Collection of Solved Problems in Physics

Properties of a state with sharp values of L² and L_z

Task number: 4576

• Hint 1

  Recall or look up how to determine the projection of the angular momentum along the direction \(\vec n\). Use the expression of the vector \(\vec n\) in spherical coordinates.

• Solution of hint 1

  The unit vector in the direction of \(\vec n\) in the spherical coordinates has the form

  \[ \vec n = \left (\sin \vartheta \cos \varphi, \, \sin \vartheta \sin \varphi, \, \cos \vartheta \right ) \, . \]

  To determine the projection of the angular momentum along this direction, we use the scalar product

  \[ \hat L_{\vec n} = \hat L \cdot \vec n = \hat L_x \sin \vartheta \cos \varphi + \hat L_y \sin \vartheta \sin \varphi + \hat L_z \cos \vartheta \, . \]

• Hint 2

  Recall or look up the relation for calculating the expectation value of a physical observable \(F\) in a state described by the normalised wavefunction \(\psi\). Next, recall or look up the definition of a linear Hermitian operator.

• Solution to hint 2

  The expectation value of a physical observable \(F\) in a state described by the normalised wavefunction \(\psi\) can be determined from \[\langle F \rangle_{\psi} = \left \langle \psi \, \Big | \, \hat F \psi \right \rangle = \psi^{\dagger} \hat F \psi \, ,\] where \(\hat F\) is the operator associated with the physical observable \(F\) and the dagger denotes the Hermitian adjoint, i.e., the transpose of the matrix followed by complex conjugation.

  Linear operator \(\hat G\) is Hermitian if for any two wavefunctions \(\varphi\) and \(\psi\) from the Hilbert space, the following holds

  \[ \left \langle \varphi \, \Big | \, \hat G \psi \right \rangle = \left \langle \hat G \varphi \, \Big | \, \psi \right \rangle \, . \]

• Hint 3

  Recall or look up what it means for the given particle to be in the state \(\psi = |lm\rangle\) with sharp values of \(L^2\) and \(L_z\).

• Solution to hint 3

  The state \(\psi = |lm\rangle\) with sharp values of \(L^2\) and \(L_z\) means that measurements of these physical observables have zero variance and the expectation values of these observables are equal to the corresponding eigenvalues, i.e., in this state we have

  \[ \left \langle L^2 \right \rangle_\psi = \hbar^2 l \left (l+1 \right ) \, , \] \[ \left \langle L_z \right \rangle_\psi = m\hbar \, . \]

  When calculating the expectation values of other components of angular momentum or their squares, it is therefore useful to express the given operator in terms of \(\hat L_z\) and \(\hat L^2\), whose expectation values are known. This can be done using both the relation between the components and the magnitude of the vector operator, as well as the uncertainty relations between the components of the angular momentum operator.

• Solution a)

  The operator for the projection of angular momentum along the direction \(\vec n\) has the form (see Hint 1)

  \[ \hat L_{\vec n} = \hat L_x \sin \vartheta \cos \varphi + \hat L_y \sin \vartheta \sin \varphi + \hat L_z \cos \vartheta \, . \]

  To determine its expectation value, we need to find the expectation values of the operators \(\hat L_x, \, \hat L_y, \, \hat L_z\). Let us now focus on the \(x\) and \(y\) components. From the commutation relations of the angular momentum components (see Commutators involving components of angular momentum, Answer), we know that

  \[ \hat L_x = \frac{1}{i\hbar}\left( \hat L_y \hat L_z - \hat L_z \hat L_y \right) \, , \] \[ \hat L_y = \frac{1}{i\hbar}\left( \hat L_z \hat L_x - \hat L_x \hat L_z \right) \, . \]

  The expectation value of \(L_x\) in the state \(\psi = |lm\rangle\) is thus given by

  \[ \left \langle L_x \right \rangle_\psi = \left \langle \frac{1}{i\hbar} \left( \hat L_y \hat L_z - \hat L_z \hat L_y \right) \right \rangle_\psi = \frac{1}{i\hbar} \left ( \left \langle \hat L_y \hat L_z \right \rangle_\psi - \left \langle \hat L_z \hat L_y \right \rangle_\psi \right ) \, . \]

  When calculating the expectation values in the parentheses above, we start from the general expression for the expectation value, into which we substitute

  \[ \left \langle \hat L_z \hat L_y \right \rangle_\psi = \left \langle \psi \, \Big | \, \hat L_z \hat L_y \psi \right \rangle \, . \]

  We now use the fact that the components of angular momentum are linear Hermitian operators, which allows us to rewrite (see Hint 2)

  \[ \left \langle \psi \, \Big | \, \hat L_z \hat L_y \psi \right \rangle = \left \langle \hat L_z \psi \, \Big | \, \hat L_y \psi \right \rangle \, . \]

  Since the particle in the state \(\psi = |lm\rangle\) has sharp values of \(L_z\), we can substitute \(\hat L_z \psi = m\hbar \, \psi\), which gives us

  \[ \left \langle \hat L_z \hat L_y \right \rangle_\psi = \left \langle m\hbar \, \psi \, \Big | \, \hat L_y \psi \right \rangle = m\hbar \left \langle \psi \, \Big | \, \hat L_y \psi \right \rangle = m\hbar \, \left \langle L_y \right \rangle_\psi \, . \]

  Analogously, we obtain

  \[ \left \langle \hat L_y \hat L_z \right \rangle_\psi = m\hbar \, \left \langle L_y \right \rangle_\psi \, . \]

  By substituting these expectation values and simplifying, we get

  \[ \left \langle L_x \right \rangle_\psi = \frac{1}{i\hbar} \left ( \left \langle \hat L_y \hat L_z \right \rangle_\psi - \left \langle \hat L_z \hat L_y \right \rangle_\psi \right ) = \frac{1}{i\hbar} \left ( m\hbar \, \left \langle L_y \right \rangle_\psi - m\hbar \, \left \langle L_y \right \rangle_\psi \right ) = 0 \, . \]

  The calculation for \(\left \langle L_y \right \rangle_\psi\) is completely analogous, so we only present the result here

  \[ \left \langle L_y \right \rangle_\psi = 0 \, . \]

  The expression for the expectation value of the angular momentum projection along the direction \(\vec n\) in the state \(\psi = |lm\rangle\) simplifies, after substituting the results obtained above, to

  \[ \langle L_{\vec n} \rangle_\psi = 0 \cdot \sin \vartheta \cos \varphi + 0 \cdot \sin \vartheta \sin \varphi + \langle L_z \rangle_\psi \cos \vartheta = \langle L_z \rangle_\psi \cos \vartheta \, . \]

  We can now again substitute \(\langle L_z\rangle_\psi = m\hbar\), giving the result

  \[ \langle L_{\vec n} \rangle_\psi = m\hbar \cos \vartheta \, . \]

• Solution b)

  Since \(\hat L^2 = \hat L_x^2 + \hat L_y^2 + \hat L_z^2 \,\), it follows that \(\left \langle L^2 \right \rangle = \left \langle L_x^2 \right \rangle + \left \langle L_y^2 \right \rangle + \left \langle L_z^2 \right \rangle\). By symmetry, we deduce that for a particle in the state \(\psi = |lm\rangle\), the expectation values \(\left \langle L_x^2 \right \rangle_\psi\) and \(\left \langle L_y^2 \right \rangle_\psi\) are equal. Hence, we can write

  \[ \left \langle L_x^2 \right \rangle_\psi = \left \langle L_y^2 \right \rangle_\psi = \frac{1}{2} \left ( \left \langle L^2 \right \rangle_\psi - \left \langle L_z^2 \right \rangle_\psi \right ) \, . \]

  Since the particle in the state \(\psi = |lm\rangle\) has sharp values of \(L_z\) and \(L^2\), we can substitute

  \[ \langle L_z\rangle_\psi = m\hbar \, , \] \[ \langle L^2\rangle_\psi = \hbar^2 \, l(l+1) \, , \]

  yielding

  \[ \left \langle L_x^2 \right \rangle_\psi = \left \langle L_y^2 \right \rangle_\psi = \frac{1}{2} \left [ \hbar^2 \, l(l+1) - m^2\hbar^2 \right ] = \frac{\hbar^2}{2} \left [ l(l+1) - m^2 \right ] \, . \]

  Note that in the state \(\psi = |lm\rangle\), the expectation values \(\left \langle L_x \right \rangle_\psi = \left \langle L_y \right \rangle_\psi = 0 \,\), while simultaneously \(\left \langle L_x^2 \right \rangle_\psi = \left \langle L_y^2 \right \rangle_\psi \neq 0 \,\). Therefore, we cannot say that the angular momentum is pointing along the \(z\)‑axis. This observation is discussed in more detail in section Comment.

• Answer

  a) In the state \(\psi = |lm\rangle\) with sharp values of \(L^2\) and \(L_z\), the expectation value of the angular momentum projection along the direction \(\vec n\) is

  \[ \langle L_{\vec n} \rangle_\psi = m\hbar \cos \vartheta \, . \]

  b) In the state \(\psi = |lm\rangle\) with sharp values of \(L^2\) and \(L_z\), the expectation values of \(L^2_x\) and \(L^2_y\) are

  \[ \left \langle L_x^2 \right \rangle_\psi = \left \langle L_y^2 \right \rangle_\psi = \frac{\hbar^2}{2} \left [ l(l+1) - m^2 \right ] \, . \]

• Comment – classical model

  From the calculations above, we know that in the state \(\psi = |lm\rangle\), the expectation values \(\left \langle L_x \right \rangle_\psi = \left \langle L_y \right \rangle_\psi = 0 \,\), while the expectation values \(\left \langle L_x^2 \right \rangle_\psi = \left \langle L_y^2 \right \rangle_\psi\) are non‑zero. The components \(L_x\) and \(L_y\) thus “average out” to zero, but we cannot say that the angular momentum points strictly along the \(z\)‑axis. A classical model for the quantum‑mechanical stationary state \(\psi = |lm\rangle\) is a set of angular momentum vectors randomly distributed on the surface of a cone with apex angle \(\vartheta_m\) (see the figure below). To characterise this model, we need to determine this apex angle.

  

  First, we verify that in this model the components \(L_x\) and \(L_y\) have zero expectation value. The projections along the \(x\)‑ and \(y\)‑axes are, according to our chosen notation,

  \[ L_x = \tilde{L} \cos \varphi \, , \] \[ L_y = \tilde{L} \sin \varphi \, , \]

  where \(\varphi \in \langle 0, \, 2\pi \rangle\).

  The expectation value for \(L_x\) is then

  \[ \langle L_x\rangle = \int_0^{2\pi} \tilde{L} \cos \varphi \, \mathrm{d}\varphi \, . \]

  Here we integrate the cosine function over its full period. The value of the definite integral is therefore zero, i.e.,

  \[ \langle L_x\rangle = 0 \, . \]

  Analogously, we find

  \[ \langle L_y\rangle = 0 \, . \]

  Next, we determine the expectation values \(\left \langle L_x^2 \right \rangle\) and \(\left \langle L_y^2 \right \rangle\). We illustrate the calculation for \(\left \langle L_x^2 \right \rangle\)

  \[ \left \langle L_x^2 \right \rangle = \int_0^{2\pi} \tilde{L}^2 \cos^2 \varphi \, \mathrm{d}\varphi = \tilde{L}^2 \int_0^{2\pi} \frac{1 + \cos 2\varphi}{2} \, \mathrm{d}\varphi = \tilde{L}^2 \int_0^{2\pi} \frac{1}{2} + \frac{\cos 2\varphi}{2} \, \mathrm{d}\varphi \, . \]

  This time we integrate the cosine function over two periods. This part of the definite integral is again zero. Overall, we then obtain

  \[ \left \langle L_x^2 \right \rangle = \pi \tilde{L}^2 \, . \]

  Analogously, we find

  \[ \left \langle L_y^2 \right \rangle = \pi \tilde{L}^2 \, . \]

  In this way, we have verified that the cone model satisfies \(\left \langle L_x \right \rangle = \left \langle L_y \right \rangle = 0 \,\) while the expectation values \(\left \langle L_x^2 \right \rangle = \left \langle L_y^2 \right \rangle\) are non‑zero.

  To determine the apex angle of the cone, we use the results obtained in Solution b). Focusing on half of the apex angle \(\vartheta_m\), we can employ the cosine function and express this angle as

  \[ \cos \frac{\vartheta_m}{2} = \frac{\langle L_z \rangle}{\sqrt{\langle L^2 \rangle}} = \frac{\langle L_z \rangle} {\sqrt{\langle L_x \rangle^2 + \langle L_y \rangle^2 + \langle L_z \rangle^2}} = \] \[ = \frac{m\hbar}{\sqrt{m^2\hbar^2 + \hbar^2 \left [ l \left (l+1 \right ) - m^2 \right ]}} = \frac{m}{\sqrt{m^2 + l \left (l+1 \right ) - m^2 }} = \frac{m}{\sqrt{l \left (l+1 \right )}} \, . \]

  In summary, we can state that the classical model of a quantum‑mechanical stationary state \(|lm\rangle\) is a set of angular momenta randomly distributed on the surface of a cone with apex angle \(\vartheta_m\), for which

  \[ \cos \frac{\vartheta_m}{2} = \frac{m}{\sqrt{l \left (l+1 \right )}} \, . \]