Pauli Hamiltonian
Task number: 4577
The Pauli Hamiltonian for a particle of charge \(q\), mass \(m\), and spin \(\frac{1}{2}\) in a conservative field with potential energy \(V(\vec r)\) and in an electromagnetic field with potentials \(\vec A\) and \(\varphi\) is given by
\[ \hat H^P = \left [ \frac{1}{2m} \left (\hat{\vec p} - q \, \vec A \right )^2 + q \, \varphi + V \right ] \, \mathbb{E} + \mu_B \, \hat{\vec \sigma} \cdot \vec B + \hat H_{SO} \, , \]where \(\hat{\vec p}\) is the momentum operator, \(\mathbb{E}\) is the \(2 \times 2\) identity matrix, \(\mu_B = \frac{e\hbar}{2m_e}\) is the Bohr magneton, \(\hat{\vec \sigma}\) are the Pauli matrices, \(\vec B\) is the magnetic induction vector, and \(\hat H_{SO} \sim \hat{\vec L} \cdot \hat{\vec S}\) is the so‑called spin‑orbit interaction.
Expand the Pauli Hamiltonian in matrix form.
Hint
Recall or look up the mathematical structure of the individual quantities in the Pauli Hamiltonian (what is a scalar, what is a vector, what is a matrix).
Solution
When expanding into matrices, we examine each term in the Pauli Hamiltonian separately.
In the first summand (inside the square brackets), the second and third terms are scalars. In the first term (inside the parentheses), we see a combination of a vector and a vector differential operator. The entire expression in parentheses is squared, which can be interpreted as the scalar product of the corresponding vector with itself, or as the square of the magnitude of this vector. Thus, multiplying this summand by the identity matrix yields a \(2 \times 2\) diagonal matrix, whose elements contain differential operators. A more detailed discussion of this summand is provided at the end of this problem in section Comment.
In the middle summand, we see the scalar \(\mu_B\) and the scalar product \(\hat{\vec \sigma} \cdot \vec B\). Here, we multiply a three‑component vector, whose components are \(2 \times 2\) matrices, by the three‑component magnetic induction vector. This product can be calculated as follows
\[ \hat{\vec \sigma} \cdot \vec B = \left ( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \, \begin{pmatrix} 0 & −i \\ i & 0 \end{pmatrix}, \, \begin{pmatrix} 1 & 0 \\ 0 & −1 \end{pmatrix} \right ) \cdot \left ( B_x, \, B_y, \, B_z \right ) = \] \[ = \begin{pmatrix} 0 & B_x \\ B_x & 0 \end{pmatrix} + \begin{pmatrix} 0 & − i \, B_y \\ i \, B_y & 0 \end{pmatrix} + \begin{pmatrix} B_z & 0 \\ 0 & − B_z \end{pmatrix} = \begin{pmatrix} B_z & B_x − i \, B_y \\ B_x + i \, B_y & - B_z \end{pmatrix} \, . \]Therefore, the entire middle summand is
\[ \mu_B \, \begin{pmatrix} B_z & B_x − i \, B_y \\ B_x + i \, B_y & - B_z \end{pmatrix} \, . \]For the last summand, we know it is proportional to the scalar product \(\hat{\vec L} \cdot \hat{\vec S}\). Here, we multiply a three‑component vector, whose components are differential operators, by a three‑component vector, whose components are \(2 \times 2\) matrices. This product can be calculated analogously to the previous summand
\[ \hat{\vec L} \cdot \hat{\vec S} = \left ( \hat L_x, \, \hat L_y, \, \hat L_z \right ) \cdot \frac{\hbar}{2} \, \left ( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \, \begin{pmatrix} 0 & −i \\ i & 0 \end{pmatrix}, \, \begin{pmatrix} 1 & 0 \\ 0 & −1 \end{pmatrix} \right ) = \] \[ = \frac{\hbar}{2} \left [ \begin{pmatrix} 0 & \hat L_x \\ \hat L_x & 0 \end{pmatrix} + \begin{pmatrix} 0 & − i \, \hat L_y \\ i \, \hat L_y & 0 \end{pmatrix} + \begin{pmatrix} \hat L_z & 0 \\ 0 & − \hat L_z \end{pmatrix} \right ] = \frac{\hbar}{2} \begin{pmatrix} \hat L_z & \hat L_x − i \, \hat L_y \\ \hat L_x + i \, \hat L_y & - \hat L_z \end{pmatrix} \, . \]The resulting expression is thus a \(2 \times 2\) matrix, whose elements are differential operators. For clarity, we continue to denote the spin‑orbit term as \(\hat H_{SO}\), even though we now know it is also a \(2 \times 2\) matrix.
Overall, the Pauli Hamiltonian can be expressed in matrix form as
\[ \hat H^P = \left [ \frac{1}{2m} \left (\hat{\vec p} - q \, \vec A \right )^2 + q \, \varphi + V \right ] \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \mu_B \, \begin{pmatrix} B_z & B_x − i \, B_y \\ B_x + i \, B_y & - B_z \end{pmatrix} + \hat H_{SO} \, . \]Answer
The Pauli Hamiltonian can be expressed in matrix form as
\[ \hat H^P = \left [ \frac{1}{2m} \left (\hat{\vec p} - q \, \vec A \right )^2 + q \, \varphi + V \right ] \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \mu_B \, \begin{pmatrix} B_z & B_x − i \, B_y \\ B_x + i \, B_y & - B_z \end{pmatrix} + \hat H_{SO} \, . \]Comment – detailed discussion of the first summand
In the first summand of the Pauli Hamiltonian, we immediately see that the part \(\left ( q \, \varphi + V \right ) \mathbb{E}\) is a \(2 \times 2\) matrix, since it is simply a scalar multiplied by the \(2 \times 2\) identity matrix.
In the part \(\left [ \frac{1}{2m} \left (\hat{\vec p} - q \, \vec A \right )^2 \right ] \mathbb{E}\), we first expand the parentheses, being careful with the order of multiplication due to the presence of operators
\[ \left [ \frac{1}{2m} \left (\hat{\vec p} - q \, \vec A \right )^2 \right ] \mathbb{E} = \frac{1}{2m} \left [ \hat{\vec p}^2 - q \, \hat{\vec p} \cdot \vec A - q \, \vec A \cdot \hat{\vec p} + q^2 \vec A^2 \right ] \mathbb{E} = \] \[ = \frac{\hat{\vec p}^2}{2m} \mathbb{E} - \frac{q}{2m} \left (\hat{\vec p} \cdot \vec A + \vec A \cdot \hat{\vec p} \right ) \mathbb{E} + \frac{q^2 \vec A^2}{2m} \mathbb{E} \, . \]In the first term we have the scalar operator \(\hat{\vec p}^2\) divided by the scalar \(2m\). This is simply the kinetic energy of the particle. After multiplying by the identity matrix, we therefore obtain a diagonal matrix, whose entries are differential operators.
The middle term cannot be further simplified without explicitly knowing the vector potential \(\vec A\). We have assumed a general electromagnetic field, and it may happen that a given component of the momentum operator does not commute with the corresponding component of the vector potential \(\vec A = \vec A \left( \vec r \right)\). As a consequence, the first term in this sum does not necessarily coincide with the second one. Nevertheless, we can say that after multiplying this term by the identity matrix, we again obtain a diagonal \(2 \times 2\) matrix.
In the last term we have \(\vec A^2\), which is a scalar, multiplied by the scalar factor \(\frac{q^2}{2m}\). After multiplying by the identity matrix, we again obtain a diagonal matrix whose entries are precisely \(\frac{q^2 \vec A^2}{2m}\).
With the first summand expanded in this way, the Pauli Hamiltonian takes the form
\[ \hat H^P = \begin{pmatrix} \frac{\hat{\vec p}^2}{2m} & 0 \\ 0 & \frac{\hat{\vec p}^2}{2m} \end{pmatrix} - \begin{pmatrix} \frac{q}{2m} \left (\hat{\vec p} \cdot \vec A + \vec A \cdot \hat{\vec p} \right ) & 0 \\ 0 & \frac{q}{2m} \left (\hat{\vec p} \cdot \vec A + \vec A \cdot \hat{\vec p} \right ) \end{pmatrix} + \begin{pmatrix} \frac{q^2 \vec A^2}{2m} & 0 \\ 0 & \frac{q^2 \vec A^2}{2m} \end{pmatrix} + \] \[ + \begin{pmatrix} q \, \varphi + V & 0 \\ 0 & q \, \varphi + V \end{pmatrix} + \mu_B \, \begin{pmatrix} B_z & B_x − i \, B_y \\ B_x + i \, B_y & - B_z \end{pmatrix} + \hat H_{SO} \, . \]
