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

• ergospherical
In summary, the expression ##<x|P|x'>## represents the momentum operator in one dimension when the value of Planck's constant is set to 1. It can be evaluated using integration by parts, resulting in the expression ##-i\frac{\partial}{\partial x}\delta(x-x')##.
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*}

ergospherical said:
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

Demystifier
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

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

topsquark, malawi_glenn and ergospherical

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