Position and momentum operators acting on wave functions

How do the position and momentum operators act on wave functions?

📚 In this video we learn how these two operators act in both the position and momentum representations of quantum mechanics. The position operator multiplies a wave function by "x", and the momentum operator calculates the derivative of a wave function.

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Director and writer: BM
Producer and designer: MC

[JohnSmall]

I love this series. Such clear explanations. I wish it had been available when I first learned QM.

[anindyarastogi]

I am currently reviewing basic quantum mechanics for one of my exams, and your series is greatly helpful in revising a lot of this stuff. Your explanations are very clear and concise.

[diegoalejandrocoronagomez]

Thanks dude, you are saving my life in QM

[wadelamble]

Thank you for the incredible content. Is the position/momentum commutator value derivable? If so, have you derived it one of the previous videos?

[Maxwells_demon_]

Hi Prof M, Thanks for this video. After watching your videos on tensor products, i am revisiting this video to try and understand how the action of the momentum operator in the position representation generalises to the gradient operator in 3D. While I understand that representing r as a tensor product and applying px to the x ket and so forth works (as seen in an earlier comment), I do not see how the spatial unit vectors appear. The gradient operator comprises of these unit vectors and I don't know how they pop up by following this logic.

Thanks in advance.

[jokerbatman]

Sir, my question is that this relation <x|P|psi> =-ihd(psi)/dx --->1 is true but then from this relation writing P operator as P=-ihd/dx is wrong fundamentally as in original equation(1) psi is in derivative and therefore cannot be cancelled from LHS and RHS therefore I think when you write P=-ihd/dx it is always implied that in LHS there is psi ket on right and x bra on left ??, as eq 1 I think is more correct fundamentally.

[BruinChang]

Is an action of the momentum operator p_hat on a state when written in position representation equivalent to "projection of the consequence of performing a momentum operation on a state onto the eigen-states of a position operator" ?

[한두혁]

I have a question. Following the logic from your video(postulate of QM vid no6: changing basis in QM); matrix representation of operator X in position space is <x|X|x’>. In momentum space however it will be <p|X|p’>. Now I put identity operator as you did s.t <p|X|p’>=<p|x><x|X|x’><x’|p’> (I will omit integral over x and x’) since <x|X|x’> is just a delta function and I know <p|x>, so this integral should give me the momentum space representation of X but actually this doesn’t. But what is wrong with this equation? Thankyou.

[Upgradezz]

How does the generalisation to three dimensions follow from the individual representations in x, y, z?

[jam]

Can you tell me the correct order to watch these videos.?

[soumapriyamondal]

Do position and momentum operators commute in position space?

[foxcote]

The crewmates have to collect coins around the map