# Sinusoidal vs exponential wave functions with Schrodinger's equation

When solving Schrodinger's equation, we end up with the following differential equation:

$$\frac{{d}^{2}\psi}{dx^2} = -\frac{2m(E - V)}{\hbar}\psi$$

As I understand it, the next step is to guess the wave function, so let $\psi = {e}^{i\kappa x}$ or let $\psi = \sin(\kappa x)$, both of which I understand as they satisfy the differential equation, but when would you use one over the other? My textbook seems to use both in different scenarios, but I can't seem to figure out what the conditions are for using exponential over sinusoidal and vice versa.

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The condition is that the quantity multiplying $\psi$ on the right is positive or negative. If it is positive, then you have an exponential, if it is negative, a sinusoid. The sinusoids are complex exponentials. – Ron Maimon Dec 7 '11 at 3:35

## 2 Answers

This is really more of a mathematical question than a physical one, but in any case, here's a simple explanation:

Hopefully you are familiar with the idea of a basis of a vector space. And hopefully you also know that for any given vector space, you can choose many different possible bases. For example, if the 2D plane is your vector space, you could choose the standard unit vectors $\hat{x}$ and $\hat{y}$, or you could choose $(\hat{x} \pm \hat{y})/\sqrt{2}$. Let's call these latter two $\hat{u}$ and $\hat{v}$.

You can always express the members of one basis as linear combinations of the members of any other basis. In the 2D plane example:

$$\begin{align}\hat{u} &= \frac{\hat{x} + \hat{y}}{\sqrt{2}} & \hat{x} &= \frac{\hat{u} + \hat{v}}{\sqrt{2}} \\ \hat{v} &= \frac{\hat{x} - \hat{y}}{\sqrt{2}} & \hat{y} &= \frac{\hat{u} - \hat{v}}{\sqrt{2}}\end{align}$$

Since every vector can be expressed as a linear combination of basis vectors, if you want to understand what happens to the whole vector space under some transformation (like rotation), then you can just look at the effect of the transformation on the basis vectors. You don't need to examine the effect of the transformation on every vector individually. That's why basis vectors are useful. As an example, consider what happens when you rotate the 2D plane by some angle. You wouldn't (and couldn't) draw the rotated form of every single vector in the plane; you just draw the rotated $x$ and $y$ axes, and that's enough for you to understand what happened to the plane and figure out the effect of the rotation on any other vector.

This is exactly what's going on with the solutions to the Schroedinger equation. All the possible solutions to the equation form a vector space, called a Hilbert space. And like any other vector space, you can choose any of an infinite number of possible bases for the space. The set of all complex exponential functions

$$\{e^{ikx}|k\in\mathbb{R}\}$$

is one possible basis. You can construct any function by making a suitable linear combination of these exponential functions,

$$\psi(x) = \int\underbrace{\frac{1}{\sqrt{2\pi}}\psi(k)}_{\text{coefficient}}\underbrace{\vphantom{\frac{1}{\sqrt{2\pi}}}e^{ikx}}_{\text{basis vector}}\mathrm{d}k$$

(You may recognize this as the inverse Fourier transform.)

Another possible basis is the set of all sine and cosine functions,

$$\{\sin(kx),\cos(kx)|k\in\mathbb{R}_+\}$$

You can also express any function as a linear combination of sine and cosine functions:

$$\psi(x) = \int\biggl(\underbrace{i\sqrt{\frac{2}{\pi}}\psi_s(k)}_{\text{coefficient}}\underbrace{\vphantom{\frac{1}{\sqrt{2\pi}}}\sin(kx)}_{\text{basis vector}} + \underbrace{\sqrt{\frac{2}{\pi}}\psi_c(k)}_{\text{coefficient}}\underbrace{\vphantom{\frac{1}{\sqrt{2\pi}}}\cos(kx)}_{\text{basis vector}}\biggr)\mathrm{d}k$$

Just as with the 2D plane vectors, you can express the elements of either basis in terms of the other basis:

$$\begin{align}\sin(kx) &= \frac{e^{ikx} - e^{-ikx}}{2i} & e^{ikx} &= \cos(kx) + i\sin(kx) \\ \cos(kx) &= \frac{e^{ikx} + e^{-ikx}}{2} & e^{-ikx} &= \cos(kx) - i\sin(kx)\end{align}$$

Depending on the particular situation you're considering, it may be more convenient to use one basis or the other. For example, if you have a potential that goes to infinity at some point (e.g. infinite square well), you know that the wavefunction becomes zero there, and thus it's to your advantage to choose a basis where the basis functions actually do go to zero somewhere: the sine/cosine basis, rather than the exponential one. But in general, any problem can be done with either basis. It's simply a choice of convenience.

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The most general solution is $Ae^{-ikx} + Be^{ikx}$

Depending on the coefficients $A$ and $B$ this can be equal to either of the example wave functions you gave, and these coefficients will be determined by boundary conditions.

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