Why is the operator <p> considered real in calculus?

[chingcx]

<p> is the integral of the product of a real function and its first derivative, multiplied by an imaginary number. But why <p> is real?

 

[Ben Niehoff]

If the wavefunction is purely real, then <p> = 0:

$$\langle p \rangle = \int_{-\infty}^{\infty} \psi^* \left(-i\hbar \frac{d\psi}{dx} \right) \; dx$$

If $\psi^* = \psi$, then this can be written

$$\langle p \rangle = -i\hbar \left. \int_{-\infty}^{\infty} \psi \frac{d\psi}{dx} \; dx = -i\hbar \psi(x) \right|_{-\infty}^{\infty} = 0$$

where we have used the boundary condition that the wavefunction must vanish as $x \to \pm \infty$. On the other hand, if we can write

$$\psi(x) = e^{ikx} R(x)$$

where R(x) is real, then

$$\langle p \rangle = \int_{-\infty}^{\infty} \psi^* \left(-i\hbar \frac{d\psi}{dx} \right) \; dx = -i\hbar \int_{-\infty}^{\infty} \left( e^{-ikx} R(x) (ik) e^{ikx} R(x) + e^{-ikx} R(x) e^{ikx} \frac{dR}{dx} \right) \; dx = \hbar k \int_{-\infty}^{\infty} R(x)^2 \; dx = \hbar k$$

So, to have a nonzero momentum, the wavefunction needs a phase that varies with x.

 

[jamesmaxwell]

Because of hermitian conjugate

 

[dextercioby]

If $\hat{p}$ is selfadjoint, then its expectation value in any state is real.

 

[alexgs]

As in the last two posts, the most elegant and general way to see that <p> comes from the fact that p is a hermitian operator.

But if you only know about p in position space (i.e. p = -ih d/dx) then you can see that <p> is real by showing that <p> = <p>* (is this statement obvious?).

Do this by starting with the first equation in Ben Niehoff's post. Take the complex conjugate of both sides, i's become -i's, and psi -> psi* and psi* -> psi. Then with some manipulations (including an integration by parts) you can show that this expression is identical to the formula you started with, so <p>*=<p>.

Hope that helps a little.