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The expression $\int | \Psi\left(x\right)|^2dx$ gives the probability of finding a particle at a given position.

If wave function gives the probabilities of positions, why do we calculate "expectation value of position"?

I don't understand the conceptual difference, we already have a wave function of a position. Expectation value is related to probabilities.

So what is the differences between them? And why do we calculate expectation value for position, although we have a function for probability of finding a particle at a given position?

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5 Answers 5

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In position-space (that is, when your functions are functions of x), the function $\int|\Psi|^2$ gives the probability of finding the particle in a given range. The expectation value of x is where you'd expect to find the particle. It is often essentially the weighted average of all the positions where the probability density, $|\Psi|^2$, is the weighting function (that's not exactly what it is, but it's a useful analogy). Similarly, you can find the expectation value for any measurable quantity. In this space, the difference between the two is that the expectation value is a number that represents the expected average position of the particle over many measurements whereas the probability is a number that gives you the probability for finding the particle within the limits of integration.

However, you can use any different basis. For example, you could choose momentum-space, $\left|\Psi\right>$ is $\Psi(p)$ (quantum physicists please don't kill me for that affront to notation). In momentum space, the integral $\int|\Psi|^2$ is now the probability of the particle having a given range of momenta. However, the expectation value of x is still the average measurement of x. What, you ask, is the point? The expectation value is a number that can be found in any basis that represents the "on-average" value of a measurement. The probability found by $\int|\Psi|^2$ is the probability that a particle will be found existing within a specified range of values for the basis you are using.

$\int_{x_1}^{x_2}|\Psi|^2dx$ is "there is #% chance that the particle will be found between $x_1$ and $x_2$"

$\left<\Psi\right|x\left|\Psi\right>$ is "the expected average position of the particle over a large number of sample measurements is at $x$=#"

$|\Psi|^2(x)$ is a function "the probability per unit length of finding the particle at this position is #%"

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Expectation value is a different concept from probability. In fact, you can have an expectation value of energy, angular momentum, etc., not just for position.

An expectation value of an observable for a given state $\Psi$ is the average value of a large number of measurements of that observable, assuming each measurement is made on the same state $\Psi$. For example, if you have probability 0.5 to measure energy $E_o$ and probability 0.5 to measure $-E_o$, the expectation value is $0.5\times E_o + 0.5\times(-E_o) = 0$. As this example shows, the expectation value doesn't have to be one of the "allowed" measurements. This also shows that knowing the probabilities is a different concept than knowing the average value of a large number of measurements.

The same goes for position. You might know what the probability density is at a particular position, but you would need to do extra computations to figure out what the average value of many position measurements would be.

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Let $\Omega\subseteq \mathbb{R}^n$; then $\int_\Omega \lvert\psi(x)\rvert^2dx$, for a normalized function $\psi\in L^2(\mathbb{R}^n)$ gives the probability that the particle is in the region of space $\Omega$, but does not give any further information on its position. If you want to obtain a quantitative information on the latter (within the limits of quantum indeterminacy), you have to calculate the expectation value $\int_{\mathbb{R}^n} x_j\lvert\psi(x)\rvert^2dx$, for each component $x_j$.

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The expectation value (of position) represents the average value (position) for the particle (it has units of length in this case) which is different from the actual location of the particle (also units of length). For example, take an electron on a hydrogen atom; the expectation value for all energy levels is at the nucleus even though many of the energy levels have 0 probability of being there.

The wave function represents a distribution of possible values and it must become unit-less (technically, units of portion of a particle) after we square and integrate it with respect to whichever basis we're looking at. In one dimension, it has units of length$^{-1/2}$.

As important as that is, most of the quantities we calculate first (expectation value, physical space wave functions, etc...) provide a simple and intuitive introduction to the formalism and practice using different bases. More valuable (easily measurable) quantities might include the expectation value of the polarization, energy, momentum and various uncertainties.

Physicists have a horrible habit of using ad-hoc notations and Jim does a great job of explaining the notations for different physical values and the choice of basis, though apologizing to physicists for abusing notation is hilarious.

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I have 5 bags labelled 1 to 5, and I have randomly dropped the letters A to J into the bags. You choose a letter at random and you win as many Francs as the number on the bag containing your letter.

If I have distributed the letters evenly, then there should be 2 letters in each bag, so we could say that ψ(bagnumber) = ψ = sqrt(2).

But if we want to know how many Francs we expect to win on average, then we say:

E[Francs] = [1 Franc x 2 + 2 Francs x 2 + ... ]/[2 + 2 + ... ] = 3

ψ is not the value you expect, but the distribution over the available values. Thus to find the expected average value, you have to do a weighted average using the values and the probabilities.

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protected by Qmechanic Aug 4 at 20:54

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