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So I thought that

$$\hat{p}~=~-i\hbar \frac{\partial}{\partial x}~$$

and

$$\hat{x}~=~i\hbar \frac{\partial}{\partial p}~$$

but I have just encountered the next problem:

"Find the wave function for the fundamental state of

$$H = \frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2$$

in the position base."

So they are asking me for: $\psi_0(x)= ⟨x|0⟩ $

I did as follows: I wanted to apply the condition:

$$a|0⟩ = 0$$

With $a$ being:

$$a=\sqrt{\frac{m\omega}{2\hbar}}(x+ip)$$

I figured that since

$$\hat{x}|x⟩=x|x⟩$$

and

$$\hat{p}|x⟩=-i\hbar \frac{\partial}{\partial x}|x⟩$$

I could just plug them in the equation for a and ended up with:

$$ \sqrt{\frac{m\omega}{2\hbar}} \bigg( x+\hbar \frac{\partial}{\partial x} \bigg) \psi_0 = 0$$

Which led me to:

$$\frac{d\psi_0}{\psi_0}=-\frac{x}{\hbar}dx$$

But checking the answers, they stated that:

$$\hat{p}|x⟩=\frac{-i\hbar}{\sqrt{m\omega\hbar}} \frac{\partial}{\partial x}|x⟩$$

And I don't understand where does the square root diving $-i\hbar$ is coming from.

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    $\begingroup$ What is the meaning of $⟨x|\hat{x}⟩=x$? $\endgroup$ – Hector Aug 20 '18 at 21:10
  • $\begingroup$ @Hector They mean that the position operator in the position basis is just x. $\hat x \psi=x\psi$ $\endgroup$ – BioPhysicist Aug 20 '18 at 21:13
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    $\begingroup$ @AaronStevens No, "position operator in the position basis is just x" is written like this: $\hat x |x ⟩ = x | x ⟩$. $\endgroup$ – Hector Aug 20 '18 at 21:15
  • $\begingroup$ @Hector I was just telling you what I thought the OP meant. $\endgroup$ – BioPhysicist Aug 20 '18 at 21:18
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    $\begingroup$ Agree with @Hector here: notation is absolutely in need of clarification. $\endgroup$ – ZeroTheHero Aug 20 '18 at 21:53
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I think this is a problem between different notations. First of all, indeed you have $ \widehat{x} \left | \Psi \right > = x\left | \Psi \right > $ and $ \widehat{p} \left | \Psi \right > = -i\hbar\frac{\partial\left | \Psi \right >}{\partial x} $. But the annihilation operator $\widehat{a}$ is equal to $\frac{\widehat{X}+i\widehat{P}}{\sqrt{2}}$ where $ \widehat{X}$ and $\widehat{P}$ are dimensionless operators.

And it happens that you have : $\widehat{X}=\sqrt{\frac{m\omega}{\hbar}}\widehat{x}$ and $\widehat{P} = \frac{\widehat{p}}{\sqrt{m\hbar \omega}}$ Hence the expression written in the answer where the $\widehat{p}$ should be interpreted as $\widehat{P}$ I guess.

And by the way if you use these expressions you will find that $\widehat{a}=\sqrt{\frac{m\omega}{2\hbar}}\left (\widehat{x}+\frac{i\widehat{p}}{m\omega} \right )$

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Your annihilation operator is wrong. It is, $$ a= (\sqrt{\frac{m\omega}{2}}x + i\frac{p}{\sqrt{2m\omega}})$$ To find $\langle{x'}|0⟩$ you just have to solve this equation: $$ \langle{x'}|a|0⟩=0$$ $$ \langle{x'}|(\sqrt{\frac{m\omega}{2}}x + i\frac{p}{\sqrt{2m\omega}})|0⟩=0$$ Since $p=-i\partial_{x'}$, $$\sqrt{\frac{m\omega}{2}}x'\langle{x'}|0⟩+\frac{\partial_{x'}\langle{x'}|0⟩}{\sqrt{2m\omega}}=0$$ On solving this differential equation you will get $\psi_0(x')$. To calculate any $\psi_n(x')$ just apply the creation operator n times on $|0⟩$ and take the inner product $⟨x'|n⟩$.

Note: I have used $\hbar =1$.

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