I have found 2 different forms of $⟨x|\hat{p}|x′⟩$ and I have no idea which one is the true form. Can anyone help please?
$⟨x|\hat{p}|x′⟩ = (i\hbar)\frac{d\delta(x − x′)}{dx′}$
$⟨x|\hat{p}|x′⟩ = -(i\hbar)\frac{d\delta(x − x′)}{dx}$
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$\begingroup$ You'll get confused to no end if you do not stick to the canonical representation picture, $\hat p = \int dx |x\rangle \frac{\hbar}{i} \partial_x \langle x|$, from which you prove what you choose. $\endgroup$– Cosmas ZachosCommented Jan 31, 2020 at 16:12
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1$\begingroup$ Related: physics.stackexchange.com/q/76299/2451 $\endgroup$– Qmechanic ♦Commented Jan 31, 2020 at 16:30
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$\begingroup$ Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis. $\endgroup$– VoulkosCommented Jan 31, 2020 at 17:46
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1 Answer
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Since $\langle \psi | A | \varphi \rangle = \langle \varphi | A^{\dagger} | \psi \rangle^*$, you have to conjugate your first line. You get $$\langle x | p | x' \rangle = -i\hbar \frac{\partial}{\partial x}\delta(x-x')$$ or $$\langle x | p | x' \rangle = \langle x' | p | x\rangle^* = [-i\hbar \frac{\partial}{\partial x'}\delta(x-x')]^* = i\hbar \frac{\partial}{\partial x'}\delta(x-x') = -i\hbar \frac{\partial}{\partial x}\delta(x-x')$$ so both are the same result now.