Harmonic Oscillator Ground State Energies
For a one-dimensional quantum harmonic oscillator with frequency $\nu$, the total energy in the ground state (n=0) is given by the formula: $E_0 = \frac{1}{2}h\nu$ where $h$ is the Planck constant.
Average Energies using Virial Theorem
A key result for the harmonic oscillator potential ($V(x) = \frac{1}{2}kx^2$) is derived from the Virial Theorem. It states that the average kinetic energy ($\langle K \rangle$) is equal to the average potential energy ($\langle V \rangle$) over any energy eigenstate.
- $\langle K \rangle = \langle V \rangle$
The total energy $E_0$ is the sum of the average kinetic and potential energies: $E_0 = \langle K \rangle + \langle V \rangle$
Calculating Ground State Averages
Substituting $\langle K \rangle = \langle V \rangle$ into the energy equation, we get: $E_0 = 2 \langle V \rangle$
Therefore, the average potential energy in the ground state is: $\langle V \rangle_0 = \frac{E_0}{2} = \frac{1}{2} \left( \frac{1}{2}h\nu \right) = \frac{h\nu}{4}$
Because $\langle K \rangle = \langle V \rangle$, the average kinetic energy in the ground state is also: $\langle K \rangle_0 = \frac{h\nu}{4}$
Conclusion
The average kinetic energy and the average potential energy for a one-dimensional harmonic oscillator in its ground state are both $\frac{h\nu}{4}$.
This matches the values stated in Option 3.
Important Questions from Harmonic Oscillator
The lowest energy of a quantum mechanical one-dimensional simple harmonic oscillator is $300$ cm$^{-1}$. The energy (in cm$^{-1}$) of the next higher level is _________________
- The wavefunction of a 1-D harmonic oscillator between $x =+\infty$ and $x = -\infty$ is given by $\psi(x) = N(2x^2 -1)e^{-x^2/2}$. The value of $N$ that normalizes the function $\psi(x)$ is
(Given: $\int_{-\infty}^{+\infty} x^{2n} e^{-x^2} dx = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2^n} \sqrt{\pi}$) - The wave function for a Harmonic oscillator described by $Nxexp(-ax^2/2)$ has
