An expectation value is the average result you would obtain if you measured the same observable many times on identically prepared systems. For an operator \(\hat{A}\) in a normalized state \(\psi(x)\), the expectation value is
\[ \langle A\rangle=\int_{-\infty}^{\infty}\psi^*(x)\,\hat{A}\,\psi(x)\,dx. \]
For the harmonic oscillator, we will consider a stationary state \(\psi_v(x)\) with energy \(E_v=\hbar\omega\left(v+\tfrac{1}{2}\right)\).
1) Expectation value of energy, \(\langle E\rangle\)
In an energy eigenstate, a measurement of energy always returns the corresponding eigenvalue, so
\[ \langle E\rangle = E_v=\hbar\omega\left(v+\frac{1}{2}\right). \]
2) Position average, \(\langle x\rangle\) (symmetry)
The harmonic potential is symmetric about \(x=0\), and the stationary-state probability density \(|\psi_v(x)|^2\) is an even function. The integrand for \(\langle x\rangle\) is \(x|\psi_v(x)|^2\), which is odd, so it integrates to zero:
\[ \langle x\rangle=\int_{-\infty}^{\infty} x\,|\psi_v(x)|^2\,dx = 0. \]
3) Mean-square position, \(\langle x^2\rangle\)
Use the fact that for the harmonic oscillator (a quadratic potential), the average kinetic and potential energies are equal in a stationary state: \(\langle T\rangle=\langle V\rangle=\tfrac{1}{2}\langle E\rangle\) (a special case of the virial theorem).
Since \(V(x)=\tfrac{1}{2}\mu\omega^2x^2\), we have
\[ \left\langle V\right\rangle =\frac{1}{2}\mu\omega^2\langle x^2\rangle =\frac{1}{2}\langle E\rangle =\frac{1}{2}\hbar\omega\left(v+\frac{1}{2}\right). \]
Solving for \(\langle x^2\rangle\) gives
\[ \langle x^2\rangle =\frac{\hbar}{\mu\omega}\left(v+\frac{1}{2}\right) =\frac{\hbar}{2\mu\omega}(2v+1). \]
4) Momentum average, \(\langle p\rangle\) (symmetry)
The momentum operator is \(\hat{p}=-i\hbar\,\dfrac{d}{dx}\). For stationary harmonic oscillator eigenstates, the probability density is symmetric and there is no preferred direction of motion. As a result, the average momentum must vanish:
\[ \langle p\rangle = 0. \]
5) Mean-square momentum, \(\langle p^2\rangle\)
The kinetic energy operator is \(\hat{T}=\dfrac{\hat{p}^2}{2\mu}\). Using \(\langle T\rangle=\tfrac{1}{2}\langle E\rangle\):
\[ \frac{\langle p^2\rangle}{2\mu} =\frac{1}{2}\hbar\omega\left(v+\frac{1}{2}\right) \quad\Rightarrow\quad \langle p^2\rangle =\mu\hbar\omega\left(v+\frac{1}{2}\right). \]
Minimum uncertainty of the ground state
The uncertainty in an observable \(A\) is defined as
\[ (\Delta A)^2 = \langle A^2\rangle - \langle A\rangle^2. \]
For the harmonic oscillator ground state \(v=0\), we have already shown that
- \(\langle x\rangle = 0\)
- \(\langle p\rangle = 0\)
- \(\langle x^2\rangle = \dfrac{\hbar}{2\mu\omega}\)
- \(\langle p^2\rangle = \dfrac{\mu\hbar\omega}{2}\)
Position uncertainty
\[ \Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} = \sqrt{\frac{\hbar}{2\mu\omega}}. \]
Momentum uncertainty
\[ \Delta p = \sqrt{\langle p^2\rangle - \langle p\rangle^2} = \sqrt{\frac{\mu\hbar\omega}{2}}. \]
Uncertainty product
Multiply the two uncertainties:
\[ \Delta x\,\Delta p = \sqrt{\frac{\hbar}{2\mu\omega}} \sqrt{\frac{\mu\hbar\omega}{2}} = \frac{\hbar}{2}. \]
This result exactly saturates the Heisenberg uncertainty principle,
\[ \Delta x\,\Delta p \ge \frac{\hbar}{2}. \]
Big idea: the ground-state harmonic oscillator wavefunction is a minimum-uncertainty state. No quantum state can have a smaller product of position and momentum uncertainties.
Big idea: symmetry immediately gives \(\langle x\rangle=\langle p\rangle=0\), while the quadratic form of the Hamiltonian implies \(\langle T\rangle=\langle V\rangle=\tfrac{1}{2}E_v\), leading directly to \(\langle x^2\rangle\) and \(\langle p^2\rangle\).