# Normalizing the solution to free particle Schrödinger equation

I have the one dimensional free particle Schrödinger equation

$$i\hbar \frac{\partial}{\partial t} \Psi (x,t) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi (x,t), \tag{1}$$

with general solution

$$\psi(x,t) = A e^{i(kx-\omega t)} + B e^{i(-kx-\omega t)}. \tag{2}$$

I'd expect that the solution is normalized:

$$\int_{-\infty}^\infty |\psi(x,t)|^2 dx = 1. \tag{3}$$

But

$$|\psi(x,t)|^2 = \psi(x,t)\psi^* (x,t) = A^2 + B^2 + AB ( e^{2ikx} + e^{-2ikx} ), \tag{4}$$

and the integral diverges:

$$\int_{-\infty}^\infty |\psi(x,t)|^2 dx = \frac{AB}{2ik} (e^{2ikx} - e^{-2ikx})\biggl|_{-\infty}^\infty + (A^2+B^2) x\biggl|_{-\infty}^\infty. \tag{5}$$

What is the reason for this? Can it be corrected?

• You can't normalize a plane wave. – Ryan Unger Feb 16 '15 at 13:33
• There are states which cannot be normalized, this is perfectly normal. If you want a normalization, you'll have to restrict the domain to a finite interval. – JamalS Feb 16 '15 at 13:34

Schroedinger's equations may have both normalizable and non-normalizable solutions. The function

$$\psi_k(x,t) = A e^{i(kx-\omega t)} + B e^{i(-kx-\omega t)}. \tag{2}$$

is a solution of the free-particle Schroedinger equation for any real $$k$$ and $$\omega = |k|/c$$.

As a rule, if the equation has a class of solutions parameterized by continuous parameter ($$k$$), these solutions are not normalizable to infinite space.

One purpose of wave function is to use it to calculate probability of configuration via the Born rule; the probability that the particle described by $$\psi$$ has $$x$$ in the interval $$(a,b)$$ of the line is

$$\int_a^b|\psi(x)|^2\,dx.$$

For this to work, $$\psi$$ has to be such that it has finite integral

$$\int_S |\psi(x)|^2\,dx$$

where $$S$$ is region where it does not vanish.

Plane wave (or sum of such waves) cannot be normalized for $$S=\mathbb R$$ (or higher-dimensional versions of whole infinite space), but it can be normalized for finite intervals (or regions of configuration space which similarly have finite volume).

People deal with this situation in several ways:

• instead of $$\mathbb R$$, they describe system by functions that are limited to an imaginary finitely-sized box, so all regular functions are normalizable (delta distributions will remain non-normalizable even there). The exact size of the box is assumed to be very large but it is almost never fixed to definite value, because it is assumed as it gets expanded to greater sizeits influence on the result becomes negligible.

• retain infinite space, but use only normalizable functions to calculate probability (never use non-normalizable function with the Born rule);

• retain infinite space, retain plane waves, use Dirac formalism and be aware of its drawbacks. Never work with $$\langle x|x\rangle$$ as with something sensible, do not think $$|x\rangle$$,$$|p\rangle$$ represent physical states (people call them states to simplify the language), mind that $$\langle x|$$ is a linear functional that is introduced to act on some ket, not a replacement notation for $$\psi^*$$.

• I have a question concerning the box normalization. If we put the particle in a box, wouldn't that mean that we put it inside the infinite potential well? If so, then many plane waves don't satisfy the boundary conditions. If not, does it mean that we can just consider plane waves but in a finite domain? How is this justified? – TheQuantumMan Jul 28 '16 at 11:14
• Yes, box is a short expression for infinite potential well. Plane wave $e^{i\mathbf k\cdot\mathbf r}$ indeed does not satisfy the boundary condition $\psi(wall)=0$, it only satisfies the Schr.equation. This means no such plane wave can describe particle in such a situation. Its significance is purely mathematical; one can use it to express the true solution of the problem, typically as linear combination of the plane waves. Such linear combination can be made to satisfy the boundary conditions, even if individual plane wave does not. – Ján Lalinský Jul 28 '16 at 21:55
• If you are interested, I was digging around about this matter and found out that there is a box normalization with periodic boundary conditions(not the more strict boundary conditions of the infinite well). This helps with the fact that plane wave solutions are now valid. – TheQuantumMan Jul 30 '16 at 7:12

The solutions of type $e^{i(kx - \omega t)}$ (so-called Fourier components) are not normalizable to 1. Though, they are widely used in quantum theory. They are normalized (quite improper expression) to $\delta$ Dirac,

$$\frac {1}{2\pi} \int_{-\infty}^{\infty} e^{i(k'x - \omega' t)}e^{-i(kx - \omega t)} dx = \delta (k - k') e^{i(\omega - \omega')}. \tag{i}$$

(Usually $\omega$ is function of $|k|$ s.t. $\delta (k - k') e^{i(\omega - \omega')}$ becomes $\delta (k - k')$.)

But, no worry, the Fourier components are an idealization, they don't exist in nature. What yes exist are wave-packets, of finite length and normalizable to 1. We can represent them as

$$\int_{-\infty}^{+\infty} f(k) e^{-i(kx - \omega t)} dk. \tag{ii}$$

When the wave-packet has a very tight distribution of $k$ values, then it is also very long, and in some cases we can afford approximating it by a Fourier component.

The other answers are right. But you might like some insight on why the plane wave is not normalizable, yet still useful.

The general solution is a superposition of components of the form $\psi(x,t) = A e^{i(kx-\omega t)} + B e^{i(-kx-\omega t)}$. Each component will have a different $k$ and $\omega$.

A single component is a uniform distribution over all space. The probability of finding the particle somewhere in all space is 1, but infinitesimal in any finite interval. As you found, to normalize this solution, the amplitude must be infinitesimal.

If you combine two components, the probability distribution is not uniform. The wave functions are waves with phases. They interfere. The probability is higher where they reinforce, and lower where they cancel.

With two or more components you will still get a non normalizable periodic function. But it is possible to make the sum be a periodic train of wave packets, where the amplitude is approximately 0 outside of the packets. You can make the integral over a single packet be 1 with finite component amplitudes. But the integral over the whole wave is still infinite.

By adding more and more components with k's closer and closer together, you can spread the packets farther and farther. As you do, the amplitude of each component must get smaller.

Taking this to extremes, you can spread the packets infinitely far apart by adding an infinite sum of components with infinitesimally separated k's. The amplitude of each component is infinitesimal, but you have an infinite number of them. If you add (integrate) the amplitudes over a small range of k, the sum is finite.

The wave function generated by this sum of components is a single normalized wave packet.

There is nothing wrong with this, as the solution to this equation doesn't live in the Hilbert space. That is to say that there is no eigenvector solution to the free particle equation since the Hamiltonian has only a continuous part in its spectrum. The best you can do if you want to stick with the Hilbert space is to find a sequence of vectors that approximately look like eigenvectors. Alternatively you will have to enrich your Hilbert space by including distributions, as described by the theory of rigged Hilbert spaces.

• The solution of the Cauchy problem (Schrödinger equation) $i\partial_t \psi(t)=-\Delta\psi(t)$, $\psi(0)=\psi_0$ in $L^2(\mathbb{R}^d)$ is unique and well known (because the $-\Delta$ is self-adjoint): $\psi(t)=e^{it\Delta}\psi_0$. There is not, in this case, so much need of advocating generalized eigenvectors and spectral theory ;-) – yuggib Feb 16 '15 at 13:41

Plane wave solutions to the Schrödinger equation are not normalizable because they extend to infinity with a constant amplitude. Any physical particle will be constrained to a finite space, though (at least to the visible universe), so you need to look at superpositions of plane waves. This means that your starting point is $$\psi(x, 0) = \int \mathrm dk \tilde \psi(k) e^{i( k x- \omega\cdot 0)}$$ where $\psi(k)$ is some function with a finite width. The plane wave situation is covered by taking $\tilde \psi(k) = \delta(k-k')$.