QM begins with a Born's rule which states that probability $P$ is equal to a modulus square of probability amplitude $\psi$:

$$P = \left|\psi\right|^2.$$

If I write down a wave function like this $\psi = \psi_0 e^{i(kx - \omega t)}$, I find $\psi_0$ inside.

If $\psi$ is called probability amplitude then what is $\psi_0$ called? Is it perhaps called an amplitude of a probability amplitude?

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    $\begingroup$ It's the absolute value of the probability amplitude. Why would you want an easier name for that? It's very important in quantum mechanics that the probability amplitudes are complex numbers. In the particular example you mentioned, a plane wave, the factor $\psi_0$ is moreover a totally irrelevant normalization factor. It's determined by conventions and/or by the requirement that the total probability is equal to one (which isn't possible for a plane wave in infinite space). $\endgroup$ Feb 8, 2013 at 14:53
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    $\begingroup$ It's called normalisation factor. $\endgroup$
    – Kenshin
    Feb 8, 2013 at 14:53
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    $\begingroup$ Probability amplitudes (like probabilities) are unitless. Square or multiply them and the result remains unitless. So, no worries there. $\endgroup$
    – Řídící
    Feb 8, 2013 at 16:12
  • $\begingroup$ In my oppinnion expression "wave function" suits $\psi$ way better than "probability amplitude". Later is confusing. I would name $\psi_0$ "probability amplitude" instead. These names are ridiculously confusing and should be changed! I mean in a plane wave here en.wikipedia.org/wiki/Plane_wave we have $A=A_0 e^{i(kx - \omega t)}$ where $A_0$ is called "amplitude". So if i have a $\psi = \psi_0 e^{i(kx - \omega t)}$ which in some way describes probability only logical thing would be to name $\psi_0$ "probability amplitude". $\endgroup$
    – 71GA
    Feb 8, 2013 at 17:05

2 Answers 2



$\psi_o$ is, as mentioned previously, the normalisation constant which is calculated by doing the integral $\int_V|\psi|^2dV$ and setting its value equal to 1 (hence normalization). This will give you the equation for $\psi_o$. If your interest is to find the probablity amplitude for a particle in a volume V, for example, then you get the equation

$$\int_V|\psi|^2dV = |\psi_o|^2 V = 1\quad,$$

which gives you

$$\psi_o = \frac{1}{\sqrt{V}}\quad,$$

and this is the normalization constant for the probability amplitude $\psi$. Thus you will write

$$\psi(\textbf{r})= \frac{1}{\sqrt{V}}\exp[i(\textbf{k}\cdot\textbf{r}-\omega t)]\quad.$$

For a particle in infinitely large volume, the probability amplitude is zero. I.e $\psi_0=0$. So the probability to find the particle at a particular position is zero.

I hope this helps understand the difference between these two. This question has been asked before in a different format, answers were given. In the light of your reworded question consider this:


The general solution to Schrodinger equation for a free-particle in 1-D (as in your question) is

$\psi(x)=\psi_0e^{i(kx-\omega t)}$

The point with this question is that it puts down arbitrary wave functions and asks how to normalise them. No problem with that, but to normalize the wave function you need to know the boundaries of the problem, hence the boundary conditions.

Assuming you have a ‘box’ of side L in the $x$-axis, the normalisation will give you $\psi_0=\frac{1}{\sqrt {L}}$ so that

$\psi(x)=\frac {1}{\sqrt {L}}e^{i(kx-\omega t)}$

If you now wish to see what will happen to the wave function if $L$ goes to infinity, you must take into consideration the fact that the phase factor is finite, so that it results in $\psi_0=0$ and the whole wave function is zero. This means, as said above, you have no chance of finding the particle at a particular point along the $x$-axis.

In the genuine case where the particle wave-function does occupy the whole of the $x$-axis (the whole volume in the general case,) then, the particle has well defined momentum, hence energy ( it is a stationary state,) but let us focus on the space dimension for simplicity. This means that in the momentum space, the wave function must be $\delta (p^/-p) $ function. This gives the standard $\delta$-function normalisation of the wave function as I mentioned in a previous communication. In other words, we do a Fourier transformation of the plane wave (the above wave function) with the condition that L goes to infinity, and this produces the correct sharp momentum value as indicated by the $\delta$-function

$\psi_k (x)=\frac {1}{\sqrt{2\pi}}e^{i(kx)}$

The limit L going to infinity has already been taken into account in the Fourier transform, hence the $2\pi$ in the normalisation coefficient.

It is very important to understand that the wf must reflect the uncertainty principle. For this reason, a free particle is often described by a wave packet, as mentioned in other answers, with a Gaussian profile. For a particle that is initially confined in a region of width $w_0$, and then is let free, the Gaussian profile expands, and the equation of evolution of the width is given by standard quantum mechanics (there is some good theory about it, see: Stephen Gasiorowicz page 67-70, David Bohm page: 45-47, for example.) Eventually, over time, the wave function reduces to a plane wave-a free particle wave function.

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    $\begingroup$ John, I've edited you math markup a bit. MathJax (the rendering engine we use for that) does rather a lot more than just making greek letters and roots and writing whole equation in it looks better than mixing fragments of latex with normally rendered text in a notional single equation. I hope you find the edits satisfactory. $\endgroup$ Feb 8, 2013 at 15:52
  • $\begingroup$ So we equate integral $\int_V|\psi|^2dV$ to 1 because we are certain that we will get a particle somewhere in this volume if we integrate over whole volume $V$? $\endgroup$
    – 71GA
    Feb 8, 2013 at 16:47
  • $\begingroup$ @71GA: 100% correct. $\endgroup$
    – Vibert
    Feb 8, 2013 at 20:58
  • $\begingroup$ This is a nice anwser. Is there any law we obey when we say for example $\psi_0 = 1$, $\psi_0 = 1$ or maybe $\psi_0 = 0.7$? $\endgroup$
    – 71GA
    Feb 11, 2013 at 5:42
  • $\begingroup$ @71GA Sorry for the belated reply!! The interpretation of such an event could mean that the particle is described by wave function that is a $\delta (x-x_0)$ function. That means the particle has a very well defined position $x_0$, but its momentum has an infinitely wide form. $\endgroup$
    – JKL
    Feb 16, 2013 at 20:29

$\psi_0$ is the initial amplitude and $\psi$ is the amplitude after some amount of time (or more appropriately, the amplitude after some change in spacetime coordinates). The function you have provided: $\psi=\psi_0e^{i(kx-\omega t)}$; is a function of space and time. So it tells me that given some initial function $\psi_0$ at some time $t$ or position $x$ it will have evolved into some function $\psi$. Both $\psi$ and $\psi_0$ are probability amplitudes, its just that $\psi_0$ is the probability amplitude at some initial point.

So my answer is that you call it the initial amplitude, or amplitude naught.

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    $\begingroup$ +1. I'll just add for 71GA sake, that you mean \psi_0 is the amplitude when t = 0 AND x = 0. I know this is what you implied Hal, but I'm just stating it more explicitly for 71GA. $\endgroup$
    – Kenshin
    Feb 9, 2013 at 11:33

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