Perturbation Theory Energy Shift Calculation

The first-order energy shift for a system in an unperturbed state $\psi_n$ subjected to a perturbation $V'(x)$ is calculated using the expectation value of the perturbation:

$ \Delta E_n^{(1)} = \langle \psi_n | V' | \psi_n \rangle = \int \psi_n^*(x) V'(x) \psi_n(x) dx $

Ground State Wavefunction and Perturbation

For the particle in a one-dimensional potential well ($0 \le x \le a$), the ground state corresponds to the quantum number $n=1$. The normalized ground state wavefunction $\psi_1(x)$ is:

$ \psi_1(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) $

The given perturbation potential $V'(x)$ is:

$ V'(x) = \begin{cases} V_0 \cos\left(\frac{\pi x}{a}\right) & \text{if } 0 \le x \le \frac{a}{2} \\ 0 & \text{otherwise} \end{cases} $

Calculating the Ground State Energy Shift

We need to compute the expectation value of the perturbation $V'(x)$ using the ground state wavefunction $\psi_1(x)$. The integral limits are from $0$ to $a$, but the perturbation is non-zero only from $0$ to $a/2$:

$ \Delta E_1^{(1)} = \int_0^{a/2} \psi_1^*(x) V'(x) \psi_1(x) dx $

Substitute the expressions for $\psi_1(x)$ and $V'(x)$:

$ \Delta E_1^{(1)} = \int_0^{a/2} \left( \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) \right) \left( V_0 \cos\left(\frac{\pi x}{a}\right) \right) \left( \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) \right) dx $

Simplify the expression:

$ \Delta E_1^{(1)} = \frac{2V_0}{a} \int_0^{a/2} \sin^2\left(\frac{\pi x}{a}\right) \cos\left(\frac{\pi x}{a}\right) dx $

Evaluating the Integral

Use the substitution method to evaluate the integral. Let $u = \sin\left(\frac{\pi x}{a}\right)$.

Then, the differential is $du = \frac{\pi}{a} \cos\left(\frac{\pi x}{a}\right) dx$, which means $\cos\left(\frac{\pi x}{a}\right) dx = \frac{a}{\pi} du$.

Adjust the limits of integration based on the substitution:

• When $x=0$, $u = \sin\left(\frac{\pi \cdot 0}{a}\right) = \sin(0) = 0$.
• When $x=a/2$, $u = \sin\left(\frac{\pi (a/2)}{a}\right) = \sin\left(\frac{\pi}{2}\right) = 1$.

The integral transforms into:

$ \int_0^{a/2} \sin^2\left(\frac{\pi x}{a}\right) \cos\left(\frac{\pi x}{a}\right) dx = \int_0^1 u^2 \left(\frac{a}{\pi} du\right) $

$ = \frac{a}{\pi} \int_0^1 u^2 du = \frac{a}{\pi} \left[ \frac{u^3}{3} \right]_0^1 $

$ = \frac{a}{\pi} \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = \frac{a}{\pi} \cdot \frac{1}{3} = \frac{a}{3\pi} $

Final Energy Shift

Now, substitute the evaluated integral value back into the expression for $\Delta E_1^{(1)}$:

$ \Delta E_1^{(1)} = \frac{2V_0}{a} \times \left( \frac{a}{3\pi} \right) $

$ \Delta E_1^{(1)} = \frac{2V_0}{3\pi} $

The first-order shift in the ground state energy due to the perturbation is $\frac{2V_0}{3\pi}$.