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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.

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2 Answers 2

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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.

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  • $\begingroup$ Oh my, you're right. Now it's a dumb question. I wanted to find an expression for $x^i_{nm}$. Thanks a lot! $\endgroup$ Commented May 30, 2020 at 23:06
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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 $$\langle n \vert \hat x^k \vert m \rangle= \left(\frac{\hbar}{2m\omega}\right)^{k/2} \langle n \vert\left(a+a^\dagger\right)^k\vert m \rangle$$

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  • $\begingroup$ I got that, ladder operators are no problem, and got the same result as you, but I'm asked to do it both ways to show that the calculation is the same (which actually is bothering me). $\endgroup$ Commented May 30, 2020 at 20:15
  • $\begingroup$ @ÁlvaroLuque Then I guess it's a matter of multiplication and summation. just be careful with the product of the Dirac Deltas. $\endgroup$ Commented May 30, 2020 at 21:09
  • $\begingroup$ I will. Thanks a lot! $\endgroup$ Commented May 30, 2020 at 21:19

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