# How do you find the expectation value of an operator in quantum mechanics?

The expected values of position and momentum are given by

$\displaystyle \left< x \right>= \int \Psi^* x \Psi\, dx$

$\displaystyle \left=\int \Psi^* \frac{\hbar}{i} \frac{\partial}{\partial x} \Psi\, dx$

In general, any dynamic variable, $Q$, can be expressed in terms of position and momentum, i.e., $Q(\hat{p},\hat{x})$.

In quantum theory, an experimental setup is described by the observable A to be measured, and the state σ of the system. The expectation value of A in the state σ is denoted as .

Mathematically, A is a self-adjoint operator on a Hilbert space. In the most commonly used case in quantum mechanics, σ is a pure state, described by a normalized[1] vector ψ in the Hilbert space. The expectation value of A in the state ψ is defined as

(1)      .

If dynamics is considered, either the vector ψ or the operator A is taken to be time-dependent, depending on whether the Schrödinger picture or Heisenberg picture is used. The time-dependence of the expectation value does not depend on this choice, however.

If A has a complete set of eigenvectors φj, with eigenvalues aj, then (1) can be expressed as

(2)      .

This expression is similar to the arithmetic mean, and illustrates the physical meaning of the mathematical formalism: The eigenvalues aj are the possible outcomes of the experiment,[2] and their corresponding coefficient  is the probability that this outcome will occur; it is often called the transition probability.

A particularly simple case arises when A is a projection, and thus has only the eigenvalues 0 and 1. This physically corresponds to a “yes-no” type of experiment. In this case, the expectation value is the probability that the experiment results in “1”, and it can be computed as

(3)      .

### References

http://en.wikipedia.org/wiki/Expectation_value_(quantum_mechanics)

You need the wavefunction to find the expectation value. If psi is your wavefunction, then the expectation value:

<Lx> = <psi*| Lx | psi>

If you weren’t given a wavefunction, grab your Libboff or Griffiths and look it up. Particle in a box? In a well? Quantum dot? They must have given you a clue, because you can’t find the expectation value without a state to apply it to.

Once you have a state, use the raising and lowering version of Lx to do the computation.

For Lx | l m>? OK, that’s easier. Lx translates to L+ and L-, as you already have.

First, split it up.

< l m | (1/2)(L+ + L-) | l m >
= (1/2) ( < l m | L+ | l m > + < l m | L – | l m> )

Then apply to the right first.

< l m | L+ | l m > = < l m | sqrt[ l ( l+1) – m(m+1) ]*hbar | l m+1 >
= sqrt(that stuff)*hbar <l m | l m+1> =0

and so on. Check the signs inside the sqrt, I may have reversed one. But if you’re applying it between the same l and m states, you’ll get zero.