Question

The state of a harmonic oscillator is given as Ψ =\(\frac{1}{ √3}\) 𝜓₀ -\(\frac{1}{ √6}\) 𝜓₁+\(\frac{1}{ √2}\) 𝜓₂ , where 𝜓₀, 𝜓₁ and 𝜓₂ are the normalized wave functions of ground, first excited, and second excited states, respectively. Which of the following statement(s) is/are true?

• A. A measurement of the energy of the system yields 𝐸=\(\frac{1}{2}\) ℏ𝜔 with non-zero probability
• B. A measurement of the energy of the system yields 𝐸 =\(\frac{5}{3}\) ℏ𝜔 with non-zero probability
• C. Expectation value of the energy of the system 〈𝐸〉 = \(\frac{5}{3}\) ℏ𝜔
• D. Expectation value of the energy of the system 〈𝐸〉 = \(\frac{7}{8}\) ℏ𝜔

Answer 1

The Correct Option is A, C

To determine the correct statements about the state \( \Psi =\frac{1}{\sqrt{3}} \psi_0 -\frac{1}{\sqrt{6}} \psi_1 +\frac{1}{\sqrt{2}} \psi_2 \) of a harmonic oscillator and its energy properties, let's analyze each statement one by one.

1. The system consists of different states \( \psi_0 \), \( \psi_1 \), and \( \psi_2 \), each associated with different energy eigenvalues. For a harmonic oscillator, the energies are given by:
   • \( E_0 = \frac{1}{2} \hbar \omega \) (ground state)
   • \( E_1 = \frac{3}{2} \hbar \omega \) (first excited state)
   • \( E_2 = \frac{5}{2} \hbar \omega \) (second excited state)
2. The probability of measuring energy \( E \) is given by the square of the coefficient of the corresponding \( \psi \) state:
   • Probability of \( \psi_0 \): \( \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3} \) (non-zero, so this statement is true).
   • Probability of \( \psi_1 \): \( \left( \frac{1}{\sqrt{6}} \right)^2 = \frac{1}{6} \). Although non-zero, the energy is not \( \frac{5}{3} \hbar \omega \) but \( \frac{3}{2} \hbar \omega \). 
   • Probability of \( \psi_2 \): \( \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} \).
3. To find the expectation value of the energy \( \langle E \rangle \), we compute:
   • \( \langle E \rangle = \left(\frac{1}{3}\right) \cdot \frac{1}{2} \hbar \omega + \left(\frac{1}{6}\right) \cdot \frac{3}{2} \hbar \omega + \left(\frac{1}{2}\right) \cdot \frac{5}{2} \hbar \omega \)
   • \( = \frac{\hbar \omega}{6} + \frac{3\hbar \omega}{12} + \frac{5\hbar \omega}{4} \)
   • \( = \frac{\hbar \omega}{6} + \frac{\hbar \omega}{4} + \frac{5\hbar \omega}{4} \)
   • \( = \frac{\hbar \omega}{6} + \frac{\hbar \omega}{4} + \frac{10\hbar \omega}{8} \)
   • After finding a common denominator and adding, we obtain \( \langle E \rangle = \frac{5}{3} \hbar \omega \). This matches the statement given and confirms it's true.

Based on the above analysis, the correct statements are:

• A measurement of the energy of the system yields \( E=\frac{1}{2} \hbar \omega \) with non-zero probability.
• Expectation value of the energy of the system \( \langle E \rangle = \frac{5}{3} \hbar \omega \).