Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability integral, Expectation Value and Square of Psi

  1. Jan 26, 2015 #1
    I have come across a bit of conflict in wording of some physics and chemistry textbooks about the probability of finding particles in certain places. To be more specific, I have come across 3 different statements:

    1. $$\int_a^b {| \psi(x) |^2 dx}$$

    The above integral is said to give the probability of finding a particle between x=a and x=b

    2. $$\int_{-\infty}^{+\infty} {x | \psi(x) |^2 dx}$$

    The above integral is said to yield the expectation value for finding the particle at point x. The problem is: My quantum physics book and other sources say that this expectation value and expectation values in general are what tie the calculations to actual physical observables. That is a problem because, if the first integral that I listed gives the probability of finding the particle in a certain region, then wouldn't that integral be the one that connects the calculations to the physical observable of position (and not the expectation value integral)? In other words, what exactly does this expectation value tell you? My hypothesis about the difference between the two integrals is that the first one gives you the probability of finding the particle in set region while the expectation value tells you something about a specific point. If my hypothesis is correct, what exactly does it tell you about said point? In a thread I made once before, someone stated that it was not the probability of finding the particle at said point.

    3. | \psi(x) |^2

    My chemistry book says that the square of the magnitude of the wave function evaluated at x gives you the probability of finding the particle at point x. I, having already known about the first integral that I mentioned, felt that simply squaring the wave function is way too simple. If you can get precision to a single point by just squaring the wave function, then integrating over a region seems almost useless. Does the square of the magnitude of the wave function alone really give you the probability of finding the probability at a point, or do you think that my AP chemistry book was simply simplifying it for students who have not taken calculus? Also, if I am right in saying that this is an over simplification, then what exactly does plugging a point directly into a wave function tell you anyway?
  2. jcsd
  3. Jan 26, 2015 #2


    User Avatar

    Staff: Mentor

    That is correct.

    The statement is not correct, particularly as x is not a point but a variable in the equation. The expectation value is basically the average:
    \langle \hat{A} \rangle \equiv \int_{-\infty}^{\infty} \psi^*(x) \hat{A} \psi(x) dx
    Given an ensemble of systems all prepared in the same state ##\psi##, then the average over all systems of the value of the observable ##\hat{A}## will be ##\langle \hat{A} \rangle##.

    That's sloppy language. You can check easily that ##| \psi(x) |^2## has units, thus is not a probability. It is a probability distribution function, and must be integrated as in 1. or 2. in order to yield probabilities. By itself it instructs on how the probability varies in space, and is vary useful for visualization.
  4. Jan 26, 2015 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    That's really a question of probability theory, independent of quantum mechanics.

    If you have a probability density [itex]P(x)[/itex], then that means that you will have (approximately, in the limit as [itex]\delta x \Rightarrow 0[/itex])probability [itex]P(x) \delta x[/itex] of finding a particle in the region between [itex]x[/itex] and [itex]x+\delta x[/itex].

    Now, if you rerun the same experiment many, many times, starting with the same probability distribution each time, and you measure the position where the object is found each time, then you will get a sequence of values: [itex]x_1[/itex], [itex]x_2[/itex]... Let [itex]N[/itex] be the number of times that you perform the measurement. Then the average value of [itex]x[/itex] for your measurements is

    [itex]\langle x\rangle_N = \frac{1}{N} \sum_i x_i[/itex]

    The prediction of probability theory is that in the limit as [itex]N \Rightarrow \infty[/itex],

    [itex]\langle x \rangle_N \Rightarrow \int x P(x) dx[/itex]

    Expectation values is a way to compute averages of measurables.
  5. Jan 26, 2015 #4
    1. This is the correct expression when the eigenvalues are continuous.
    2. The expectation value only coincides with the probability distribution in the idealized case of a pure state(only one value in the diagonal of the density matrix) so 1. and 2. are giving you quite different information in general.
    3. This is a simplification for a pure state probability distribution, in practice only mixed states are measured, so you either use 2. or if you are considering pure states probabilities you use the integral as in 1. for continuous eigenvalues or the sum Σi for discrete ones.
  6. Jan 26, 2015 #5
    What do you mean by pure and mixed states? Also, how would I know whether an eigenvalue was continuous or not before calculating the integral (integral 1.)?
  7. Jan 26, 2015 #6
    It is quite unfortunate that most most undergraduate QM books only deal with pure states and mention mixed states only in passing or as a footnote. A pure state is one that is in a superposition, a mixed state is a mixture of pure states. Both can be represented with a statistical operator.
    You know what observable you are going to measure right? Then you know if their iegenvalues are discrete or continuous. i.e. position is continuous, spin discrete....
  8. Jan 26, 2015 #7


    User Avatar
    Science Advisor

    General mathematics statement. Let f(x) be the probability density of some random variable X.

    [itex]P(a<X<b) = \int_a^b f(x)dx [/itex]
    [itex]E(X)=\int_{-\infty}^{\infty} xf(x)dx [/itex], where E(X) is average
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook