Confused about operator action

[dyn]

Hi. I have come across the following equation < x | p | ψ > = - iħ d/dx < x | ψ > where p is the momentum operator. I'm confused as to how p which acts to the right on | ψ >can then be taken outside the inner product so it now also acts on the bra < x | ?
Thanks

 

[hilbert2]

The inner product $\left<x|\psi\right>$ is the wave function $\psi(x)$ that corresponds to the abstract state vector $\left|\psi\right>$. That equation just says that the momentum operator in the position representation is the differential operator $-i\hbar\frac{d}{dx}$.

 

[dyn]

Still confused. What is p | ψ > then ? and why does the differential operator not act on the < x | bra ?

 

[hilbert2]

$p\left|\psi\right>$ is the abstract state vector that has a corresponding wave function $\left<x|p|\psi\right>=-i\hbar\frac{d\psi(x)}{dx}$.

 

[dyn]

Does p | ψ > = -iħ d/dx | ψ > ?

 

[hilbert2]

Does p | ψ > = -iħ d/dx | ψ > ?

No it isn't, because $\left|\psi\right>$ is not a function of $x$, it is a vector. The inner product $\left<x|\psi\right>$ is a function of $x$ and you can act on it with differential operators. There is a difference between the abstract operator $p$ that acts on state vectors, and the corresponding differential operator $-i\hbar\frac{d}{dx}$ that acts on wavefunctions.

 

[dyn]

Thanks. So does p | ψ > have any meaning on its own ? Can it be evaluated as it is ?

 

[hilbert2]

Thanks. So does p | ψ > have any meaning on its own ? Can it be evaluated as it is ?

You can't really make anything of the expression $\left|\psi\right>$ unless you represent it in some basis. The position representation, where you convert it into a function $\psi(x)$ is one possible basis. Another way would be to give the complex numbers $\left<0|\psi\right>,\left<1|\psi\right>,\left<2|\psi\right>,\dots$, where the $\left|0\right>,\left|1\right>,\left|2\right>,\dots$ are a complete set of energy eigenstates.