Question:
Consider operators \(\hat{A}\), \(\hat{B}\), and \(\hat{C}\) for three observables of a quantum system satisfying \([\hat{A},\hat{B}] = 0\), \([\hat{B},\hat{C}] = 0\), and \([\hat{A},\hat{C}] \neq 0\), with uncertainties \(\Delta A, \Delta B, \Delta C\), respectively. From the options given below, which is/are implied by the commutation relations among \(\hat{A}\), \(\hat{B}\), and \(\hat{C}\)?
Consider operators \(\hat{A}\), \(\hat{B}\), and \(\hat{C}\) for three observables of a quantum system satisfying \([\hat{A},\hat{B}] = 0\), \([\hat{B},\hat{C}] = 0\), and \([\hat{A},\hat{C}] \neq 0\), with uncertainties \(\Delta A, \Delta B, \Delta C\), respectively. From the options given below, which is/are implied by the commutation relations among \(\hat{A}\), \(\hat{B}\), and \(\hat{C}\)?
Show Hint
A zero commutator means the pair can share eigenstates, so their uncertainty product can hit zero; a non-zero commutator blocks a full common eigenbasis, so the pair's uncertainty product must stay positive.
Apply this separately to each pair among \(\hat{A}, \hat{B}, \hat{C}\).
Apply this separately to each pair among \(\hat{A}, \hat{B}, \hat{C}\).
Updated On: Jul 28, 2026
- \(\Delta A \, \Delta B > 0\)
- \(\Delta A \, \Delta C > 0\)
- \(\hat{A}, \hat{B}\) can be simultaneously diagonalized.
- \(\hat{A}, \hat{B}, \hat{C}\) can be simultaneously diagonalized.
Show Solution
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The Correct Option is B, C
Solution and Explanation
Step 1: Recall the generalized uncertainty relation.
For any two Hermitian operators \(\hat{X}\) and \(\hat{Y}\), the Robertson uncertainty relation gives \(\Delta X \, \Delta Y \geq \frac{1}{2}\left|\langle [\hat{X},\hat{Y}] \rangle\right|\). When the commutator of two operators is the zero operator, this lower bound is exactly zero for every state, so their uncertainty product is allowed to fall all the way to zero.
Step 2: Recall the simultaneous diagonalization theorem.
Two Hermitian operators can be simultaneously diagonalized, meaning they share a complete common eigenbasis, if and only if they commute. This is a standard result applied to quantum observables.
Step 3: Check statement (A).
Since \([\hat{A},\hat{B}] = 0\), the lower bound on \(\Delta A \, \Delta B\) is zero. In a common eigenstate of \(\hat{A}\) and \(\hat{B}\), both uncertainties vanish, so \(\Delta A \, \Delta B = 0\) is possible. The relation does not force this product to stay positive in every state, so (A) is FALSE.
Step 4: Check statement (B).
Here \([\hat{A},\hat{C}] \neq 0\), so \(\hat{A}\) and \(\hat{C}\) do not share a complete common eigenbasis. No state can be a simultaneous eigenstate of both, so \(\Delta A\) and \(\Delta C\) cannot both be driven to zero together. This is what the non-zero commutator between \(\hat{A}\) and \(\hat{C}\) implies, so \(\Delta A \, \Delta C > 0\) and (B) is TRUE.
Step 5: Check statement (C).
Since \([\hat{A},\hat{B}] = 0\), Step 2's theorem applies directly: \(\hat{A}\) and \(\hat{B}\) commute, so they can be simultaneously diagonalized. Statement (C) is TRUE.
Step 6: Check statement (D).
If \(\hat{A}, \hat{B}, \hat{C}\) could all be simultaneously diagonalized, they would share one common eigenbasis, and any two operators sharing a common eigenbasis must commute, including \(\hat{A}\) and \(\hat{C}\). This contradicts \([\hat{A},\hat{C}] \neq 0\). So all three cannot be simultaneously diagonalized, and (D) is FALSE.
Final Answer:
Only \(\hat{A}\), \(\hat{B}\) commute in a way that also blocks a full common eigenbasis for the trio; (B) and (C) are what the given commutation relations actually force. \[ \boxed{\text{B, C}} \]
For any two Hermitian operators \(\hat{X}\) and \(\hat{Y}\), the Robertson uncertainty relation gives \(\Delta X \, \Delta Y \geq \frac{1}{2}\left|\langle [\hat{X},\hat{Y}] \rangle\right|\). When the commutator of two operators is the zero operator, this lower bound is exactly zero for every state, so their uncertainty product is allowed to fall all the way to zero.
Step 2: Recall the simultaneous diagonalization theorem.
Two Hermitian operators can be simultaneously diagonalized, meaning they share a complete common eigenbasis, if and only if they commute. This is a standard result applied to quantum observables.
Step 3: Check statement (A).
Since \([\hat{A},\hat{B}] = 0\), the lower bound on \(\Delta A \, \Delta B\) is zero. In a common eigenstate of \(\hat{A}\) and \(\hat{B}\), both uncertainties vanish, so \(\Delta A \, \Delta B = 0\) is possible. The relation does not force this product to stay positive in every state, so (A) is FALSE.
Step 4: Check statement (B).
Here \([\hat{A},\hat{C}] \neq 0\), so \(\hat{A}\) and \(\hat{C}\) do not share a complete common eigenbasis. No state can be a simultaneous eigenstate of both, so \(\Delta A\) and \(\Delta C\) cannot both be driven to zero together. This is what the non-zero commutator between \(\hat{A}\) and \(\hat{C}\) implies, so \(\Delta A \, \Delta C > 0\) and (B) is TRUE.
Step 5: Check statement (C).
Since \([\hat{A},\hat{B}] = 0\), Step 2's theorem applies directly: \(\hat{A}\) and \(\hat{B}\) commute, so they can be simultaneously diagonalized. Statement (C) is TRUE.
Step 6: Check statement (D).
If \(\hat{A}, \hat{B}, \hat{C}\) could all be simultaneously diagonalized, they would share one common eigenbasis, and any two operators sharing a common eigenbasis must commute, including \(\hat{A}\) and \(\hat{C}\). This contradicts \([\hat{A},\hat{C}] \neq 0\). So all three cannot be simultaneously diagonalized, and (D) is FALSE.
Final Answer:
Only \(\hat{A}\), \(\hat{B}\) commute in a way that also blocks a full common eigenbasis for the trio; (B) and (C) are what the given commutation relations actually force. \[ \boxed{\text{B, C}} \]
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