# Mean value with delta function [closed]

How do I compute this matrix element $$\langle 1|\delta(\hat x-b)|1\rangle$$ that models a 1-D harmonic oscillator?

I have done the same for the ground state (by seting delta function as the Fourier integral, Sakurai exercise), $\langle 0|e^{ik\bf x}|0\rangle=exp[-k^2\langle 0|x^2|0\rangle/2]$

using the BH identity, how to continue in the first excited state?

I don't know how to calculate $\langle 1|e^{ik\bf x}|1\rangle$ does anyone have an idea?

I know how to prove that $\langle n|e^{ik\bf x}|0\rangle=\left(\frac{ik}{\sqrt{n!}}\right)^n\left(\sqrt{\frac{\hbar}{2m\omega}}\right)^ne^{-k^2/2\langle 0|x^2|0\rangle}$ is it right to say $\langle 1|e^{ik\bf x}|0\rangle\langle 0|e^{ik\bf x}|1\rangle =\langle 1|e^{ik\bf x}|1\rangle$ and use the identity but conjugate?

• What's wrong with just $\langle 1\vert \delta (\hat x-b)\vert 1\rangle = \int_{-\infty}^\infty \psi^*_1(x) \delta (x-b) \psi_1(x) dx$? – ZeroTheHero Jan 29 '18 at 16:47
• Hi thanks for the interest, I am looking for an answer through Baker-Hausdorf identity. Maybe with expanding the exponential or some tricky thing with ladder operators, any ideas on that are more than welcome. – Anastasios Jan 29 '18 at 18:53
• I really don't know what you're trying to achieve. I don't see how BCH can enter as this is to expand the exponential of operators that don't commute, and you have here a single operator. – ZeroTheHero Jan 29 '18 at 20:30
• I edit my question maybe now someone can help. – Anastasios Jan 29 '18 at 20:45

There are two fundamental questions here. The expectation value of the $\delta$ was already shown by @ZeroTheHero. Let's compute the expectation value of $\exp \left(i k \hat{x} \right)$ with $k$ a real constant and $\hat{x}$ the standard position operator. Let's make use of the expression for $\hat{x}$ in terms of the creation and annihilation operators of the 1-D Harmonic Oscillator, \begin{equation} \hat{x} = \sqrt{\frac{\hbar}{2 m \omega}} \left(a^{\dagger} + a \right) \, . \end{equation} Let's define $A = ik \sqrt{\frac{\hbar}{2 m \omega}} a^{\dagger}$ and $B = ik \sqrt{\frac{\hbar}{2 m \omega}} a$. We have then $[A,B]=k^2 \frac{\hbar}{2 m \omega}$ a constant and hence $[A,[A,B]] = 0 = [B,[A,B]]$. Because of this we can write \begin{eqnarray} \exp \left(i k \hat{x} \right) =\exp\left(A+B\right)&=&\exp\left(A\right)\exp\left(B\right)\exp\left[-\frac{1}{2}[A,B]\right] \\ &=& \exp\left(ik \sqrt{\frac{\hbar}{2 m \omega}} a^{\dagger}\right)\exp\left(ik \sqrt{\frac{\hbar}{2 m \omega}} a\right)\exp\left[-k^2 \frac{\hbar}{4 m \omega}\right] \, . \end{eqnarray} The expectation value with respect to the ground state takes only the very first terms in the expansions of the exponentials and hence $\langle0|\exp \left(i k \hat{x} \right)|0\rangle = \exp\left[-k^2 \frac{\hbar}{4 m \omega}\right] = \exp\left[- \frac{k^2}{2} \langle0|\hat{x}^2|0\rangle\right]$. For the first excited state we have two contributions, namely \begin{equation} \langle1|\exp \left(i k \hat{x} \right)|1\rangle = \exp\left[-k^2 \frac{\hbar}{4 m \omega}\right] \langle1| 1 - k^2 \frac{\hbar}{2 m \omega} a^{\dagger} a|1\rangle = \exp\left[-k^2 \frac{\hbar}{4 m \omega}\right] \left[ 1 - k^2 \frac{\hbar}{2 m \omega}\right] \, . \end{equation} By looking at the behavior we can generalize and write \begin{equation} \langle n | \exp \left(i k \hat{x} \right) |n \rangle = \exp\left[-k^2 \frac{\hbar}{4 m \omega}\right] \sum^{n}_{j=0} \left(- k^2 \frac{\hbar}{2 m \omega}\right)^j j! \, . \end{equation}