Question

Consider an operator \(\hat{A}\) which is not Hermitian. Find the possible values of \(c\) and \(d\) such that the operator \((c\hat{A} - d\hat{A}^{\dagger})\) is Hermitian.

• A. \(c = i\) and \(d = i\)
• B. \(c = 1\) and \(d = 1\)
• C. \(c = -1\) and \(d = i\)
• D. \(c = i\) and \(d = -i\)

Hint:

Take the adjoint of \((c\hat{A}-d\hat{A}^{\dagger})\) and match it back to itself; use \((c\hat{A})^{\dagger}=c^{*}\hat{A}^{\dagger}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
An operator \(\hat{X}\) is Hermitian when it equals its own adjoint, \(\hat{X}^{\dagger} = \hat{X}\). For a combination built from an operator \(\hat{A}\) and its adjoint \(\hat{A}^{\dagger}\), work out the adjoint of the whole combination and force it to match the original expression.

Step 2: Key Formula or Approach:
For a complex number \(c\) and an operator \(\hat{A}\), the adjoint follows two rules: \((\hat{A}^{\dagger})^{\dagger} = \hat{A}\), and \((c\hat{A})^{\dagger} = c^{*}\hat{A}^{\dagger}\), with \(c^{*}\) the complex conjugate of \(c\). Since \(\hat{A}\) is not Hermitian, \(\hat{A}\) and \(\hat{A}^{\dagger}\) behave as two genuinely different, independent operators, so their coefficients can be matched separately.

Step 3: Detailed Explanation:
Let \(\hat{X} = c\hat{A} - d\hat{A}^{\dagger}\). Take its adjoint term by term:
\[ \hat{X}^{\dagger} = c^{*}\hat{A}^{\dagger} - d^{*}\hat{A} \]
For \(\hat{X}\) to be Hermitian, \(\hat{X}^{\dagger} = \hat{X}\):
\[ c^{*}\hat{A}^{\dagger} - d^{*}\hat{A} = c\hat{A} - d\hat{A}^{\dagger} \]
Since \(\hat{A}\) and \(\hat{A}^{\dagger}\) are independent, match the coefficient of \(\hat{A}\) on both sides, and separately the coefficient of \(\hat{A}^{\dagger}\):
\[ \text{coefficient of } \hat{A}: \quad -d^{*} = c \]
\[ \text{coefficient of } \hat{A}^{\dagger}: \quad c^{*} = -d \]
Both statements say the same thing, one is the complex conjugate of the other, so the requirement reduces to the single condition:
\[ c = -d^{*} \]

Step 4: Check each option against \(c=-d^*\).
Option (A): \(c=i\), \(d=i\), so \(d^{*}=-i\) and \(-d^{*}=i\), which equals \(c\). This works.
Option (B): \(c=1\), \(d=1\), so \(d^{*}=1\) and \(-d^{*}=-1\), which does not equal \(c=1\). This fails.
Option (C): \(c=-1\), \(d=i\), so \(d^{*}=-i\) and \(-d^{*}=i\), which does not equal \(c=-1\). This fails.
Option (D): \(c=i\), \(d=-i\), so \(d^{*}=i\) and \(-d^{*}=-i\), which does not equal \(c=i\). This fails.

Final Answer:
Only \(c=i\) and \(d=i\) satisfy \(c=-d^{*}\), so this is the only choice that makes the operator Hermitian. \[ \boxed{c=i,\ d=i} \]