Operators in quantum mechanics

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📚We learn that operators act on quantum states to change them to other quantum states, allowing us to "do" physics. An unusual property of operators is that their product does not commute, which has profound implications in quantum theory, and we define the commutator to characterise this feature. We also discuss operators in the dual space by introducing the adjoint operator, and operators as outer products of a ket and a bra.

⏮️ BACKGROUND

⏭️ WHAT NEXT?

✏️ EXERCISES

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Director and writer: BM
Producer and designer: MC
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These are the most easily understandable quantum mechanics videos I have ever found by a long shot. Keep up the great work!

tannermiller
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Hands down everyone ! This channel is far the best I 've ever met, from a french student

oraange
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another great video :) I hope your channel takes off because, I know from first hand experience, the usual hap hazard manner that QM is introduced can often be disorienting so having a rigorous and structured approach to the math underlying QM is invaluable.

I have one small suggestion which is that you consider your presentation speed. I'm watching your videos to complement a linear algebra text that I'm working through in order to review material in greater depth before being thrown into higher level QM and I've found that, since I am trying to pay close attention and diligently take notes where helpful, I often pause your videos and have actually begun watching on 0.75 speed.

This is of course not necessarily and issue, and one of the nice aspects of youtube (unlike an in person lecture) is that you can do just that. Additionally, I of course do not know if other viewers feel the same, but it's something I figured I would at least mention. Overall, wonderful video. Thank you

workerpowernow
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These videos are so good. I think it was Richard Feynman who said if you think you understand quantum mechanics then you dont understand quantum mechanics, clearly Feynman never met Professor M does science.

richardd
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I am an undergrad, thank you so much for these vedios it provides a cripsy idea and save me from getting frustrated by those hours long vedios and all I ended up with is with frustration and embarrassment.but your videos are so good and so crispy complex topic in elegant way. Love from India keep going

sawanbhattacharyya
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Wow, you pedagogy is one of the best I've encountered so far, explanation+illustration are in very good coordination, makes it incredibly clear for such a complex topic !

Thanks !

InSaNeXmANu
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Excellent! Thank you! These early videos serve as great reviews as well as good initial content.

richardthomas
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Oh I wish I had these videos 10 years ago! There is great value in reviewing these concepts, I realize that I was confusing many things.
I remember being super confused when we did the tensor product of states to create a new basis which could express entangled states. I think those were Bell states.
My prof was calling this "le produit dyadique" which seems different to the outer product. Is it because in many-particle systems, each particle lives in its own distinct hillbert space? Reduced to working in a single hillbert space |u><v| is an operator?

MrOvipare
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Feel so lucky I came across this channel. I'm not studying anything related to quantum mechanics, but even I understood the video very well.

s..l
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Great video! Thank you! I have a quantum midterm coming up and your videos provide the perfect review :)

bryannac
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excellent videos, helped me while I was reading the book

diyabatool
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Thank you, Professor, this series is just awesome! I have a question: on 5:18 you are saying that the proposed expressions are equivalent by definition. Although intuitively it indeed seems right, could you please explain why it's rigorously true? And by definition of what? Thanks again!

wwmheat
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Thank you for taking the time to make these videos! I'm a first year uni student and this is great background knowledge for an MIT OCW course on quantum information science I'm taking.

karthikknarayanan
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Awesome, can you describe how Hilbert space and Eigenvectors come together in quantum mechanics? I am a little confused

TheWingEmpire
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In 10:45 you define the "inner product between a bra and a ket", but the inner product is between two kets, that is more or less equivalent to the action of a functional bra, acting on a ket (braket). The inner product is between the vectors of the same vector space.

pabloweigandt
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I've read in some papers the expression "The charge and flux variables are promoted to non-commuting observables". This means we put a hat to the charge and flux variables and they become operators. How does this work? I don't know how to convert a quantity to an operator. Do you just fix every eigenvalue to be the observations of, for example, charge, and then build the operator that has those eigenvalues?

DrMarcoArmenta
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Dear sir, i was taught that operator can not act on bra vector, it always act on object what sits right to it. I am confused a bit now! Can you please help me out? TIA

sayanjitb
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Hi,
I have a doubt regarding operators,

We usually expand the operators of the form, e^A,
I have stumbled across [x e^p] commutation problem,
Here normally we will expand e^p, but why it will be mistake full if we take x as,
ln(base e) e^x in commutator relation and do further calculations ??

vivekpanchal
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Hi professor M, my question is following, 1) we can write an operator in matrix form. Now lets say a matrix is acting on ket, and gives us a scalar times ket. Now my question is what will be the result if that matrix acts on bra instead ket??? Because operator or matrix can't differ wheather it is bra or ket . So what we'll get for this sandwich operation
(< Bra | operator) | ket > = ???
And < Bra |( operator | ket >) = ???

pritamroy
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Hello Professor. The videos are excellent but would you mind if you could tell us about some books to refer to in case we want some practice . I need some suggestions of books to help me start with quantum mechanics. Thank you.

moinakdey