📂Quantum Mechanics

Matrix Representation of Operators in Quantum Mechanics

Build-up

Let’s consider two unit vectors $\widehat{\mathbf{x}} = (1, 0)$ and $\widehat{\mathbf{y}} = (0, 1)$ in a 2-dimensional space. The coordinate vector of an arbitrary point $(a, b)$ in this space can be expressed as a linear combination of these two unit vectors as follows.

$$(a, b) = a(1, 0) + b(0, 1) \implies \begin{bmatrix} a \\ b \end{bmatrix} = a\begin{bmatrix} 1 \\ 0 \end{bmatrix} + b\begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

This expression is possible because the set of unit vectors $\left\{ \widehat{\mathbf{x}}, \widehat{\mathbf{y}} \right\}$ contains orthogonal vectors equal to the number of dimensions. Such a set is mathematically called a basis. In other words, if a basis is given, all vectors in the space can be represented as a linear combination of these basis vectors. The condition for being a basis is that the number of elements equals the number of dimensions and they must be composed of orthogonal vectors. Hence, they do not need to be unit vectors like $\widehat{\mathbf{x}}$ and $\widehat{\mathbf{y}}$.

For example, let’s consider two orthogonal vectors $\mathbf{v} = (-1, 2)$ and $\mathbf{u} = (2, 1)$. Then the coordinate vector of point $(a, b)$ is as follows:

$$(a, b) = \dfrac{-a + 2b}{5}(-1, 2) + \dfrac{2a+b}{5}(2, 1) \implies \begin{bmatrix} \dfrac{-a + 2b}{5} \\ \dfrac{2a+b}{5} \end{bmatrix} = \dfrac{-a + 2b}{5}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + \dfrac{2a+b}{5}\begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

It is known that the coordinates of each vector in the basis are $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$, respectively. Now, let’s assume an arbitrary matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given. To obtain the component in the first row and second column of this matrix, you multiply the coordinates of the first and second basis vectors, respectively, as follows $A$.

$$\begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix} = b$$

If we notate the first basis vector as $\ket{1}$ and the second basis vector as $\ket{2}$, the $ij$ component of the matrix $A$ can be expressed as follows:

$$[A_{ij}] = \bra{i}A\ket{j}$$

This notation is called Dirac notation. The core points above are fourfold:

1. If there are orthogonal vectors equal to the number of dimensions, all points can be expressed in coordinates as a linear combination of these vectors. (Such a set is called a basis)
2. The coordinates of a point can vary depending on the basis.
3. The coordinate vector of the $i$-th vector in the basis is as follows: $$\begin{bmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{bmatrix} \gets i\text{-th row}$$
4. If the coordinates of the $i$-th basis vector are notated as $\ket{i}$, the $ij$ component of the matrix is as follows: $$[A_{ij}] = \bra{i}A\ket{j}$$

Explanation

In quantum mechanics, operators (corresponding to different eigenvalues) have eigenfunctions that are all orthogonal. Thus, the set of eigenfunctions forms a basis. Using the coordinate vectors of these eigenfunctions, the action of the operator on these eigenfunctions can be expressed as matrix multiplication. For example, let’s consider the Hamiltonian operator $H$ and assume the following eigenvalue equation holds.

$$H\ket{1} = h_{1} \ket{1} \\ H\ket{2} = h_{2} \ket{2} \\$$

Then the above eigenvalue equation can be expressed as the following matrix multiplication.

$$\begin{bmatrix} h_{1} & 0 \\ 0 & h_{2} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} h_{1} \\ 0 \end{bmatrix} = h_{1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\[1em] \begin{bmatrix} h_{1} & 0 \\ 0 & h_{2} \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ h_{2} \end{bmatrix} = h_{2} \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

Therefore, $\begin{bmatrix} h_{1} & 0 \\ 0 & h_{2} \end{bmatrix}$ is the matrix corresponding to the Hamiltonian $H$. The method to obtain each component of this matrix is, as explained above, multiplying the eigenvectors front and back.

$$[H_{ij}] = \bra{i}H\ket{j}$$

Example

Harmonic Oscillator

• Energy Operator: $$H=\hbar w \begin{pmatrix} \frac{1}{2} & 0 & 0 & 0 & 0& \cdots \\ 0 & \frac{3}{2} & 0 & 0 &0 & \cdots \\ 0 & 0 & \frac{5}{2} & 0 & 0 & \cdots \\ 0 & 0 & 0 & \frac{7}{2} & 0 & \cdots \\ 0 & 0& 0& 0 & \frac{9}{2} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$$

• Ladder Operator:

  $$a_{+}=\begin{pmatrix} 0 & 0 & 0 & 0 & 0& \cdots \\ \sqrt{1} & 0 & 0 & 0 &0 & \cdots \\ 0 & \sqrt{2} &0 & 0 & 0 & \cdots \\ 0 & 0 & \sqrt{3} &0 & 0 & \cdots \\ 0 & 0& 0& \sqrt{4} &0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$$ $$a_{-}=\begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & 0& \cdots \\ 0 & 0 & \sqrt{2} & 0 &0 & \cdots \\ 0 & 0 & 0 & \sqrt{3} & 0 & \cdots \\ 0 & 0 & 0 & 0 & \sqrt{4} & \cdots \\ 0 & 0& 0& 0& 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$$

Angular Momentum Operator

When $\ell = 1$,

• Angular Momentum Operator:

  $$L_{z}=\hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$ $$L_{x}=\dfrac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} ,\qquad L_{y}=\dfrac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & -\i & 0 \\ \i & 0 & -\i \\ 0 & \i & 0 \end{pmatrix}$$

• Ladder Operator:

  $$L_{+}=\hbar \begin{pmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{pmatrix} ,\qquad L_{-}=\hbar \begin{pmatrix} 0 & 0 & 0 \\ \sqrt{2} &0 & 0 \\ 0 & \sqrt{2} &0 \end{pmatrix}$$