A particle is moving in a circle in the \(x y\) plane. The only coordinate of importance is the angle \(\varphi\) which can vary from 0 to \(2 \pi\) as the particle goes around the circle. We are interested in measurements of the angular momentum \(l_{z}\) of the particle. The angular momentum operator for such a system is given by \((\hbar / \mathrm{i}) \mathrm{d} / \mathrm{d} \varphi\) (a) Suppose that the state of the particle is described by the wavefunction \(\psi(\varphi)=N \mathrm{e}^{-\mathrm{i} \varphi}\) where \(N\) is the normalization constant. What values will we find when we measure the angular momentum of the particle? If more than one value is possible, what is the probability of obtaining each result? What is the expectation value of the angular momentum? (b) Now suppose that the state of the particle is described by the normalized wavefunction \(\psi(\varphi)=N\left\\{(3 / 4)^{1 / 2} \mathrm{e}^{-i \varphi}-(\mathrm{i} / 2) \mathrm{e}^{2 \text { ip }}\right\\} .\) When we measure the angular momentum of the particle, what values will we find? If more than one value is possible, what is the probability of obtaining each result? What is the expectation value of the angular momentum?

Short Answer

Expert verified
For the first case, the single possible value for angular momentum is \( l_{z} = -\hbar \) with a probability of 1 and an expectation value of \( -\hbar \). For the second case, there are two possible values: \( l_{z1} = -\hbar \) with a probability of 3/4, and \( l_{z2} = 2\hbar \) with a probability of 1/4. The expectation value is \( = -1/2 \hbar \) .

Step by step solution

01

Determine Angular Momentum Value

The angular momentum operator is \((\hbar / i) \mathrm{d} / \mathrm{d} \varphi\). Applying this to the wavefunction \(\psi=Ne^{-i\varphi}\) yields \((\hbar / i) \mathrm{d} / \mathrm{d} \varphi Ne^{-i\varphi} = -\hbar \psi\). So the angular momentum value is \(l_{z} = -\hbar\).
02

Compute Probabilities and Expectation Value

Since there is only one possible value for the angular momentum, that is -\hbar, the probability of obtaining it is 1 and the expectation value is therefore -\hbar.
03

Determine Angular Momentum Values for Second Case

For the second case, applying the angular momentum operator to the given normalized wavefunction \(\psi(\varphi)=N((3 / 4)^{1 / 2} e^{-i \varphi}-(i / 2) e^{2 i\varphi})\) yields two possible values for angular momentum: \(l_{z1} = -\hbar\) and \(l_{z2} = 2\hbar\).
04

Compute Probabilities and Expectation Value

The probability of obtaining each result is found by taking the square of the coefficient of the corresponding term in the wavefunction. This yields a probability of 3/4 for \(l_{z1} = -\hbar\) and 1/4 for \(l_{z2} = 2\hbar\). The expectation value is given by following formula: \( = \sum pi \cdot li \) , where \( pi \) is the probability and \( li \) is the angular momentum. Substituting values this yields the expectation value is \( = -1/2 \hbar \) .

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