# Quantum mechanics: Wave vector is zero?

I am learning about basic quantum mechanics and am considering the time independent Schrodinger equation in 1D, which reduces to a second order ODE:

$\frac{d^2}{dx^2}\psi (x)+[E-V(x)]\psi (x)=0$

Now for a trial form of $\psi (x)=\psi _0e^{ikx}$, the solution values of the wavevector $k$ are

$k=\pm \frac{\sqrt{2m[E-V(x)]}}{\hbar}$ which clearly can be real where $E>V$ (leading to a complex wavefunction, such as inside the potential well) or can be complex leading to exponentially decaying wavefunction where $E<V$, such as a particle penetrating a potential barrier.

But what about if $E=V$ that is $k=0$? Something strange seems to happen, since $k=0\implies \lambda = \infty$ (since $k=\frac{2\pi}{\lambda}$), and also $p=0$ ($k=\frac{p}{\hbar}$).

What is going on here? In lectures we did not cover this case, although I think I recall my lecturer glossed over it at some point and just said that the equations "break down". Does anyone have a good explanation as to what happens in this case, and why the equations are giving these unphysical results? Surely if the $k$ real and $k$ imaginary solutions are possible, then by continuity argument the case of $k=0$ also should be?

So, given a potential that depends on position, wave functions of the form $\psi=e^{ikx}$ aren't solutions (letting $k$ depend on $x$ does, however, give a decent approximation to solutions in certain circumstances, but we'll get back to that).

Let's look at a situation where this trial wave function does work, $V(x)=v$ is a constant. In this case, if $E=v$, then $k=0$, as claimed. In this case, we know exactly the momentum of the particle, and so the uncertainty principle tells us that the uncertainty in position must diverge. This doesn't signal a break-down in any way, it is simply a property of quantum mechanics.

I can't say this without pointing out that, although there is nothing (naively) unphysical about wave functions of the form $e^{ikx}$, when trying to interpret $|\psi|^2$ as a probability distribution, it's impossible to properly normalize it without introducing a length scale $L$ signifying the length of the box we put our experiments in.

Now, to the approximation I promised. Considering a trial wave function $\psi=e^{ikx}$ is useful in a general situation wherein the potential varies slowly with position. This consideration leads to the famous WKB approximation, given by

$$\psi(x)=\frac{C_1}{\sqrt{k(x)}}\exp\left\{i\int_{a}^{x}\mathrm{d}x'k(x')\right\}+\frac{C_2}{\sqrt{k(x)}}\exp\left\{-i\int_{a}^{x}\mathrm{d}x'k(x')\right\},\hspace{0.5cm}E>V(x)$$

Where $k(x)=\sqrt{2m(E-V(x))/\hbar^2}$. This is an excellent approximation in most cases and consequences of it can be used to find the energy levels of the Harmonic Oscillator and the Hydrogen Atom exactly. This approximation, however, has a flaw. Near the boundary, the wave function diverges. Thus, we have to consider, very carefully, the behavior of the wavefunction near a barrier ($k=0$). This leads to a set of equations called the connection formulas, which are used to correctly find energy levels of systems with barriers.

In summary, there is nothing "unphysical" about $E=v$ in the Schrodinger equation. It is simply when the solutions become wave functions of definite momentum. The wavelength becoming infinite doesn't signal a breakdown of physics. However, in certain approximations of solutions to the Schrodinger equation, the point $k=0$ signals a breakdown of the approximation, and this consideration leads to interesting physics.

I hope this is helpful!

The solution is $\psi(x)= C$ with the constant $C$ being any non-zero complex number. The particle has no kinetic energy, so it just sits there, but having definite momentum (zero) its position is completely uncertain, so it sits everywhere at once. Just like the case $\psi(x) =C e^{ikx}$ the wavefunction is not normalizable when the system in infinite. If the system had length $L$ with periodic boundary consitions you could take $C=1/{\sqrt L}$ to get a normalized wavefunction.

(My answer only applies to the case where $V$ is position independent. The answer by Bob is better for when E=V(x) at a sinle point)

• Thank you for your reply! I can see now how this case can relate to a delocalised particle, but I am still wondering about the fact that there is only one solution. Is it not the case that any second order differential equation needs two solutions? In this case, where did the other one go? (the wavefunction would become $\psi=A+Bx$? I suppose it could be that $B=0$, however I do not see what would motivate such a move since we are allowing the wavefunction to be non-normalisable anyway; why would it be an additional problem (assuming I have the right idea) that $\psi->\infty$ as $x->\infty$? – Meep Mar 31 '17 at 16:26
• @21joanna12: To get a well definied eigenvalue problem one needs boundary conditions. To allow a solution like $e^{ikx}$ they must be periodic (or periodic with a twist). Such boundary conditions do not alow $\psi(x)=bx$. – mike stone Apr 1 '17 at 17:38