How Do Position and Momentum Operators Act in Quantum Mechanics?

[Joao Victor]

Yesterday, I was solving an exercise from Cohen-Tannoudji's book - Quantum Mechanics -, but then I got stuck on the second question that the exercise brings. I wonder if you guys could help me, and here is the exercise:

"Using the relation <x|p> = (2πħ)^-½ e^ipx/ħ, find the expressions <x|XP|ψ> and <x|PX|ψ> in terms of ψ(x). Can these results be found directly by using the fact that in the { |x> } representation, P acts like h/i d/dx ?"

I have found the expressions, but I don't know how to answer the question in bold (I do know that P acts like h/i d/dx in the x position representation, but I haven't figured out how to use this information on the exercise.

Here are my solutions:

<x|XP|ψ> = (<x|X)(P|ψ>) = x <x|P|ψ> = x ∫dp <x|p><p|P|ψ> =
= x ∫dp (2πħ)^-½ e^ipx/ħ p <p|ψ> = x(2πħ)^-½ ∫dp e^ipx/ħpψ(p)
= x ħ/i d/dx (ψ(x))
⇒ <x|XP|ψ> = x ħ/i ψ'(x)

<x|PX|ψ> = ∫dp <x|p><p|PX|ψ> = ∫dp (2πħ)^-½ e^ipx/ħ p <p|X|ψ>
= (2πħ)^-½ ∫ e^ipx/ħp dp ∫dx <p|x><x|X|ψ>
= (2πħ)^-½ ∫ e^ipx/ħp dp ∫dx (2πħ)^-½ e^-ipx/ħ x ψ(x)
= (2πħ)^-½ ∫ dp e^ipx/ħ iħ pψ'(p)
= -ħ/i (2πħ)^-½ ∫dp e^ipx/ħ pψ'(p)

After some integration by parts...

= ħ/i [ψ(x) + xψ'(x)]
⇒ <x|PX|ψ> = x ħ/i ψ'(x) + ħ/i ψ(x)

I hope you can help me - and I apologize for the horrible format above.

​

 

[nrqed]

Yesterday, I was solving an exercise from Cohen-Tannoudji's book - Quantum Mechanics -, but then I got stuck on the second question that the exercise brings. I wonder if you guys could help me, and here is the exercise:

"Using the relation <x|p> = (2πħ)^-½ e^ipx/ħ, find the expressions <x|XP|ψ> and <x|PX|ψ> in terms of ψ(x). Can these results be found directly by using the fact that in the { |x> } representation, P acts like h/i d/dx ?"

I have found the expressions, but I don't know how to answer the question in bold (I do know that P acts like h/i d/dx in the x position representation, but I haven't figured out how to use this information on the exercise.

Here are my solutions:

<x|XP|ψ> = (<x|X)(P|ψ>) = x <x|P|ψ> = x ∫dp <x|p><p|P|ψ> =
= x ∫dp (2πħ)^-½ e^ipx/ħ p <p|ψ> = x(2πħ)^-½ ∫dp e^ipx/ħpψ(p)
= x ħ/i d/dx (ψ(x))
⇒ <x|XP|ψ> = x ħ/i ψ'(x)

<x|PX|ψ> = ∫dp <x|p><p|PX|ψ> = ∫dp (2πħ)^-½ e^ipx/ħ p <p|X|ψ>
= (2πħ)^-½ ∫ e^ipx/ħp dp ∫dx <p|x><x|X|ψ>
= (2πħ)^-½ ∫ e^ipx/ħp dp ∫dx (2πħ)^-½ e^-ipx/ħ x ψ(x)
= (2πħ)^-½ ∫ dp e^ipx/ħ iħ pψ'(p)
= -ħ/i (2πħ)^-½ ∫dp e^ipx/ħ pψ'(p)

After some integration by parts...

= ħ/i [ψ(x) + xψ'(x)]
⇒ <x|PX|ψ> = x ħ/i ψ'(x) + ħ/i ψ(x)

I hope you can help me - and I apologize for the horrible format above.

​

It is probably too late for answering your question but what they meant is that your final results are equivalent to making the following substitutions:

$$\langle x | XP | \psi \rangle \rightarrow x (\frac{\hbar}{i} \frac{d}{dx} ) \psi(x)$$

and

$$\langle x | PX | \psi \rangle \rightarrow (\frac{\hbar}{i} \frac{d}{dx} ) x \psi(x)$$