Normalisation factor $\psi_0$ for wave function $\psi = \psi_0 \sin(kx-\omega t)$

Question

I know that if I integrate probabilitlity $|\psi|^2$ over a whole volume $V$ I am supposed to get 1. This equation describes this.

$$\int \limits^{}_{V} \left|\psi \right|^2 \, \textrm{d} V = 1\\$$

How would we calculate a normalisation factor $\psi_0$ for two simple wave functions like:

$$ \begin{split} \psi &= \psi_0 \sin(kx-\omega t)\\ \psi &= \psi_0 e^{i(kx-\omega t)} \end{split} $$

I know this is quite a basic question, but I have to straighten this out so I can continue reading QM.

Answer 1

Jerry Schirmer's answer applies in an infinite space. If you put the system in a box there is no problem: the normalized wavefunction is $\mathrm{e}^{ikx}/\sqrt{V}$. This is the usual theoretical device to make everything nice and well behaved. Then we take the limit $V\rightarrow\infty$ at the end of the day and, if we've done our job correctly, all of the $V$ dependence should drop out of the final answer. It's not too unreasonable either, since we don't actually know that the universe is infinite. Infinite for all practical purposes might as well just be a big box.

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EDIT in response to comments:

Assume we operate in a box of volume $V$ (in any number of dimensions - interpret as length, area or volume as appropriate). We want to find the normalisation factor $N$ for the wavefunction $\psi = N \mathrm{e}^{i\vec{k}\cdot\vec{x}}$. So we calculate the norm of the wavefunction, which must be equal to one:

$$ \begin{array}{lcl} 1 &=& \int\mathrm{d}x\ \psi^\star \psi \\ &=& \int\mathrm{d}x\ N^\star N \\ &=& \left| N \right|^2 V \end{array}$$

So $N$ has a magnitude of $1/\sqrt{V}$ and an arbitrary phase which can be chosen to make it real and positive.

Answer 2

You can't, really. In the continuum, the "free particle wavefunction" is an abstraction to make the math easier. In a real context, you'd talk about a Wave envelope $\phi(x) = \int dk A(k)e^{ikx}$, where $A(k)$ is a bounded function that guarantees that $\int\phi^{*}\phi$ is finite and normalizable. Note that this is the generalization to the continuum of saying that $\phi(x) = \sum a_{i}\phi_{i}$, where the $\phi_{i}$ are the eigenstates of the Hamiltonian. It's just that, in the continuum, we have an infinite number of eigenstates.

Answer 3

This is mostly a comment but I feel it's important enough that it warrants some space as an answer. Note carefully that the wavefunction $\psi=\psi_0 \sin(kx-\omega t)$ is not a solution to the free-particle Schrödinger equation.

This is easy to see mathematically. The kinetic term, $-\frac{\partial^2}{\partial x^2}$, will return a sine term, while the single time derivative in the total energy term will return a cosine. These can never be proportional.

More physically, you can decompose that wavefunction into two complex exponentials, as $\psi=\frac1{2i}\psi_0(e^{i(kx-\omega t)}-e^{-i(kx-\omega t)})$. The first term is OK, but the second one simply isn't: it's a plane wave going to the left (which is ok) multiplied by the phase $e^{+i\omega t}$. This phase represents a negative energy, which is not physical in a free particle with purely positive kinetic energy.

Sine-like wavefunctions do happen in quantum theory, most notably for the infinite square well. There the energy eigenfunctions are of the form $\sin(n\pi x/L)$ to vanish at $x=0$ and $x=L$, but the time dependence is $$\psi_n(x,t)=A_n\sin\left(\frac{n\pi}{L}x\right)e^{-i\omega_n t},$$ and the total energy is positive. Beware!