction which would also be eigenfunction. Thus this complexity does have physical significance, and you can't avoid it. –  Ruslan Dec 26 '13 at 11:10

Rajesh: That means time independent Schrodinger equation does not demand complex numbers if one chooses to avoid them. right?

NOTE: This answer is sort of cheating due to the degeneracy, and thus doesn't really answer the OP's question. Hopefully someone with better knowledge can answer Rajesh's answer with real rigor.

Here's an example of a physical system which admits eigenstates which are nontrivially complex-valued. In molecular physics in the limit of the separability of vibrational motion from the other degrees of freedom, polyatomics with doubly-degenerate vibrational eigenbases have stationary states which can be represented via their occupation numbers in $\nu$ and $l$, the total vibrational quantum number and the vibrational angular momentum quantum number, respectively.

In polar coordinates with radius $\rho$ and angle $\theta$, the stationary states of $H$ are given via their expansion in terms of the generating function $$G=\pi^{-1/2}\rho^{1/2}\mbox{exp}\left[-\frac{1}{2}\rho^2+\frac{1}{\sqrt{2}}\rho e^{i\theta }su+\frac{1}{\sqrt{2}}\rho e^{-i\theta}s u^{-1}-\frac{1}{2}s^2\right]$$ $$=\sum_{\nu,l}\frac{\left<\rho,\theta|\nu,l\right>s^\nu u^l}{\left\{2^\nu[\frac{1}{2}(\nu+l)]![\frac{1}{2}(\nu-l)]!\right\}^{1/2}}$$ where $l\in\{-\nu,-\nu+2,...,\nu-2,\nu\}$.

For example, here is a complex plot of the function $|\nu=3,l=1\rangle$ state:

enter image description here

Here is a plot of the $|\nu=6,l=0\rangle$ state:

enter image description here

And here is a plot of the $|\nu=7,l=3\rangle$ state:

enter image description here

The plots are generated in a Hue colorspace, with the brightness proportional to amplitude, and color cyclically coded according to phase angle in the complex plane.

With the exception of zero vibrational angular momentum states, it is visually apparent that none of the wavefunctions can be represented as a real function multiplied by a phase factor $e^{i\alpha}$. Hopefully you find this sufficient as an example where a system has stationary states which are non-trivially complex-valued. I'm sure there are lots of other examples of cases where the time-independent Schrodinger equation has eigenbases which are nontrivially complex-valued, but this is the first example which came to mind.

share|improve this answer
    
Yeah, they are complex-valued... too bad they are also degenerate. You can choose such linear combinations of them (e.g. $\left|+m\right>+\left|-m\right>$ where $m$ is magnetic quantum number) which will become real and still remain eigenstates. Of course, they won't have some observables (like $L_z$) definite, but they will have others. –  Ruslan Dec 25 '13 at 20:02
1  
Not that it's worth changing here but I leave this comment any time I see pictures like this -- choose different color maps! If it has red and green in it, particularly close together, those of us who are color-blind have a really tough time with it. Spectrum, rainbow, etc are all poor choices for displaying most data, despite them being the default in most viz. software! Single color changes, saturation changes, and so on make better color maps for everybody, but especially those who are color-blind. Just some friendly advice :) –  tpg2114 Dec 25 '13 at 20:04
    
@Ruslan: Haha, yeah I figured somebody would point that out. Still though, that leaves the OP's question a bit unanswered, ie, whether non-degeneracy implies that the wavefunction can be represented as a real function times a constant factor. I'm curious if there's a theorem which supports or negates this. –  DumpsterDoofus Dec 25 '13 at 20:24

Before starting, let me be pedantic and define the phrase "real up to an overall phase." I will say a function $f(x)$ is real up to an overall phase if I can find a real number $\alpha$ such that $e^{i\alpha}f(x)$ is real everywhere. [Incidentally, note that if a wave function is real up to an overall phase, it is also imaginary up to an overall phase].

Now that we've established that terminology, I claim that the question of whether or not the wave function is real up to a phase must be considered in two cases:

  1. If the energy eigenfunction $\Psi(x)$ is associated with a non-degenerate energy eigenvalue will be real up to an overall phase.
  2. If the energy eigenfunction $\Psi(x)$ is living in the energy eigenspace associated with a degenerate energy eigenvalue will generically be complex. While there will always be special eigenfunctions living in this space that are real up to an overall phase, the general eigenfunction is complex and you cannot restrict yourself to the "real subspace" without losing the ability to describe some physical systems.

The examples you gave (particle in a box, harmonic oscillator) fall under case 1: there are no degenerate energy levels, so the eigenfunctions are real up to an overall phase. However I can easily give you physical examples that fall under case 2 (and I will later, and DumpsterDoofus and DarenW have already done so).

Before going into some gory detail, I should just say that whether or not the wave function is real up to an overall phase is sort of irrelevant, physically speaking. You should think of the wave function as being a complex valued quantity. For example, if the state is a momentum eigenfunction--which is the case in particle accelerators, which measure the momenta of particles very precisely--then the wave function is complex $e^{ipx}$ and is certainly not real up to an overall phase. In certain cases it happens to turn out that the wave function is real up to an overall phase, but this is frankly something of a mathematical accident and doesn't have any interesting physical consequences. You certainly should not make the leap from "there exist wave functions that are real up to an overall phase" to "all physical situations can be described by a wave function that is real up to an overall phase."


With that said, let's discuss case 1. We are looking for eigenfunctions $\psi(x)$ that obey the energy eigenvalue equation [working in the position representation] \begin{equation} \hat{H}\psi=E\psi \end{equation}

where $\hat{H}=-\frac{\hbar^2}{2m}\nabla^2+V(x)$ is the Hamiltonian operator.

Since the Hamiltonian must be a hermitian operator, we have that $(\hat{H}f)^*=\hat{H}f^*$ for any function $f(x)$. Meanwhile, $E=E^*$ since $\hat{H}$ is hermitian so it has real eigenvalues.

As a result, $\psi^*(x)$ is also an eigenfunction of $\hat{H}$ with eigenvalue $E$.

However, we are assuming that the energy eigenvalue $E$ is non-degenerate. In other words, there is only one eigenfunction with eigenvalue $E$, up to an overall factor. So it must be that there is some $\lambda$ such that

\begin{equation} \psi(x)=\lambda \psi^*(x) \end{equation}

Clearly this equation is only consistent if $|\lambda|=1$, so we may write $\lambda=e^{-2 i\alpha}$ for a real $\alpha$ without any loss of generality. But then $\psi(x)$ is real up to an overall phase, because $\phi(x)=e^{i\alpha}\psi(x)$ obeys $\phi(x)=\phi^*(x)$.

That establishes case 1.

Before going on to case 2 it is interesting to see why the particle in a box falls under this case.

The hamiltonian for a particle in a box is the same as the hamiltonian for a free particle \begin{equation} \hat{H}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}, \end{equation} with the boundary conditions \begin{equation} \psi(0)=\psi(L)=0. \end{equation} Since the hamiltonian is the same as for a free particle the eigenfunction must have the form \begin{equation} \psi(x)=Ae^{ikx}+Be^{-ikx} \end{equation} where $A$ and $B$ are complex coefficients. You can see that the general solution is not real up to an overall phase! This is because there are TWO solutions with energy $E$: one is $e^{ikx}$ and the other is $e^{-ikx}$. It is only after we impose the "particle in a box" boundary conditions that we find that the eigenfunction must have the form $\psi(x)=N\sin\left(\frac{\sqrt{2mE}}{\hbar}x\right)$. The boundary conditions essentially "project out" one of the two eigenfunctions with the energy level $E$. We are left with one linearly independent solution that can satisfy the boundary conditions, and by the general argument I gave above this eigenfunction has to be real. (And indeed, explicit calculation shows it is--it is the sine function). If you want me to go into more detail about this part, I would be happy to update the answer to be more precise.

The 1D harmonic oscillator is slightly different. There are two solutions for a given energy eigenvalue $E$. However only one will be normalizable. So for each energy level $E$ there is only 1 energy eigenfunction, and thus this eigenfunction is real up to an overall phase.


OK, case 2. It is easy to see that in general a member of a degenerate eigenspace won't be real. $\psi(x)$ and $\psi^*(x)$ are still both solutions since $\hat{H}$ is hermitian. However there is no reason that $\psi(x)=\lambda\psi^*(x)$.

Even more to the point, since the energy eigenspace is degenerate it means that there are at least two distinct, orthogonal eigenfunctions $\psi(x), \phi(x)$ that have the same energy $E$. Any linear combination of them will also have the energy $E$. So it's very easy to construct a complex energy eigenfunction living in this subspace, even if $\psi$ and $\phi$ happen to be real: I just take $a \psi+b\phi$ with complex $a,b$.

On the other hand it's also easy to see that there are always special eigenfunctions in the eigenspace that are real. For example, $\psi(x)+\psi^*(x)$ is real.

It's useful to give an example. A good example is actually the particle in a periodic box. This is just like the particle in a box in that the Hamiltonian is the same as for a free particle. However, instead of imposing the boundary conditions $\psi(0)=\psi(L)=0$ we will impose the periodic boundary conditions \begin{array} \ \psi\left( 0 \right)&=&\psi \left( L \right) \\ \psi'\left(0\right)&=&\psi'\left(L\right) \end{array} where $\psi'(x)=d\psi/dx$.

These boundary conditions are appropriate in many situations, for example:

  1. Electrons in a solid will experience periodic conditions because of the periodic nature of a crystal
  2. In extra dimensional models where the extra dimension is compact, the wave function has to be periodic
  3. If you work in polar coordinates in 2D or spherical coordinates in 3D, the angle $\theta$ will obey periodic boundary conditions

Starting with the general solution $\psi(x)=A e^{ikx}+Be^{-ikx}$, the boundary conditions amount to the single condition $e^{ikL}=1$, so $k=2\pi n/L$ (which quantizes the energy levels). Crucially, the boundary conditions do not impose any conditions on the coefficients $A$ and $B$. So it for any allowed energy level $E$, there are TWO allowed energy eigenfunctions. Thus in general, a state with energy $E$ is not real up to an overall phase.

There are two special states with energy $E$ that are real. They are \begin{equation} \psi_c(x)=\frac{1}{2}\left(e^{ikx}+e^{-ikx}\right)=\cos kx, \psi_s(x)=\frac{1}{2i}\left(e^{ikx}-e^{-ikx}\right)=\sin kx \end{equation} However there is no reason to think that nature will "prefer" these states over other possible states with the same energy with complex valued wave functions. If we set up a periodic particle in a box in the lab and fixed the energy to be $E$, in general we would get a state that wasn't $\psi_s$ or $\psi_c$.

As I said, there are many interesting examples that fall under case 2. Another one is the 2D harmonic oscillator, with $V(x,y)=m\omega^2(x^2+y^2)$. In general, as you go to dimensions greater than 1, the solvable problems tend to have degenerate eigenvalues. So even though you start off with examples that are of the form of case 1, you should expect to see many more examples in case 2 as you advance in quantum mechanics.

share|improve this answer
    
Thanks for the great answer. But point 2 being complex solution is still not a worry for me. –  Rajesh D Dec 26 '13 at 10:59
    
@Rajesh D Do you mind elaborating on what you mean? I don't mind expanding my answer. I think point 2 is very important. For example, all scattering problems fall under point 2. In scattering it is actually very important that the wave function is complex, and not just real up to a phase. A related statement is that it is very important to use the Feynman propagator, and not the usual retarded propagator, when doing scattering in quantum mechanics. –  Andrew Dec 26 '13 at 11:27
    
i fully agree that it has to be complex. I am working on ways to make $\psi(x)$ a real function all the time by tweaking the theory a bit. Crazy thing to do, but anyway its not relevant to mention here. sorry about that. –  Rajesh D Dec 26 '13 at 11:40
    
I see. It sounds like an interesting thing to try--even if you ultimately fail you will probably learn a lot. One thing you might find interesting if you haven't heard of it is the "wick rotation"--you use analytic continuation to deal with real quantities to do quantum calculations. –  Andrew Dec 26 '13 at 23:51

For example, you can start with any real solution and multiply it by a phase factor $\exp(i\alpha)$, where $\alpha$ is constant and real, but is not a multiple of $\pi$.

share|improve this answer
    
but that has no physical significance! –  Rajesh D Dec 25 '13 at 17:50
1  
@RajeshD Yes, and that means that having strictly real solutions also has no physical significance and is preferred only because most people are more comfortable in $\mathbb{R}$ than in $\mathbb{C}$. –  dmckee Dec 25 '13 at 17:59
    
@dmckee : That means time independent Schrodinger equation does not demand complex numbers if one chooses to avoid them. right? –  Rajesh D Dec 25 '13 at 18:01
1  
I'm still thinking about that. –  dmckee Dec 25 '13 at 18:03
    
@RajeshD : If you have some degeneracy for a energy $E_n$, you could have different and linearily independent real solutions $\psi_n^i$. Now, because of the linearity of QM, any complex linear combination of the $\psi_n^i$ is also a solution, for instance $\psi_n^1 + i \psi_n^2$ is a complex solution. –  Trimok Dec 25 '13 at 18:49

In one dimension, a running wave, as eq. (2) in that question. There isn't any more one can do in 1D that won't come out seeming contrived.

In two dimensions, a good example is a circular box, or any spherically symmetric potential. The wavefunction can be written as a radial factor times an angular factor. The angular factor may be a linear combination of $\sin(n\theta)$ and $\cos(n\theta)$ or, equivalently, of $\exp(i n \theta)$ and $\exp(-i n \theta)$. The latter are nicer to deal with in terms of eigenvalues and angular momenta. It's the same as in 3D, without 'z', so let's go there.

In three dimension, in atomic physics, the 'm' quantum number for orbitals. You can have an electron in, for example, some mix of $2p_x$, $2p_y$, and $2p_z$ orbitals, or you can use $2p_+$ $2p_-$ and $2p_z$, where $2p_\pm = 2p_x \pm i2p_y$. (likewise for n=3, 4, ... and similar combinations for the d, f ...) The $\pm$ orbitals are better when dealing with magnetic fields, spin-orbit coupling, conservation of angular momentum in scattering, and so on.

share|improve this answer
1  
I don't understand the later part. What is your answer? 2-d and 3-d demand complex $\Psi(x,y,z)$? –  Rajesh D Dec 25 '13 at 18:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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