Let's solve the **Virial Theorem** problem step by step. --- ### **(a) Show that for any energy eigenstate, the expectation value of the momentum operator \(\langle \hat{p} \rangle\) is always zero.** #### **Step 1: Hamiltonian and eigenstates** The Hamiltonian is: \[ \hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{q}) \] Energy eigenstates satisfy: \[ \hat{H} |\psi_k\rangle = E_k |\psi_k\rangle \] #### **Step 2: Consider the commutator \([\hat{q}, \hat{H}]\)** \[ [\hat{q}, \hat{H}] = \left[\hat{q}, \frac{\hat{p}^2}{2m} + V(\hat{q})\right] = \frac{1}{2m} [\hat{q}, \hat{p}^2] + [\hat{q}, V(\hat{q})] \] But \([\hat{q}, V(\hat{q})] = 0\) (since both are functions of \(\hat{q}\)), so: \[ [\hat{q}, \hat{H}] = \frac{1}{2m} [\hat{q}, \hat{p}^2] \] Recall: \([\hat{q}, \hat{p}^2] = \hat{q}\hat{p}^2 - \hat{p}^2\hat{q}\) We use the identity: \([A, BC] = [A, B]C + B[A, C]\) \[ [\hat{q}, \hat{p}^2] = [\hat{q}, \hat{p}]\hat{p} + \hat{p}[\hat{q}, \hat{p}] = (i\hbar)\hat{p} + \hat{p}(i\hbar) = 2i\hbar \hat{p} \] So, \[ [\hat{q}, \hat{H}] = \frac{1}{2m} (2i\hbar \hat{p}) = \frac{i\hbar}{m}\hat{p} \] #### **Step 3: Take expectation value in energy eigenstate** \[ \langle\psi_k|[\hat{q}, \hat{H}]|\psi_k\rangle = \frac{i\hbar}{m}\langle\psi_k|\hat{p}|\psi_k\rangle \] But for any Hermitian operator \(A\) and eigenstate \(|\psi_k\rangle\) of \(H\), \(\langle\psi_k|[A,H]|\psi_k\rangle = 0\), because: \[ \langle\psi_k|[A,H]|\psi_k\rangle = \langle\psi_k|A H|\psi_k\rangle - \langle\psi_k|H A|\psi_k\rangle = E_k \langle\psi_k|A|\psi_k\rangle - E_k \langle\psi_k|A|\psi_k\rangle = 0 \] Therefore, \[ \frac{i\hbar}{m}\langle\psi_k|\hat{p}|\psi_k\rangle = 0 \implies \langle\psi_k|\hat{p}|\psi_k\rangle = 0 \] **Final Answer (a):** \[ \boxed{\langle \hat{p} \rangle = 0} \] This means the average momentum for any bound state is zero. This is expected, as the particle has no preferred direction in a bound state. --- ### **(b) Show the virial theorem for \(V(\hat{q}) = a\hat{q}^n\): \(\langle T \rangle = \frac{n}{2} \langle V \rangle\)** #### **Step 1: Consider the commutator \([\hat{p}\hat{q}, \hat{H}]\)** But more generally, we use \([\hat{p}\hat{q}, \hat{H}]\) or \([\hat{p}\hat{q}, \hat{H}]\), but the hint suggests \([\hat{p}\hat{q}, \hat{H}]\). Let’s use the operator \(A = \hat{p}\hat{q}\): \[ \langle\psi_k|[\hat{p}\hat{q}, \hat{H}]|\psi_k\rangle = 0 \] #### **Step 2: Compute the commutator** \[ [\hat{p}\hat{q}, \hat{H}] = [\hat{p}\hat{q}, \frac{\hat{p}^2}{2m}] + [\hat{p}\hat{q}, V(\hat{q})] \] First term: \[ [\hat{p}\hat{q}, \hat{p}^2] = \hat{p}[\hat{q}, \hat{p}^2] + [\hat{p}, \hat{p}^2]\hat{q} \] But \([\hat{p}, \hat{p}^2] = 0\), so: \[ [\hat{p}\hat{q}, \hat{p}^2] = \hat{p}[\hat{q}, \hat{p}^2] \] We already have \([\hat{q}, \hat{p}^2] = 2i\hbar\hat{p}\), so: \[ [\hat{p}\hat{q}, \hat{p}^2] = \hat{p}(2i\hbar\hat{p}) = 2i\hbar\hat{p}^2 \] So, \[ [\hat{p}\hat{q}, \frac{\hat{p}^2}{2m}] = \frac{1}{2m} \cdot 2i\hbar \hat{p}^2 = \frac{i\hbar}{m}\hat{p}^2 \] Second term: \[ [\hat{p}\hat{q}, V(\hat{q})] = \hat{p}[\hat{q}, V(\hat{q})] + [\hat{p}, V(\hat{q})]\hat{q} \] But \([\hat{q}, V(\hat{q})] = 0\), and \([\hat{p}, V(\hat{q})] = -i\hbar V'(\hat{q})\), so: \[ [\hat{p}\hat{q}, V(\hat{q})] = 0 + (-i\hbar V'(\hat{q}))\hat{q} = -i\hbar \hat{q} V'(\hat{q}) \] So the total commutator: \[ [\hat{p}\hat{q}, \hat{H}] = \frac{i\hbar}{m}\hat{p}^2 - i\hbar \hat{q} V'(\hat{q}) \] #### **Step 3: Take expectation value in eigenstate** \[ \langle\psi_k|[\hat{p}\hat{q}, \hat{H}]|\psi_k\rangle = 0 \] \[ \frac{i\hbar}{m} \langle \hat{p}^2 \rangle - i\hbar \langle \hat{q} V'(\hat{q}) \rangle = 0 \] \[ \frac{1}{m} \langle \hat{p}^2 \rangle = \langle \hat{q} V'(\hat{q}) \rangle \] But kinetic energy: \[ T = \frac{\hat{p}^2}{2m} \implies \langle \hat{p}^2 \rangle = 2m \langle T \rangle \] So: \[ 2 \langle T \rangle = \langle \hat{q} V'(\hat{q}) \rangle \] For \(V(\hat{q}) = a\hat{q}^n \implies V'(\hat{q}) = an\hat{q}^{n-1}\): \[ \hat{q} V'(\hat{q}) = a n \hat{q}^n = n V(\hat{q}) \] So: \[ 2\langle T \rangle = n \langle V \rangle \] \[ \boxed{\langle T \rangle = \frac{n}{2} \langle V \rangle} \] --- ## **Summary of Final Answers** ### (a) \[ \boxed{\langle \hat{p} \rangle = 0} \] For any energy eigenstate, the expectation value of momentum is zero. ### (b) \[ \boxed{\langle T \rangle = \frac{n}{2} \langle V \rangle} \] For \(V(\hat{q}) = a\hat{q}^n\), the kinetic and potential energies satisfy the virial theorem.

Certainly! Let's go through each part carefully, providing detailed calculations, explanations, and the final answers. --- ## **Part (a): Show that for any energy eigenstate, \(\langle \hat{p} \rangle = 0\)** ### **Step 1: Understand the Hamiltonian and eigenstates** Given the Hamiltonian: \[ \hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{q}) \] and the eigenstates: \[ \hat{H} |\psi_k\rangle = E_k |\psi_k\rangle \] ### **Step 2: Use the commutator \([\hat{q}, \hat{H}]\)** The goal is to relate the expectation value of \(\hat{p}\) to some commutator. Recall: \[ [\hat{q}, \hat{H}] = \hat{q}\hat{H} - \hat{H}\hat{q} \] ### **Step 3: Calculate \([\hat{q}, \hat{H}]\)** Since \(\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{q})\), and \(\hat{q}\) commutes with \(V(\hat{q})\), we have: \[ [\hat{q}, \hat{H}] = [\hat{q}, \frac{\hat{p}^2}{2m}] + [\hat{q}, V(\hat{q})] \] \[ = [\hat{q}, \frac{\hat{p}^2}{2m}] + 0 \] because \([\hat{q}, V(\hat{q})] = 0\). Now, focus on \([\hat{q}, \hat{p}^2]\): \[ [\hat{q}, \hat{p}^2] = \hat{q}\hat{p}^2 - \hat{p}^2 \hat{q} \] Using the property: \[ [\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}] \] we get: \[ [\hat{q}, \hat{p}^2] = [\hat{q}, \hat{p}]\hat{p} + \hat{p}[\hat{q}, \hat{p}] \] Recall the canonical commutation relation: \[ [\hat{q}, \hat{p}] = i \hbar \] which is a c-number and commutes with \(\hat{p}\), so: \[ [\hat{q}, \hat{p}^2] = (i\hbar)\hat{p} + \hat{p}(i\hbar) = 2 i \hbar \hat{p} \] Therefore, \[ [\hat{q}, \hat{H}] = \frac{1}{2m} \times 2 i \hbar \hat{p} = \frac{i \hbar}{m} \hat{p} \] ### **Step 4: Take the expectation value in an eigenstate** Now, consider the expectation value: \[ \langle \psi_k | [\hat{q}, \hat{H}] | \psi_k \rangle = \frac{i \hbar}{m} \langle \psi_k | \hat{p} | \psi_k \rangle \] But, for any eigenstate \(|\psi_k\rangle\): \[ \langle \psi_k | [\hat{q}, \hat{H}] | \psi_k \rangle = 0 \] because: \[ \langle \psi_k | [\hat{q}, \hat{H}] | \psi_k \rangle = \langle \psi_k |\hat{q} \hat{H} - \hat{H} \hat{q} |\psi_k \rangle = E_k \langle \psi_k |\hat{q}| \psi_k \rangle - E_k \langle \psi_k |\hat{q}| \psi_k \rangle = 0 \] which is generally zero since the expectation value of a commutator in an eigenstate is zero for a Hermitian operator. Hence, \[ 0 = \frac{i \hbar}{m} \langle \hat{p} \rangle \] implying: \[ \boxed{ \langle \hat{p} \rangle = 0 } \] ### **Final interpretation:** **The average momentum in any bound eigenstate is zero.** This makes physical sense: in a bound state, the particle oscillates around a point with no net directional bias, so the average momentum cancels out to zero. --- ## **Part (b): Show that for \(V(q) = a q^n\), the expectation values satisfy \(\langle T \rangle = \frac{n}{2} \langle V \rangle\)** ### **Step 1: Use the commutator involving \(\hat{p}\hat{q}\)** Define the operator: \[ A = \hat{p} \hat{q} \] We will evaluate \(\langle [A, \hat{H}] \rangle\). Since the eigenstates are stationary, the expectation value of the commutator in an eigenstate is zero: \[ \langle [A, \hat{H}] \rangle = 0 \] ### **Step 2: Compute \([\hat{p} \hat{q}, \hat{H}]\)** Recall: \[ \hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{q}) \] We evaluate: \[ [\hat{p} \hat{q}, \hat{H}] = [\hat{p} \hat{q}, \frac{\hat{p}^2}{2m}] + [\hat{p} \hat{q}, V(\hat{q})] \] --- ### **Part 2a: Compute \([\hat{p} \hat{q}, \frac{\hat{p}^2}{2m}]\)** Using the property: \[ [\hat{A}\hat{B}, \hat{C}] = \hat{A}[\hat{B}, \hat{C}] + [\hat{A}, \hat{C}]\hat{B} \] we get: \[ [\hat{p} \hat{q}, \hat{p}^2] = \hat{p} [\hat{q}, \hat{p}^2] + [\hat{p}, \hat{p}^2]\hat{q} \] We already know: \[ [\hat{q}, \hat{p}^2] = 2 i \hbar \hat{p} \] and \[ [\hat{p}, \hat{p}^2] = 0 \] Thus, \[ [\hat{p} \hat{q}, \hat{p}^2] = \hat{p} (2 i \hbar \hat{p}) + 0 = 2 i \hbar \hat{p}^2 \] Dividing by \(2m\): \[ [\hat{p} \hat{q}, \frac{\hat{p}^2}{2m}] = \frac{1}{2m} \times 2 i \hbar \hat{p}^2 = \frac{i \hbar}{m} \hat{p}^2 \] --- ### **Part 2b: Compute \([\hat{p} \hat{q}, V(\hat{q})]\)** Since: \[ [\hat{p} \hat{q}, V(\hat{q})] = \hat{p} [\hat{q}, V(\hat{q})] + [\hat{p}, V(\hat{q})] \hat{q} \] \[ [\hat{q}, V(\hat{q})] = 0 \] and \[ [\hat{p}, V(\hat{q})] = - i \hbar V'(\hat{q}) \] Therefore, \[ [\hat{p} \hat{q}, V(\hat{q})] = 0 + (- i \hbar V'(\hat{q})) \hat{q} = - i \hbar \hat{q} V'(\hat{q}) \] --- ### **Step 3: Sum the two parts** \[ [\hat{p} \hat{q}, \hat{H}] = \frac{i \hbar}{m} \hat{p}^2 - i \hbar \hat{q} V'(\hat{q}) \] --- ### **Step 4: Expectation value in an eigenstate** Since the expectation value of the commutator is zero: \[ 0 = \left\langle [\hat{p} \hat{q}, \hat{H}] \right\rangle = \frac{i \hbar}{m} \langle \hat{p}^2 \rangle - i \hbar \left\langle \hat{q} V'(\hat{q}) \right\rangle \] Dividing through by \(i \hbar\): \[ 0 = \frac{1}{m} \langle \hat{p}^2 \rangle - \left\langle \hat{q} V'(\hat{q}) \right\rangle \] Recall: \[ \langle T \rangle = \frac{1}{2m} \langle \hat{p}^2 \rangle \implies \langle \hat{p}^2 \rangle = 2 m \langle T \rangle \] Substitute: \[ 0 = \frac{1}{m} \times 2 m \langle T \rangle - \left\langle \hat{q} V'(\hat{q}) \right\rangle \] \[ 0 = 2 \langle T \rangle - \left\langle \hat{q} V'(\hat{q}) \right\rangle \] \[ \boxed{ 2 \langle T \rangle = \left\langle \hat{q} V'(\hat{q}) \right\rangle } \] --- ### **Step 5: For \(V(q) = a q^n\)** Calculate \(V'(\hat{q})\): \[ V'(\hat{q}) = a n \hat{q}^{n-1} \] and hence: \[ \hat{q} V'(\hat{q}) = a n \hat{q}^n = n V(\hat{q}) \] Taking expectation values: \[ \left\langle \hat{q} V'(\hat{q}) \right\rangle = n \langle V \rangle \] So, \[ 2 \langle T \rangle = n \langle V \rangle \] which gives the **virial theorem**: \[ \boxed{ \langle T \rangle = \frac{n}{2} \langle V \rangle } \] --- ## **Final Summary:** ### **(a)** \[ \boxed{ \langle \hat{p} \rangle = 0 } \] ### **(b)** \[ \boxed{ \langle T \rangle = \frac{n}{2} \langle V \rangle } \] This relation holds for a potential \(V(q) = a q^n\), illustrating the virial theorem for such systems. --- Let me know if you'd like further clarification!