# Solving the time independent Schrodinger equation: Does a complex solution make sense?

In my notes, I have the Time Independent Schrodinger equation for a free particle $$\frac{\partial^2 \psi}{\partial x^2}+\frac{p^2}{\hbar^2}\psi=0\tag1$$

The solution to this is given, in my notes, as $$\Large \psi(x)=C e^\left(\frac{ipx}{\hbar}\right)\tag2$$

Now, since (1) is a second order homogeneous equation with constant coefficients, given the coefficients we have, we get a pair of complex roots:$$r_{1,2}=\pm \frac{ip}{\hbar}\tag3$$

Thus, the most general solution looks something like:$$\psi(x)=c_1 \cos \left(\frac{px}{\hbar}\right)+c_2 \sin \left(\frac{px}{\hbar}\right)\tag4$$

However, instead of writing the solution as a cosine plus a sin, the professor seems to have taken a special case of the general solution (with $c_1=1$ and $c_2=i$) and converted the resulting $$\psi(x)=\cos \left(\frac{px}{\hbar}\right)+ i\sin \left(\frac{px}{\hbar}\right)\tag5$$ into exponential form, using $$e^{i\theta}=\cos \theta + i\sin \theta \tag6$$ to get (2).

The main question I have concerning this is: shouldn't we be going after real solutions, and ignoring the complex ones for this particular situation? According to my understanding $\Psi(x,t)$ is complex but $\psi(x)$ should be real. Thanks in advance.

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The wavefunction needn't and shouldn't be real. – Chris Feb 8 at 10:38
There are cases where you can get away with a real wavefunction, but the complex case is more general and fundamental. The free particle Hamiltonian $\hat{H}$ commutes with reflection $x\rightarrow -x$,$p\rightarrow -p$, so states with momenta $\pm p$ are both solutions. In equation (2) they have chosen the solution which is an eigenvalue of the momentum operator $\hat{p}$ with a plus sign $+$. The other sign is also a solution, representing a wave going in the opposite direction. Your real solution contains both left moving and right moving waves. – Michael Brown Feb 8 at 11:05
If you look at the particle current $\vec{j}\propto \psi^\star \nabla \psi - \psi \nabla \psi^\star$ you'll see that real wavefunctions correspond to states where there is no net current, so you can only really expect them to turn up when you have bound states. If there is nothing to reflect a particle back the way it came then it is free to move off to infinity and the current can't vanish, so the wavefunction can't be real. – Michael Brown Feb 8 at 11:10
Related: The book of Griffiths, Intro to QM, Problem 2.1b, p.24; and this Phys.SE post. – Qmechanic♦ Feb 8 at 15:54

## 1 Answer

There is no need for the solution $\psi(x)$ to be real. What must be real is the probability density that is "carried" by $\psi(x)$. In some loose and imprecise intuitive way, you may think about a TV image carried by electromagnetic waves. The signal that travels is not itself the image, but it carries it, and you can recover the image by decoding the signal properly.

Somewhat similarly, the complex wave function that is found by solving Schrödinger equation carries the information of "where the particle is likely to be", but in an indirect manner. The information on the probability density $P(x)$ of finding the particle is recovered from $\psi(x)$ simply by multiplying it times its complex conjugate:

$$\psi(x)^*\psi(x) = P(x)$$

that gives a real function as a result. Note that it is a density: what you compute eventually is the probability of finding the particle between $x=a$ and $x=b$ as $\int_{a}^{b} P(x) dx$

As you know, when you multiply a complex number(/function) times its complex conjugate, the information on the phase is lost:

$$\rho e^{i \theta}\rho e^{-i \theta}=\rho^{2}$$

For that reason, in some places one can (not quite correctly) read that the phase has no physical meaning (see footnote), and then you may wonder "if I eventually get real numbers, why did not they invent a theory that directly handles real functions?".

The answer is that, among other reasons, complex wave functions make life interesting because, since the Schrödinger equation is linear, the superposition principle holds for its solutions. Wave functions add, and it is in that addition where the relative phases play the most important role.

The archetypical case happens in the double slit experiment. If $\psi_{1}$ and $\psi_{2}$ are the wave functions that represent the particle coming from the hole number $1$ and $2$ respectively, the final wave function is $$\psi_{1}+\psi_{2}$$ and thus the probability density of finding the particle after it has crossed the screen with two holes is found from $$P_{1+2}= (\psi_{1}+\psi_{2})^{*}(\psi_{1}+\psi_{2})$$

That is, you shall first add the wave functions representing the individual holes to have the combined complex wave function, and then compute the probability density. In that addition, the phase informations carried by $\psi_{1}$ and $\psi_{2}$ play the most important role, since they give rise to interference patterns.

Comment: Feynman is quoted to have said "One of the miseries of life is that everybody names things a little bit wrong, and so it makes everything a little harder to understand in the world than it would be if it were named differently." It is quite similar here. Every book says that the phase of the wave function has no physical meaning. That is not 100% correct, as you see.

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