Physics Problems & Solutions: Hermitian

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Quantum Mechanics - Hermitian Operator

The eigenvalues of a Hermitian operator are always

A. Real
B. Imaginary
C. Degenerate
D. Linear
E. Positive

(GR0177 #27)

Solution:

The eigenvalues are always real and hence observable.
Proof:
$\begin{aligned} A|\psi\rangle&=a|\psi\rangle \\A^\dagger|\psi\rangle&=a^*|\psi\rangle \\(A-A^\dagger)|\psi\rangle&=(a-a^*)|\psi\rangle \end{aligned}$

Hermitian operator: $A^\dagger=A$ .
So, $a-a^*=0\rightarrow a=a^* \rightarrow$ real

Answer: A

Quantum Mechanics - Hermitian Matrix

The matrix $A= \left| \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right|$ has 3 eigenvalues $\lambda_i$ defined by $Av_i=\lambda_i v_i$ . Which of the following statements is NOT true?

A. $\lambda_1 + \lambda_2 + \lambda_3 = 0$
B. $\lambda_1, \lambda_2,$ and $\lambda_3$ are all the real numbers
C. $\lambda_1\lambda_2 = + 1$ for some pair roots
D. $\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_3\lambda_1= 0$
E. $\lambda_i^3 =+1, i =1,2,3$

(GR9277 #98)

Solution:

Every real symmetric matrix is Hermitian, and therefore all its eigenvalues are real.

A symmetric matrix is a square matrix that is equal to its transpose.
$A=A^T \rightarrow a_{ij} = a_{ji}$

Matrix $A = \left| \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right|$

is not symmetric, since $a_{ij} \neq a_{ji}$ . Therefore, its eigenvalues are NOT real.

Answer: B

Quantum Mechanics – Ladder Operator

The operator, $\hat a = \sqrt{\frac{m \omega_0}{2 \hbar}} \big(\hat x+i\frac{\hat p}{m \omega_0}\big)$ when operating on a harmonic energy eigenstate ψ_n with energy E_n, produces another energy eigenstate whose energy is E_n− ħω₀. Which of the following is true?

I. $\hat a$ commutes with the Hamiltonian.

II. $\hat a$ is a Hermitian operator and therefore an observable.

III. The adjoint operator $\hat a^{\dagger}\neq \hat a$

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

(GR9677 #100)

Solution:

I. FALSE
Commutes if [H, a] = 0
But a is a ladder operator, a raises the energy level so that [H, a] = − ħωa

II. FALSE
Hermitian operator if $\hat a=\hat a^{\dagger}$

$\hat a = \sqrt{\frac{m \omega_0}{2 \hbar}} \big(\hat x+i\frac{\hat p}{m \omega_0}\big)$

$\hat a^{\dagger} = \sqrt{\frac{m \omega_0}{2 \hbar}} \big(\hat x-i\frac{\hat p}{m \omega_0}\big)\neq\hat a$

III. TRUE
See II.

Answer: C