Why $\langle x|p \rangle$ is momentum eigenstate of which eigenvalue is $p$ at position basis

Question

I don't understand why $\langle x|p \rangle$ is momentum eigenstate of which eigenvalue is p at position basis. (It appear at Griffiths, Quantum mechanics 3rd, example 3.9)

I'm not sure, but I try thinking about following:

At (time independent) Schrodinger equation and eigen equation ($\hat{H}ψ=Eψ$), $ψ$ is wave function and eigenfunction

At $Ψ(x,t)=\langle x|S(t) \rangle$, $Ψ$ is position space wave function and position component of S(t) at position basis.

Let p is momentum state, then $\langle x|p \rangle$ is momentum eigenstate at position basis.

At $\langle \hat{p}|x|p \rangle=\langle p|x|p \rangle$, $\langle x|p \rangle$ is momentum eigenfunction of which eigenvalue is p.

I think this is too far-fetched.

It's very strange and I think I wrote it wrong since I don't know Dirac notation well.

Answer 1

What he means is that $|p\rangle$ is the momentum eigenstate with eigenvalue $p$, i.e. $\hat{p}|p\rangle=p|p\rangle$, where $\hat{p}$ is the momentum operator and $p$ the eigenvalue. Then $\langle x|p\rangle$ is the $|p\rangle$ eigenstate expressed in the position basis.

Hope this helps.

Answer 2

Although it would be great if you were comfortable with Dirac notation still I will try my best.

Eigenvalue equation for momentum operator look like: $$P|p\rangle =p|p\rangle$$ You can compare it with the form: $$\text{Operator - Vector }\rightarrow \text{Eigenvalue - Vector}$$

If we want to project this on a position basis (That is to say above is written in abstract form we want to write on a position basis) : $$\langle x|P|p\rangle =p\langle x|p\rangle $$

or $$-i\hbar \frac{d\psi_p(x)}{dx}=p\psi_p(x)$$ where $\psi_p(x)=\langle x|p\rangle$ is momentum eigen function in position basis with eigenvalue $p$. You can solve above to find $$\psi_p(x)=\frac{1}{\sqrt{2\pi \hbar}}e^{ipx/\hbar}$$

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Note : \begin{align*} \langle x|P|p\rangle &= \int \langle x|P|x'\rangle \langle x'|p\rangle dx' \\ &= \int [-i\hbar \delta'(x-x')]\psi_p(x')dx' \\ &= -i\hbar \frac{d\psi_p(x)}{dx} \end{align*}

which just demonstrates How the momentum operator looks on in position basis.