Question

The second-order correction to the energy of the ground state is always:

• A. Zero
• B. Imaginary
• C. Positive
• D. Negative

Hint:

A useful physical insight: "The ground state always gets pushed down by second-order perturbations." Because the denominator \((E_0^{(0)} - E_m^{(0)})\) forces every term to be negative, \(E_0^{(2)}\) can never be positive.

Answer 1

The Correct Option is D

Concept: In time-independent non-degenerate perturbation theory, we look at a total Hamiltonian \(H = H_0 + H'\), where \(H_0\) is the unperturbed, solvable Hamiltonian and \(H'\) represents a small perturbing potential. The exact energy eigenvalues can be expanded as an infinite power series in a small parameter. The second-order energy correction formula for a specific non-degenerate unperturbed energy level \(E_n^{(0)}\) is given by: \[ E_n^{(2)} = \sum_{m \neq n} \frac{|\langle \psi_m^{(0)} | H' | \psi_n^{(0)} \rangle|^2}{E_n^{(0)} - E_m^{(0)}} \]

Step 1: Applying the generic correction formula to the ground state energy level.
Let the index \(n = 0\) denote the baseline, absolute lowest energy level, which represents the ground state of the system. Let \(E_0^{(0)}\) be its unperturbed energy. Substituting \(n = 0\) into the second-order perturbation expansion formula yields: \[ E_0^{(2)} = \sum_{m \neq 0} \frac{|\langle \psi_m^{(0)} | H' | \psi_0^{(0)} \rangle|^2}{E_0^{(0)} - E_m^{(0)}} \] Here, the summation index \(m\) covers all possible excited states of the quantum system (\(m = 1, 2, 3, \ldots\)).

Step 2: Evaluation of the sign of the components within the sum.
Let us separately examine the mathematical nature of both the numerator and the denominator inside the summation:
• The Numerator: The numerator contains the term \(|\langle \psi_m^{(0)} | H' | \psi_0^{(0)} \rangle|^2\). This is the absolute square of a complex matrix element. By the absolute definition of any squared modulus of a complex number, this term must be strictly greater than or equal to zero: \[ |\langle \psi_m^{(0)} | H' | \psi_0^{(0)} \rangle|^2 \geq 0 \] Assuming there is non-vanishing coupling between the ground state and the excited states, this term is strictly positive.
• The Denominator: The denominator contains the difference between the unperturbed ground state energy and an unperturbed excited state energy, expressed as \((E_0^{(0)} - E_m^{(0)})\). By the definition of the ground state, \(E_0^{(0)}\) is the lowest possible energy eigenvalue of the operator \(H_0\). Therefore, any excited state energy \(E_m^{(0)}\) (where \(m \neq 0\)) must be strictly greater than \(E_0^{(0)}\): \[ E_m^{(0)} > E_0^{(0)} \quad \Rightarrow \quad E_0^{(0)} - E_m^{(0)} < 0 \] Thus, the denominator is always strictly negative.

Step 3: Concluding the sign of the overall summation.
Every single individual fraction in the summation series consists of a positive (or zero) numerator divided by a guaranteed negative denominator. Therefore, each individual term being summed is less than or equal to zero: \[ \text{Term}_m = \frac{\text{Positive}}{\text{Negative}} < 0 \] Summing up an entire collection of strictly negative terms will inevitably yield a net negative value: \[ E_0^{(2)} < 0 \] Hence, the second-order perturbation correction to the ground state energy is always negative. It pulls the unperturbed ground state energy downward.