# Momentum operator in position representation and viceversa

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.

Thank you.

• What is the meaning of $⟨x|\hat{x}⟩=x$? – Hector Aug 20 '18 at 21:10
• @Hector They mean that the position operator in the position basis is just x. $\hat x \psi=x\psi$ – Aaron Stevens Aug 20 '18 at 21:13
• @AaronStevens No, "position operator in the position basis is just x" is written like this: $\hat x |x ⟩ = x | x ⟩$. – Hector Aug 20 '18 at 21:15
• @Hector I was just telling you what I thought the OP meant. – Aaron Stevens Aug 20 '18 at 21:18
• Agree with @Hector here: notation is absolutely in need of clarification. – ZeroTheHero Aug 20 '18 at 21:53

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 $\widehat{a}=\sqrt{\frac{m\omega}{2\hbar}}\left (\widehat{x}+\frac{i\widehat{p}}{m\omega} \right )$