Discussion
A particle is in the normalized state \(|\psi\rangle\). \(|\psi\rangle\) is a superposition of energy eigen states \(|E_{0} = 10\ \text{eV}\rangle\) and \(|E_{1} = 30\ \text{eV}\rangle\). The average value of energy of the particle in the state \(|\psi\rangle\) is 20 eV. The state \(|\psi\rangle\) is :
Answer: D
The state is \(\frac{1}{\sqrt{2}}|E_{0}\rangle - \frac{1}{\sqrt{2}}|E_{1}\rangle\) (option D).
Write \(|\psi\rangle = c_{0}|E_{0}\rangle + c_{1}|E_{1}\rangle\). Normalisation and the given average energy require:
\[ \begin{aligned} |c_{0}|^{2} + |c_{1}|^{2} &= 1 \\ 10|c_{0}|^{2} + 30|c_{1}|^{2} &= 20 \end{aligned} \]
so \(|c_{0}|^{2} = |c_{1}|^{2} = 1/2\), and \(|c_{0}| = |c_{1}| = 1/\sqrt{2}\). The relative sign is a phase and does not affect the average energy, so option D fits. Options A and C are not normalised (\(1/4 + 3/16 = 7/16\)), and option B is normalised but gives \(10/3 + 20 = 23.3\) eV.
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