📂Quantum Mechanics

Matrix Representation of Harmonic Oscillator Operators

Explanation

Harmonic

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The Schrödinger equation is a linear equation, so a linear combination of several wave functions that satisfy the equation also satisfies the equation. If we denote the eigenfunctions of each state of the harmonic oscillator as $\ket{\psi_{0}}$, $\ket{\psi_{1}}$, $\cdots$, $|\psi_{n}>$, then the linear combination of these eigenfunctions $\ket{\psi}=c_{0}\ket{\psi_{0}} + c_{1}\ket{\psi_{1}} + \cdots + c_{n}|\psi_{n}>$ is also a solution of the Schrödinger equation. (This does not mean it is an eigenfunction. In general, a linear combination of eigenfunctions is not an eigenfunction.) Let the basis $\mathrm{basis}$ be $ (\psi_{0},\ \psi_{1},\cdots , \psi_{n})$ and order them sequentially starting from the ground state. Then the eigenfunctions starting from the ground state can be expressed as the following matrices.

$$ \ket{\psi_{0}} = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \ket{\psi_{1}} = \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix}, |\psi_{n}> = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} $$

If you are not familiar with linear algebra, consider how each unit vector is expressed in the Cartesian coordinate system. $ \hat{\mathbf{x}} = (1,\ 0,\ 0)$, $\hat{\mathbf{y}} = (0,\ 1,\ 0), \hat{\mathbf{z}} = (0,\ 0,\ 1) $$

Therefore, the linear combination of the eigenfunctions $\ket{\psi}$ can be expressed as the following matrix.

$$ \ket{\psi}=c_{0}\ket{\psi_{0}} + c_{1}\ket{\psi_{1}} + \cdots + c_{n}|\psi_{n}> = \begin{pmatrix} c_{0} \\ c_{1} \\ \vdots \\ c_{n} \end{pmatrix} $$

Before finding the matrix representation of the ladder operators, let me explain the principle with a simple example. Consider an arbitrary $2\times 2$ matrix $A$. Since we do not know its components, let us express it as follows. $A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ The way to extract the row 1, column 1 component of this matrix is to multiply it front and back by the two matrices $\begin{pmatrix} 1 & 0 \end{pmatrix}$, $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. $\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} a \\ c \end{pmatrix} = a $ The process of finding each component using the same method is as follows.

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

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

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

Once you understand this principle, let us now earnestly find the matrix representation of the harmonic oscillator’s ladder operators. From now on, we will simply denote the eigenfunctions of the harmonic oscillator as $\psi_{n}> \equiv |n>$. Then the relation between the eigenfunctions and the ladder operators can be written as follows. $a_{+}|n>= \sqrt{n+1}|n>$ Using this, we can find each component of $a_{+}$. Note that the states of the harmonic oscillator start from $n=0$. Also, using $\braket{m | n}=\delta_{mn}$, we find that a matrix component exists only when $\braket{n+1 | a_{+}|n}$. All other components are $0$. For example, $\braket{1 | a_{+}|1}=\braket{1 | \sqrt{2}|2}=\sqrt{2}\braket{1 | 2}=0 $$ \braket{2 | a_{+}|1}=\braket{2 | \sqrt{2}|2}=\sqrt{2}\braket{2 | 2}=\sqrt{2}$ Finding only the components that are not $0$, $\braket{1 | a_{+}|0}=1 $$ \braket{2 | a_{+}|1}=\sqrt{2} $$ \braket{3 | a_{+}|2}=\sqrt{3}$ $\vdots $$ \braket{n+1 | a_{+}|n}=\sqrt{n+1}$ Therefore, the matrix of $a_{+}$ has the nonzero value $\sqrt{n+1}$ at row $(n+2)$, column $(n+1)$. (Remember once again that the states of the harmonic oscillator start from $n=0$.) $a_{+}=\begin{pmatrix} 0 & 0 & 0 & 0 & 0& \cdots \\ \sqrt{1} & 0 & 0 & 0 &0 & \cdots \\ 0 & \sqrt{2} &0 & 0 & 0 & \cdots \\ 0 & 0 & \sqrt{3} &0 & 0 & \cdots \\ 0 & 0& 0& \sqrt{4} &0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$ In the same way, we can also find the matrix representation of $a_{-}$. The process is omitted, so try it yourself. $a_{-}=\begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & 0& \cdots \\ 0 & 0 & \sqrt{2} & 0 &0 & \cdots \\ 0 & 0 & 0 & \sqrt{3} & 0 & \cdots \\ 0 & 0 & 0 & 0 & \sqrt{4} & \cdots \\ 0 & 0& 0& 0& 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$ We can see that the matrix representation also satisfies $(a_{+})^{\ast}=a_{-}$. The Hamiltonian $H$ of the harmonic oscillator can also be expressed as a matrix. Since we know that $H|n>=(n+\frac{1}{2})\hbar w|n>$, using the same method as above, $H=\hbar w\begin{pmatrix} \frac{1}{2} & 0 & 0 & 0 & 0& \cdots \\ 0 & \frac{3}{2} & 0 & 0 &0 & \cdots \\ 0 & 0 & \frac{5}{2} & 0 & 0 & \cdots \\ 0 & 0 & 0 & \frac{7}{2} & 0 & \cdots \\ 0 & 0& 0& 0 & \frac{9}{2} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$ The process here is also omitted, so try it yourself.