I am having difficulty normalizing the wavefunction for a free particle in 1 dimension. I have sent an attachment of my working out.

My attempt at the normalisation

Free electron:

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

\begin{align} \psi(x) &= A\sin(kx) + B\cos(kx) \\ \psi(x) &= \psi^*(x) \end{align}

$$\therefore \int^\infty_{-\infty} \left[ A\sin(kx) + B \cos(ks)\right]^2 = 1$$

$$\int^\infty_{-\infty}\left[A^2\sin^2(kx) + AB\sin(kx)\cos(kx) + AB\cos(kx)\sin(kx) +B^2\cos^2(kx)\right]\mathrm dx = 1$$

Since, $\sin^2 x + \cos^2 x = 1$,

$$\implies \int^\infty_{-\infty}\left(A^2 + B^2 + 2AB\sin(kx)\cos(kx\right)\mathrm dx) = 1$$

Since, $\sin(2x) = \sin x\cdot\cos x$,

$$\int^\infty_{-\infty}\left(A^2 + B^2 + 2AB\sin(2kx)\right)\mathrm dx = 1$$

$$\int^\infty_{-\infty}\left(A^2 + B^2\right) + AB\sin(2kx)dx = 1$$

\begin{align} A^2 + B^2 + AB\int^\infty_{-\infty}&\sin(2kx)dx \\ &=A^2+ B^2 +AB\left[-\frac{1}{2}cos(2kx)\right]^\infty_{-\infty} \neq 1 \end{align}

Could someone please let me know of my mistake?

  • 5
    $\begingroup$ @ToddMinehardt The question explicitly says "for a free particle in 1 dimension" and "free electron" so there is no box, and the integral really is over all space. But I do agree the bounded cases is much easier, in fact the integral that is being attempted here is not well defined - see physics.stackexchange.com/questions/165373/… $\endgroup$
    – Ian Bush
    Commented May 3, 2021 at 18:05

2 Answers 2


To answer the question, over and above the errors noted in the other answer the integral is just not well defined - it is infinite. There are two ways to see this

  1. The "common sense" method. Consider $$ \int^\infty_{-\infty} \left[ A\sin(kx) + B \cos(kx)\right]^2dx$$ The quantity under the integral is positive semi-definite - it is always greater than or equal to zero. It describes oscillations all of which are positive. Each oscillation bounds a finite positive area between it and the x axis. As the bounds of the integral are all space there are an infinite number of oscillations. An infinite number of positive finite areas evaluates to an infinite area. Thus your integral is not well defined.

  2. Maths - assuming A and B are real (but removing this restriction makes no difference to the final conclusion)

$$\left[ A\sin(kx) + B \cos(kx)\right]^2 = \left[A^2\sin^2(kx) + 2AB\sin(kx)\cos(kx) +B^2\cos^2(kx)\right]$$

The way to evaluate the integral is to use the identities $$ 2sin(\theta)cos(\theta) = sin(2\theta) $$ $$ sin^2(\theta) = {{(1-cos(2\theta))}\over2}$$ $$ cos^2(\theta) = {{(1+cos(2\theta))}\over2}$$ Thus we can write the integral as $$ \int^\infty_{-\infty} \left[A^2{{(1-cos(2kx))}\over2}+B^2{{(1+cos(2kx))}\over2}+ABsin(2kx)\right]dx $$

The first term contains $$ \int^\infty_{-\infty} \left[ A^2\right]dx $$

This is infinite. Similarly the second term for $B^2$. Thus the integral is not well defined [yes this is slightly hand-waving but avoids a discussion of $cos(\infty)$ and similar]. So what you are doing wrong is trying to evaluate an integral that doesn't have an answer.

What you are running into is the fact that not all wavefunctions are normalisable. See Normalizing the solution to free particle Schrödinger equation in the physics stack exchange forum for more details


Your attempt has multiple issues, most of which are very basic.

The first issue that I noticed was how you expanded and simplified your integral.

\begin{align} \int^\infty_{-\infty}[A^2\sin^2(kx) &+ AB\sin(kx)\cos(kx) \\ &+ AB\cos(kx)\sin(kx) +B^2\cos^2(kx)]\mathrm dx = 1 \tag{1} \end{align}

Since, $\sin^2 x + \cos^2 x = 1$,

$$\implies \int^\infty_{-\infty}\left(A^2 + B^2 + 2AB\sin(kx)\cos(kx\right)\mathrm dx) = 1$$

Error in equation($1$): $A^2\sin^2 x + B^2\cos^2 x \neq A^2 + B^2$

If you were to simplify equation (1), you can not use the above statement, for a very simple reason. That is because even if $\sin^2 x + \cos^2 x = 1$:

$$A^2\sin^2 x + B^2\cos^2 x \neq A^2 + B^2$$

Instead it should be:

$$\int^\infty_{-\infty}\left[A^2\sin^2(kx) + B^2\cos^2(kx) + 2AB\sin(kx)\cos(kx)\right]\mathrm dx= 1$$

The second issue is in the actual solving of the integral itself:

$$\int^\infty_{-\infty}\left(A^2 + B^2 + AB\sin(2kx)\right)\mathrm dx = 1$$ $$\implies A^2 + B^2 + AB\int^\infty_{-\infty}\sin(2kx)dx = 1$$

Two issues here: $A^2 + B^2$ are not being multiplied and so cannot be just taken outside the integral, this means that the solution is wrong, also a $2$ went missing. Notice where?

$$\int^\infty_{-\infty}\left(A^2 + B^2 + \color{red}{(2?)}AB\sin(2kx)\right)\mathrm dx = 1$$

These are just the two issues in the solution. Hope this is enough.


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