Question

The correct option for the average value of kinetic energy and the average value of potential energy of a one-dimensional harmonic oscillator with frequency \(\nu\) in its ground state is:

• A. \(\dfrac{h\nu}{2}\) and \(\dfrac{h\nu}{4}\), respectively
• B. \(\dfrac{h\nu}{2}\) and \(\dfrac{h\nu}{2}\), respectively
• C. \(\dfrac{h\nu}{4}\) and \(\dfrac{h\nu}{4}\), respectively
• D. \(\dfrac{h\nu}{4}\) and \(\dfrac{h\nu}{2}\), respectively

Hint:

Ground state energy is \(h\nu/2\); the virial theorem for a quadratic potential gives \(\langle T\rangle=\langle V\rangle\), so each is half of \(h\nu/2\).

Answer 1

The Correct Option is C

Step 1: Write the ground-state energy of the 1D quantum harmonic oscillator.
For a 1D quantum harmonic oscillator with frequency \(\nu\), the allowed energies are \(E_n=\left(n+\frac{1}{2}\right)h\nu\), \(n=0,1,2,\ldots\). In the ground state, \(n=0\):
\[ E_0 = \frac{1}{2}h\nu \]

Step 2: Recall the virial theorem for a quadratic potential.
The potential is \(V(x)=\frac{1}{2}kx^2\), a pure quadratic (power \(m=2\)) in \(x\). The quantum virial theorem for \(V(x)\propto x^m\) states \(2\langle T\rangle=m\langle V\rangle\). For \(m=2\), \(2\langle T\rangle=2\langle V\rangle\), so:
\[ \langle T\rangle=\langle V\rangle \]
This holds for the harmonic oscillator in every stationary state, not just the ground state.

Step 3: Split the total energy equally.
Since \(E_0=\langle T\rangle+\langle V\rangle\) and \(\langle T\rangle=\langle V\rangle\), each equals half of \(E_0\):
\[ \langle T\rangle=\langle V\rangle=\frac{E_0}{2}=\frac{1}{2}\left(\frac{h\nu}{2}\right)=\frac{h\nu}{4} \]

Step 4: Check the other options.
Options (A) and (D) give unequal values for \(\langle T\rangle\) and \(\langle V\rangle\), correct only for an asymmetric or anharmonic potential, not the harmonic oscillator, where the virial theorem forces exact equality. Option (B) gives \(h\nu/2\) for both, which would make \(\langle T\rangle+\langle V\rangle=h\nu\), twice the actual ground-state energy \(h\nu/2\).

Final Answer:
Both averages equal \(h\nu/4\), option (C). \[ \boxed{\langle T\rangle=\langle V\rangle=\dfrac{h\nu}{4}} \]