What is the Wave Function for a Particle in One Dimension in Dirac Formalism?

[ergospherical]

What is $<x|P|x'>$? (for particle in 1d, and $\hbar = 1$)?\begin{align*}
<x|P|x'> &= \int dp' <x|P|p'><p'|x'> \\
&= \int dp' \ p' <x|p'> <p'|x'> \\
&= \int dp' \ p' \frac{1}{\sqrt{2\pi}} e^{ip'x} \frac{1}{\sqrt{2\pi}} e^{-ip'x'} \\
&= \frac{1}{2\pi} \int dp' \ p' e^{ip'(x-x')}
\end{align*}

 

[topsquark]

What is $<x|P|x'>$? (for particle in 1d, and $\hbar = 1$)?\begin{align*}
<x|P|x'> &= \int dp' <x|P|p'><p'|x'> \\
&= \int dp' \ p' <x|p'> <p'|x'> \\
&= \int dp' \ p' \frac{1}{\sqrt{2\pi}} e^{ip'x} \frac{1}{\sqrt{2\pi}} e^{-ip'x'} \\
&= \frac{1}{2\pi} \int dp' \ p' e^{ip'(x-x')}
\end{align*}

I'm not sure what the question is? So far so good. Now integrate by parts.

-Dan

 

[ergospherical]

it's supposed to evaluate to \begin{align*}
-i \frac{\partial}{\partial x} \delta(x-x')
\end{align*}but even integrating by parts I'm not sure how to get this

 

[vanhees71]

What is $<x|P|x'>$? (for particle in 1d, and $\hbar = 1$)?\begin{align*}
<x|P|x'> &= \int dp' <x|P|p'><p'|x'> \\
&= \int dp' \ p' <x|p'> <p'|x'> \\
&= \int dp' \ p' \frac{1}{\sqrt{2\pi}} e^{ip'x} \frac{1}{\sqrt{2\pi}} e^{-ip'x'} \\
&= \frac{1}{2\pi} \int dp' \ p' e^{ip'(x-x')}
\end{align*}

From this you get
$$\langle x|\hat{P}|x' \rangle=\frac{1}{2 \pi} (-\mathrm{i} \partial_x) \int_{\mathbb{R}} \mathrm{d} p' \exp[\mathrm{i} p' (x-x')]=-\mathrm{i} \partial_x \delta(x-x').$$