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?

  • 8
    $\begingroup$ This is a probability density, not a probability. $\endgroup$ – ACuriousMind Jul 28 '15 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 '15 at 16:29
  • 3
    $\begingroup$ 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 '15 at 16:59

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 '15 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 '15 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 '15 at 21:48
  • $\begingroup$ Yes, that's right. $\endgroup$ – Bill N Jul 28 '15 at 22:31
  • 4
    $\begingroup$ This has no relation to the uncertainty principle. It's just how a general probability distribution works, in any area of statistics/physics. $\endgroup$ – Jahan Claes Jul 29 '15 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 '15 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". $\endgroup$ – Andre Maizel Jul 29 '15 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.