Question 2.5: Find the expectation value of the potential energy in the nt...

\left\langle V\right\rangle =\left\langle\frac{1}{2}m\omega ^{2}x^{2} \right\rangle =\frac{1}{2}m \omega ^{2}\int_{-\infty }^{\infty }{\psi ^{*}_{n}} x^{2} \psi _{n}dx

There’s a beautiful device for evaluating integrals of this kind (involving powers of x or \hat{p} : Use the definition (Equation 2.48) [\hat{a} _{\pm}\equiv \frac{1}{\sqrt{2\hbar m\omega } } (\mp i\hat{p}+m\omega x )] to express x and in terms of the raising and lowering operators:

x=\sqrt{\frac{\hbar }{2m \omega } }(\hat{a} _{+}+\hat{a} _{-}) ;\hat{p}=i\sqrt{\frac{\hbar m\omega }{2} } (\hat{a} _{+}-\hat{a} _{-})               (2.70)

In this example we are interested in  x^{2} :

x^{2}=\frac{\hbar }{2m \omega }\left[(\hat{a} _{+})^{2}+(\hat{a} _{+}\hat{a} _{-})+(\hat{a} _{-}\hat{a} _{+})+(\hat{a} _{-})^{2}\right]

So

\left\langle V\right\rangle =\frac{\hbar \omega }{4}\int_{-\infty }^{\infty }{\psi ^{*}_{n}} \left[( \hat {a} _{+})^{2}+(\hat{a} _{+}\hat{a} _{-})+(\hat{a} _{-}\hat{a} _{+})+(\hat{a} _{-} )^{2}\right] \psi _{n}dx

But (\hat{a} _{+})^{2}\psi _{n} is (apart from normalization) \psi _{n+2} , which is orthogonal to \psi _{n} , and the same goes for (\hat{a} _{-})^{2}\psi _{n}, which is proportional to \psi _{n-2} . So those terms drop out, and we can use Equation 2.66 [\hat{a} _{+}\hat{a} _{-}\psi _{n}=n\psi _{n},\hat{a} _{-}\hat{a} _{+}\psi _{n}=(n+1)\psi _{n}] to evaluate the remaining two:

\left\langle V\right\rangle =\frac{\hbar \omega }{4}(n+n+1)=\frac{1}{2} \hbar \omega\left(n+\frac{1}{2} \right)

As it happens, the expectation value of the potential energy is exactly half the total (the other half, of course, is kinetic). This is a peculiarity of the harmonic oscillator, as we’ll see later on (Problem 3.37).