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I am working through an exercise to show that $\langle p|x \rangle = Ae^{-ipx} $ using $[\hat{x},\hat{p}] = i $ alone. The first part of the exercise is to use the commutator and show that $$ e^{-ia\hat{p}} |x\rangle = |x+a\rangle $$

which I have been able to do. It then states that from this I can deduce what $ \langle p | x \rangle $ is. I have used the fact that $\hat{p}|p\rangle = p|p\rangle$, to show that

$$ \langle p |e^{-ia\hat{p}} |x\rangle = \langle p |e^{-iap} |x\rangle =e^{-iap}\langle p |x\rangle $$

and $$ \langle p |e^{-ia\hat{p}} |x\rangle = \langle p |x +a\rangle$$

to get

$$ e^{-iap}\langle p |x\rangle = \langle p |x +a\rangle $$

If I set $a = x$ I find

$$ e^{-ixp}\langle p |x\rangle = \langle p |2x\rangle $$

Now I don't know where to move from here. Any hints would be greatly appreciated!

Qmechanic
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Hermitian_hermit
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    Your third equality seems to suggest that the function $\langle p|x\rangle$ you are looking for turns addition into multiplication. Perhaps you can use this fact to conclude something about the functional form of $x\mapsto \langle p|x\rangle$. – Phoenix87 Aug 13 '17 at 13:26
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    Possible duplicate: https://physics.stackexchange.com/q/41880/2451 – Qmechanic Aug 13 '17 at 13:27
  • If I let $f_p(x) = \langle p|x \rangle $ and using the fact that $e^{-iap}\langle p | x \rangle = \langle p | x+a \rangle $, I find $ e^{-iap}f_p(x) = f_p(x+a) $. The only function that can satisfy this is$ f_p(x) = Ae^{-ixp} $. Is this correct reasoning? – Hermitian_hermit Aug 13 '17 at 13:53
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    Yes, that's correct! –  Aug 13 '17 at 14:01
  • It's not 100% obvious to me that your reasoning leads to a unique solution. The way I like to derive the eigenstate is by Taylor expanding $|x+a\rangle$, then solving the resulting differential equation for $\langle p |x \rangle$. – astronautgravity Aug 13 '17 at 18:57

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