Collection of Solved Problems in Physics

Matrix formalism for l = 1

Task number: 4578

• Hint 1

  Recall or look up how to determine whether two states are orthogonal.

• Solution to hint 1

  For two mutually orthogonal states \(\psi_j, \, \psi_k\) in the coordinate representation, the following must hold

  \[ \left \langle \psi_j \, | \, \psi_k \right \rangle = \int \psi_j^* \, \psi_k \, \mathrm{d}V = 0 \, , \]

  where the asterisk denotes complex conjugation and the integral is over the entire space.

  For two mutually orthogonal states \(\psi_j, \, \psi_k\) in the matrix representation, the analogous condition is

  \[ \left \langle \psi_j \, | \, \psi_k \right \rangle = \psi_j^\dagger \, \psi_k = 0 \, , \]

  where the dagger denotes the Hermitian adjoint, i.e., the transpose of the vector followed by complex conjugation.

• Hint 2

  Recall or look up how to determine a matrix element using the vectors of the corresponding canonical basis.

• Solution to hint 2

  We consider an operator \(\hat F\) represented by a square matrix \(F\). If we denote its elements by \(F_{jk}\), then

  \[ F_{jk} = \left \langle \vec {e_j} \, \Big | \, F_k \right \rangle = \left \langle \vec {e_j} \, \Big | \, F \, \Big | \, \vec {e_k} \right \rangle \, , \]

  where \(\vec {e_j}\) is the \(j\)‑th vector of the corresponding canonical basis, and \(F_k = F \, \vec {e_k}\) is the product of the matrix \(F\) with the \(k\)‑th vector of the corresponding canonical basis.

• Solution a)

  Since we have three basis vectors, we immediately see that the dimension of this Hilbert space is \(3\).

  Now let us verify the orthogonality of the given states in the coordinate representation. If two states \(\psi_j, \, \psi_k\) are mutually orthogonal, then

  \[ \left \langle \psi_j \, | \, \psi_k \right \rangle = \int \psi_j^* \, \psi_k \, \mathrm{d}V = 0 \, . \]

  We will integrate over the entire space of the variables \(\theta\) and \(\varphi\), i.e., over the sphere. The “volume” element \(\mathrm{d}V\) therefore takes the form

  \[ \mathrm{d}V = \sin \theta \, \mathrm{d}\theta \, \mathrm{d}\varphi \, . \]

  The final step before the explicit calculation is to set the integration limits. Since we are integrating over the entire space, \(\theta\) ranges from \(0\) to \(π\) and \(\varphi\) takes values from \(0\) to \(2π\).

  We perform the calculation for the pair of states \(\psi_1, \, \psi_0\)

  \[ \left \langle \psi_1 \, | \, \psi_0 \right \rangle = \int \psi_1^* \, \psi_0 \, \mathrm{d}V = \] \[ = \int_0^{2\pi} \int_0^\pi \left( -\frac{1}{2} \sqrt{\frac{3}{2\pi}} \right) \sin \theta \, e^{-i \varphi} \, \left( \frac{1}{2} \sqrt{\frac{3}{\pi}} \right) \cos \theta \, \sin \theta \, \mathrm{d}\theta \, \mathrm{d}\varphi = \] \[ = - \frac{3}{4\pi\sqrt 2} \int_0^{2\pi} \int_0^\pi \sin^2 \theta \, \cos \theta \, e^{-i\varphi} \, \mathrm{d}\theta \, \mathrm{d}\varphi \, . \]

  Since \(e^{-i\varphi}\) can be rewritten as \(\cos \varphi - i\sin \varphi\), it is clear that the integral over \(\varphi\) involves sine and cosine functions over their full periods. These definite integrals are equal to zero, so we have

  \[ \left \langle \psi_1 \, | \, \psi_0 \right \rangle = \int \psi_1^* \, \psi_0 \, \mathrm{d}V = 0 \, . \]

  For the remaining two pairs of states, the calculation is analogous. In both cases, we would integrate an integer power of the function \(e^{i\varphi}\) over \(\varphi\) from zero to \(2\pi\). As above, these definite integrals are equal to zero. Altogether, we have

  \[ \left \langle \psi_1 \, | \, \psi_0 \right \rangle = \left \langle \psi_1 \, | \, \psi_{-1} \right \rangle = \left \langle \psi_{-1} \, | \, \psi_0 \right \rangle = 0 \, , \]

  meaning that all the given states in the coordinate representation are mutually orthogonal.

  Next, we verify the orthogonality of the given states in the matrix representation. For two states \(\psi_j, \, \psi_k\) to be orthogonal, the following must hold

  \[ \left \langle \psi_j \, | \, \psi_k \right \rangle = \psi_j^\dagger \, \psi_k = 0 \, . \]

  We perform the calculation for the pair of states \(\psi_1, \, \psi_0\)

  \[ \left \langle \psi_1 \, | \, \psi_0 \right \rangle = \psi_1^\dagger \, \psi_0 = \begin{pmatrix} 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = 0 \, . \]

  For the remaining two pairs of states, the calculation is analogous. Thus, we have verified that all given states in the matrix representation are mutually orthogonal.

• Solution b)

  An operator acts on a state and produces another state. In this representation, a state is represented by a three‑component vector. To produce another state from it, we must act on it with a \(3 \times 3\) matrix.

  We will show the calculation of the individual matrix elements for the operator \(\hat L_z\). The calculation for the remaining operators is analogous.

  Let us denote the matrix elements of this operator as

  \[ \hat L_z = \begin{pmatrix} A & D & G \\ B & E & H \\ C & F & I \end{pmatrix} \, . \]

  To determine the individual matrix elements, let us act with this operator on a chosen eigenstate and expand the result as a linear combination of all eigenstates. Let us now apply the operator \(\hat L_z\) to the eigenstate \(|1, \, 1\rangle\)

  \[ \hat L_z |1, \, 1\rangle = \begin{pmatrix} A & D & G \\ B & E & H \\ C & F & I \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} A \\ B \\ C \end{pmatrix} \, . \]

  We now expand this state as a linear combination of all eigenstates

  \[ \hat L_z |1, \, 1\rangle = \begin{pmatrix} A \\ B \\ C \end{pmatrix} = A |1, \, 1\rangle + B |1, \, 0\rangle + C |1, \, -1\rangle \, . \]

  To determine the element \(A\), we calculate the scalar product

  \[ A = \left \langle 1, \, 1 \, \left | \, \hat L_z \, \right | \, 1, \, 1 \right \rangle \, . \]

  We calculate this scalar product in the coordinate representation, where we know both the state and the operator \(\hat L_z\). Since the states are expressed in spherical coordinates, we also use the representation of \(\hat L_z\) in these coordinates

  \[ \hat L_z = -i\hbar \frac{\partial}{\partial \varphi} \, . \]

  Before computing the matrix element \(A\), we first apply \(\hat L_z\) to the state \(|1, \, 1\rangle\)

  \[ \hat L_z |1, \, 1\rangle = -i\hbar \frac{\partial}{\partial \varphi} \left [ \left( -\frac{1}{2} \sqrt{\frac{3}{2\pi}} \right) \sin \theta \, e^{i \varphi} \right ] = - \frac{\hbar}{2} \sqrt{\frac{3}{2\pi}} \, \sin \theta \, e^{i \varphi} \, . \]

  We substitute this result into the scalar product

  \[ A = \left \langle 1, \, 1 \, \left | \, \hat L_z \, \right | \, 1, \, 1 \right \rangle = \int \psi_1^* \, \hat L_z \psi_1 \, \mathrm{d}V \, . \]

  The integration will be carried out over the entire range of the variables \(\theta\) and \(\varphi\). The “volume” element \(\mathrm{d}V\) in spherical coordinates is

  \[ \mathrm{d}V = \sin \theta \, \mathrm{d}\theta \, \mathrm{d}\varphi \, . \]

  The final step before the integration is to set the integration limits. The coordinate \(\theta\) ranges from zero to \(π\), and the coordinate \(\varphi\) ranges from zero to \(2π\).

  Now we perform the direct calculation of the matrix element \(A\)

  \[ A = \left \langle 1, \, 1 \, \left | \, \hat L_z \, \right | \, 1, \, 1 \right \rangle = \] \[ = \int_0^{2\pi} \int_0^\pi \left [ \left( -\frac{1}{2} \sqrt{\frac{3}{2\pi}} \right) \sin \theta \, e^{- i \varphi} \right ] \left [ - \frac{\hbar}{2} \sqrt{\frac{3}{2\pi}} \, \sin \theta \, e^{i \varphi} \right ] \, \sin \theta \, \mathrm{d}\theta \, \mathrm{d}\varphi = \] \[ = \frac{3\hbar}{8\pi} \int_0^{2\pi} \int_0^\pi \sin^3 \theta \, e^{- i \varphi} \, e^{i \varphi} \, \mathrm{d}\theta \, \mathrm{d}\varphi = \frac{3\hbar}{8\pi} 2\pi \int_0^\pi \sin^3 \theta \, \mathrm{d}\theta = \] \[ = \frac{3\hbar}{4} \int_0^\pi \left( 1 - \cos^2 \theta \right) \sin \theta \, \mathrm{d}\theta \, . \]

  Let us perform the substitution \(\cos \theta = t\), which gives \(- \sin \theta \, \mathrm{d}\theta = \mathrm{d} t\). The integration limits transform as \(\theta = 0 \rightarrow t = 1\) and \( \, \theta = \pi \rightarrow t = -1\).

  After substitution, we get

  \[ A = \frac{3\hbar}{4} \int_1^{-1} - (1 - t^2) \, \mathrm{d} t = \frac{3\hbar}{4} \int_{-1}^1 1 - t^2 \, \mathrm{d} t = \frac{3\hbar}{4} \left [ t - \frac{t^3}{3} \right ]_{-1}^1 = \] \[ = \frac{3\hbar}{4} \frac{4}{3} = \hbar \, . \]

  We now determine the matrix element \(B\) in an analogous way

  \[ B = \left \langle 1, \, 0 \, \left | \, \hat L_z \, \right | \, 1, \, 1 \right \rangle = \] \[ = \int_0^{2\pi} \int_0^\pi \left [ \left( \frac{1}{2} \sqrt{\frac{3}{\pi}} \right) \cos \theta \right ] \left [ - \frac{\hbar}{2} \sqrt{\frac{3}{2\pi}} \, \sin \theta \, e^{i \varphi} \right ] \, \sin \theta \, \mathrm{d}\theta \, \mathrm{d}\varphi = \] \[ = - \frac{3\hbar}{4\pi\sqrt2} \int_0^{2\pi} \int_0^\pi \sin^2 \theta \, \cos \theta \, e^{i \varphi} \, \mathrm{d}\theta \, \mathrm{d}\varphi \, . \]

  Here we integrate the function \(e^{i \varphi}\) from zero to \(2\pi\). This definite integral is equal to zero (for a detailed explanation, see Solution a)).

  Hence, we have \(B = 0\).

  For all off‑diagonal elements, the integral either involves an integer power of the function \(e^{i \varphi}\) or is directly zero (since \(\hat L_z |1, \, 0\rangle = 0\)). Therefore, all off‑diagonal elements equal zero. The diagonal element \(E\) is also zero, and for the element \(I\) we obtain the same as for \(A\) but with the opposite sign.

  Overall, we have

  \[ \hat L_z = \begin{pmatrix} \hbar & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & - \hbar \end{pmatrix} = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \, . \]

  The matrix representation of all the operators from the task assignment can be found at the end of this problem in Comment.

• Answer

  a) The dimension of this Hilbert space is \(3\). All the given states are mutually orthogonal in both representations..

  b) In the matrix representation, the operators are represented by \(3 \times 3\) matrices. The individual matrix elements of the operators can be determined by decomposing into the basis of eigenstates. The operator \(\hat L_z\) has the form

  \[ \hat L_z = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \, . \]

• Comment – forms of the operators in the matrix representation

  Using the same procedure as in the solution, we could compute for all the operators

  \[ \hat L_x = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \, , \qquad \hat L_y = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} \, , \] \[ \hat L_z = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \, , \qquad \hat L^2 = 2 \hbar^2 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \, . \]