# Expectation values and operators.

1. May 17, 2008

i'm just not sure on this little detail, and its keeping me from finishing this problem.

take the arbitrary operator $$\tilde{p}^{n}\tilde{y}^{m}$$ where p is the momentum operator , and x is the x position operator

the expectation value is then <$$\tilde{p}^{n}\tilde{y}^{m}$$ >

is this the same as <$$\tilde{p}^{n}$$> <$$\tilde{y}^{m}$$>?

if not, how would i go about calculating <$$\tilde{p}^{n}\tilde{y}^{m}$$ >?

Last edited: May 17, 2008
2. May 18, 2008

### CompuChip

In general, they are not the same. The expectation value of an operator $\hat A$ is
$$\int \Psi^*(x) \hat A(x) \Psi(x) \, \mathrm{d}x$$
where $\Psi(x)$ is your wavefunction (assuming you are talking QM here).
In this case,
$$\int \Psi^*(x) \hat p^n \hat y^m \Psi(x) \, \mathrm{d}x \neq \left( \int \Psi^*(x) \hat p^n \Psi(x) \, \mathrm{d}x \right) \left( \int \Psi^*(x) \hat y^m \Psi(x) \, \mathrm{d}x \right).$$
You could write out $\hat p$ in the position basis and work out what $\hat p^n(y^m \Psi)$ looks like.