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I'm asked to compute the matrix elements $\left( x^i \right)_{nm}$ where $i=2,3,4$ and $x$ represents the position operator for the quantum harmonic oscillator. I know that for $i=1$: \begin{equation} (x)_{nm}=\sqrt{\frac{\hbar}{2m\omega}}(\sqrt{m}\delta_{n,m-1}+\sqrt{m+1}\delta_{n,m+1}) \end{equation}

I have already solved the problem using the expression of $\hat{x}$ in terms of ladder operators $a_+,a_-$, but I'm asked to do it with matrix multiplication aswell. Is this the correct calculation with this method? \begin{equation} (x^2)_{nm}=\sum_{k=0}^{\infty}x_{nk}x_{km} \end{equation}

I tried to compute the upper expression but the answer I got is different from that obtained through the ladder operators method, which is: \begin{equation} (x^2)_{nm}=\frac{\hbar}{2m\omega}\left[ (m+1)\delta_{n,m+2}+m\delta_{n,m-2}+(2m+1)\delta_{nm} \right] \end{equation} Thank you in advance for any answer.

Álvaro Luque
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2 Answers2

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I think you didn't quite get what this exercise is asking for. You are asked to do it with matrix multiplication. Therefore you need to begin by spelling out the matrix elements $(x)_{nm}$:

$$\hat{x}=\sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \cdots\\ \sqrt{1} & 0 & \sqrt{2} & 0 & \cdots \\ 0 & \sqrt{2} & 0 & \sqrt{3} & \cdots\\ 0 & 0 & \sqrt{3} & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$$

Then you get the matrix of $\hat{x}^2$ by matrix multiplication.

$$\begin{align} \hat{x}^2&=\hat{x}\cdot\hat{x} \\ &=\sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \cdots\\ \sqrt{1} & 0 & \sqrt{2} & 0 & \cdots \\ 0 & \sqrt{2} & 0 & \sqrt{3} & \cdots\\ 0 & 0 & \sqrt{3} & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} \cdot\sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \cdots\\ \sqrt{1} & 0 & \sqrt{2} & 0 & \cdots \\ 0 & \sqrt{2} & 0 & \sqrt{3} & \cdots\\ 0 & 0 & \sqrt{3} & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} \\ &=\frac{\hbar}{2m\omega} \begin{pmatrix} 1 & 0 & \sqrt{1\cdot 2} & 0 & 0 & \cdots \\ 0 & 3 & 0 & \sqrt{2\cdot 3} & 0 & \cdots \\ \sqrt{1\cdot 2} & 0 & 5 & 0 & \sqrt{3\cdot 4} & \cdots \\ 0 & \sqrt{2\cdot 3} & 0 & 7 & 0 & \cdots \\ 0 & 0 & \sqrt{3\cdot 4} & 0 & 9 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} \end{align}$$

And likewise you get the matrices of $\hat{x}^3$ and $\hat{x}^4$. You should be able to do that now.

Thomas Fritsch
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Hint: Use the fact $$\hat x = \sqrt{\frac{\hbar}{2m\omega}} \left(a+a^\dagger\right)$$ and similarly, use $$\hat x^k = \left(\frac{\hbar}{2m\omega}\right)^{k/2} \left(a+a^\dagger\right)^k$$ which will give $$\langle n \vert \hat x^k \vert m \rangle= \left(\frac{\hbar}{2m\omega}\right)^{k/2} \langle n \vert\left(a+a^\dagger\right)^k\vert m \rangle$$

SchrodingersCat
  • 4,994