For a particle in a one dimensional infinite potential well of width $L$ the probability function is:

$$P_n(x)=\left(\frac{2}{L}\right)\sin^2\left(\frac{n\pi x}{L}\right)$$

for $0\leq x\leq L$. The probability function reaches maximum when the squared sine reaches 1, that is for

$$x = \frac{L}{2n} \, .$$

For these value of x $P = 2 / L$.

$L$ is typically very small, so that would make $2/L \gg 1$. However, $P$ is a probability and thus $0\leq P\leq 1$. How can $P$ be much larger than 1?

  • 9
    $\begingroup$ This is a probability density, not a probability. $\endgroup$
    – ACuriousMind
    Jul 28, 2015 at 16:17
  • 1
    $\begingroup$ Welcome to Physics Stack Exchange! Please note that clarity is very important in questions. I edited this question to improve its clarity in a few ways. Please note that good titles are important. See this meta post for tips. Also note that in English we only capitalize proper nouns and the first word of each sentence. $\endgroup$
    – DanielSank
    Jul 28, 2015 at 16:29
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    $\begingroup$ The distinction between probability density and probability trips up a lot of people learning probability. It works just like regular density and mass: if I have an object with mass 1 kg, it doesn't mean that it's impossible for its density to be more than 1 kg/m^3 at any point. The two statements are totally unrelated. $\endgroup$
    – knzhou
    Jul 28, 2015 at 16:59

2 Answers 2


It must not be greater than 1.

To find the probability function you must integrate the probability density, $\psi^* \psi$, over the region in which you want to calculate the probability: $$P_n(x_1,x_2)=\int_{x_1}^{x_2} \psi^*(x)\psi(x)\,dx \, .$$

  • $\begingroup$ $P_n(x_1,x_2)=\int_{x_1}^{x_2} \psi^2(x)\,dx$, for Real wave functions, is that correct? Probability density, not mere probability. $\endgroup$
    – Gert
    Jul 28, 2015 at 19:17
  • $\begingroup$ If $\psi (x)$ is real, then $\psi^*(x)=\psi(x)$. So, yes. I don't understand what your last fragment means. $\endgroup$
    – Bill N
    Jul 28, 2015 at 19:23
  • $\begingroup$ Sorry, forget that last fragment. $P_n(x_1,x_2)=\int_{x_1}^{x_2} \psi^2(x)\,dx$ means also that for x1 = x2, P = 0. So the actual probability in a point (x1 = x2) is zero? The particle can only be found in an interval Δx, so P > 0. Is that correct? (I believe so). And for the interval x = 0, x = L, P = 1 (Normalisation). $\endgroup$
    – Gert
    Jul 28, 2015 at 21:48
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Bill N
    Jul 28, 2015 at 22:31
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    $\begingroup$ This has no relation to the uncertainty principle. It's just how a general probability distribution works, in any area of statistics/physics. $\endgroup$ Jul 29, 2015 at 3:11

Your integration is wrong. The probability density function measures the probability of finding the particle between $x$ and $x+dx$. If you integrate over $0 \leq x \leq L$ you don't get a function of $x$ but a number instead (one for this interval). If you state that $P_n$ is the integrated density, then it should depend on the interval in which you did the integration. If it is around the maximum density point then the probability would actually be the difference between this $P_n$ function, which would be small.

If it is the infinite well as you state, the solution is

$\Psi_n(x) = \sqrt{\frac{2}{L}}\sin\left( \frac {n \pi x}{L}\right)$

which would have probabilities given by

$P_n = \int_{x_1}^{x_2} \Psi_n^2 (x)dx =\frac{2}{L}\int_{x_1}^{x_2} \sin^2\left( \frac {n \pi x}{L}\right)dx$

which is normalized, since

$\frac{2}{L}\int_{0}^{L} \sin^2\left( \frac {n \pi x}{L}\right)dx = \frac{2}{L} \int_{0}^{L} \frac{1}{2}\left(1 - \cos(2\frac {n \pi x}{L})\right)dx = 1 $

  • $\begingroup$ But it's correct to call $\Psi_n^2(x)$ the probability density or probability distribution, right? $\endgroup$
    – Gert
    Jul 29, 2015 at 13:57
  • $\begingroup$ Yes, it is correct to call $\Psi^2(x)$ probability density or probability distribution. You just cannot call it probability since it is a continuous case, which means you always have to state "what is the probability of the particle being between this point and this other point" , it make no sense to ask "what is the probability of the particle being in this point exactly". $\endgroup$ Jul 29, 2015 at 14:00

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