Question

Show by symmetry arguments that the expectation value of the momentum for an even- \(n\) state of a one-dimensional harmonic oscillator is zero.

Short answer

Question: Show that the expectation value of the momentum for an even-\(n\) state of a one-dimensional harmonic oscillator is zero. Answer: The wavefunction for even-$n$ states of a one-dimensional harmonic oscillator is symmetric and real, which leads to the product of the derivative of the wavefunction and the wavefunction itself being even. By symmetry, the integral over this even product becomes zero, implying that the expectation value of the momentum for even states is also zero (`$`\langle p \rangle = 0`$`).

Step-by-step solution

Recall the wavefunction of a one-dimensional harmonic oscillator

The wavefunction for the \(n\)-th state of a one-dimensional harmonic oscillator can be written as: \(\psi_n(x) = N_n H_n(\alpha x) e^{-\frac{1}{2}\alpha^2 x^2}\), where \(N_n\) is the normalization constant, \(H_n(\alpha x)\) is the Hermite polynomial of order \(n\), and \(\alpha\) is a constant depending on the mass of the particle and the oscillation frequency.

Analyze the symmetry property of the Hermite polynomials for even-\(n\) states

For even-\(n\) states, the Hermite polynomial satisfies \(H_{2k}(-x) = H_{2k}(x)\), where \(k \in \{0, 1, 2, ... \}\). This means that for even-\(n\) states, the wavefunction satisfies: \(\psi_{2k}(-x) = N_{2k} H_{2k}(-\alpha x) e^{-\frac{1}{2}\alpha^2 x^2} = N_{2k} H_{2k}(\alpha x) e^{-\frac{1}{2}\alpha^2 x^2} = \psi_{2k}(x)\). So, the wavefunction is even for even-\(n\) states.

Calculate the expectation value of the momentum operator for even-\(n\) states

To find the expectation value of the momentum \(\langle p \rangle\), we apply the momentum operator (\(-i\hbar \frac{d}{dx}\)) on the wavefunction \(\psi_{2k}\) and then compute the integral \(\langle p \rangle = \int_{-\infty}^{\infty} \psi^*_{2k}(x) (-i \hbar \frac{d}{dx}) \psi_{2k}(x) dx\). Now, since \(\psi^*_{2k}(x) = \psi_{2k}(x)\) (as the wavefunction is real), we can rewrite the integral as: \(\langle p \rangle = -i\hbar \int_{-\infty}^{\infty} \psi_{2k}(x) \frac{d\psi_{2k}(x)}{dx} dx = -i\hbar \left[ \psi_{2k}(x) \psi_{2k}(x) \Big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{d\psi_{2k}(x)}{dx} \psi_{2k}(x) dx \right]\).

Show that the expectation value of the momentum is zero

The first term of the integral will vanish at the limits because the wavefunction goes to zero at \(-\infty\) and \(\infty\). The second term of the integral can be written as follows: \(-i\hbar \int_{-\infty}^{\infty} \frac{d\psi_{2k}(x)}{dx} \psi_{2k}(x) dx\). Since the wavefunction \(\psi_{2k}(x)\) is even for even states, this product will also be even (derivative of a even function is odd, and product of odd and even functions is even). Then, by symmetry, we have that the integral over an even function will be zero: \(\int_{-\infty}^{\infty} \left[ -i\hbar \frac{d\psi_{2k}(x)}{dx} \psi_{2k}(x) \right] dx = 0\). Thus, the expectation value of the momentum for even states is zero: \(\langle p \rangle = 0\).