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$\quad \;$For a one-dimensional harmonic oscillator of mass $\;M\;$ and angular frequency $\;\omega\;$ calculate $\left\langle k \vert \hat{x} \vert n\right\rangle$, where $\left\vert k\right\rangle$ and $\left\vert n\right\rangle$ are eigenstates of the harmonic oscillator, and show that it vanishes unless $k=n\!\pm\!1$.

I am currently doing a question on Harmonic Oscillators, I (sort of) understand the notation until it gets to the part where delta is included. Why is it there?

\begin{equation} \hat{a}^{\dagger}\left\vert n\right\rangle=\sqrt{n\!+\!1}\left\vert n\!+\!1 \right\rangle\:, \qquad \hat{a}\left\vert n\right\rangle=\sqrt{n}\left\vert n\!-\!1 \right\rangle \tag{01} \end{equation}

\begin{align} \left\langle k \vert \hat{x} \vert n\right\rangle & =\sqrt{\dfrac{\hbar}{2m\omega}}\biggl(\left\langle k \left\vert \hat{a}\vphantom{\hat{a}^{\dagger}}\right\vert n \right\rangle +\left\langle k \left\vert \hat{a}^{\dagger} \right\vert n \right\rangle\biggr) \nonumber\\ &=\sqrt{\dfrac{\hbar}{2m\omega}}\Bigl(\sqrt{n}\left\langle k \vert n\!-\!1 \right\rangle +\sqrt{n\!+\!1}\left\langle k \vert n\!+\!1 \right\rangle \Bigr) \nonumber\\ &=\sqrt{\dfrac{\hbar}{2m\omega}}\Bigl(\sqrt{n}\,\delta_{k,n-1} +\sqrt{n\!+\!1}\,\delta_{k,n+1} \Bigr) \tag{02} \\ &= \begin{cases} \sqrt{\dfrac{\hbar n}{2m\omega}} & \: k=n\!-\!1 \vphantom{\sqrt{\dfrac{\hbar n}{2m\omega}}^{\frac12}}\\ \sqrt{\dfrac{\hbar\left(n\!+\!1\right)}{2m\omega}} & \: k=n\!+\!1 \vphantom{\sqrt{\dfrac{\hbar n}{2m\omega}}^{\frac12}}\\ \qquad 0 & \: k \ne n\!\pm\!1 \vphantom{\sqrt{\dfrac{\hbar n}{2m\omega}}^{\frac12}} \end{cases} \nonumber \end{align}

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  • $\begingroup$ Hint: the states form an orthonormal basis, i.e. $\langle m|n\rangle=\delta_{m,n}$ $\endgroup$ – Soba noodles Aug 4 '17 at 22:41
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That is the Kronecker delta, a function $\delta_{ij}$ such that $\delta_{ij}=0$ if $i\neq j$ and $\delta_{ij}=1$ if $i=j$. It arises as a consequence of orthonormality of the $|n\rangle$ basis.

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  • $\begingroup$ Thanks! I wonder though, what gives one the intuition to use the Kronecker delta to solve a question like this? Is it solely based on the question asking to show that the eigenstate vanishes? $\endgroup$ – Munir Malik Aug 4 '17 at 23:01
  • $\begingroup$ @MunirMalik The Kronecker delta is essentially a shorthand for several cases rolled into one. You often see it when talking about families of commutators, or when working with orthonormal bases. Note that the Kronecker delta describes the value of the dot product of any two arbitrary basis vectors. $\endgroup$ – probably_someone Aug 4 '17 at 23:07

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