Lecture 7 | Modern Physics: Quantum Mechanics (Stanford)

[samjenkinssam]

brilliant video

[grunder20]

More like the Einstein of the modern world.

[Andalfe]

" Is analytical reduction of states analogous to this?"

Some of these questions people ask honestly jesus wept

[qftftw]

@Andalfe Agreed. I cringe every time I hear "Michael"'s whiny voice with yet another trivial observation or nonsensical question. And he's with us all the way through the field theory lectures...

[dirac17]

Call me nuts but I do not care for his lecturing style. It is somewhat disorganized and more abstract than it needs to be. Would prefer if he did some applied problems to tie the lecture to reality.

[jamma246]

He says that the Hermitian operators have a complete set of orthogonal eigenvectors, and that this definition is equivalent to the one saying that it must be equal to its conjugate transpose. But what about, for example, a matrix with a "1" in the top left corner and "0" elsewhere. This satisfies one of the criteria but not the other. I think that he needs to add something like "invertible" also to his definition. Why do people in the lecture only ask dumb questions :(

[MrHenshin]

@jamma246 The matrix you describe does indeed have a complete set of orthogonal eigenvectors; the second one can be the [0 1] vector with eigenvalue zero. Of course, you could also pick [0 -1], and that would work equally well. The basis exists, but that doesn't necessarily mean it is unique.

[jamma246]

@MrHenshin Ah, never though about it like that before, thanks. So the zero matrix is Hermitian xD

[bhigr]

He is really focused on the basic conceptual ideas. What is a state, what is a pure or prepared state? What is an observable? What is the expected value (average value) of a measurement. But, is the average value really an "expected value"? Well the average value may deviate from any possible measurement value (the average value for flipping coins (+1,-1) is zero, but zero is not a possible observable value) In this case the average value is not an expected outcome, otherwise it is...

[airwanifs]

These lectures are the greatest! I couldn't find any sources on modern physics until I found your SR series. If I win any prizes in the future, I'll be sure to acknowledge you.

[xXTaNodel408Xx]

i wish i had a brain like his......even though i dont get it it fasinates me

[ravaabyee]

these lectures are better than james cameron and steven spielberg put together

[luzzie9]

Professor Susskind suggested that a course in quantum field theory would follow. What happened to it?

[foketesz]

I may be loco, but I enjoy these lectures more than recent blockbusters...

[AmanAttitude]

@wawens

@Mazmatickz

@ArjenDijksman

Dear friends, as you all must be knowing that the magnitude of electric field vector is constant in Circular polarized light, can anybody of you tell how does physically rotating linear polarized light source ensure the above condition??? So, i think it is not possible to obtain a circular polarized light the way "wawens" has suggested (as linear polarized light source will give e. field with varying magnitude). we can discuss further, i may not be correct..

[mosswalker2]

GEEKS ! !  ;-)

[loopantenna]

@mosswalker2 Dumbo! :P

[cursorminer]

maybe all us youtube viewers should club together to buy Prof Susskind a bunch of pens??

[listminusv]

lol i'm up for that....we should like post it to him and write down...."from your faithful you tube physicists" !!!

[wawens]

Excellent video. Thanks.

If I were to physically rotate my linear polarized light source would I then obtain an equivalent circular polarized photon stream in the same way as an ordinary cicular polarized source?

[Mazmatickz]

Well, I think it would be almost impossible to rotate a linear polarized light source manually to match the speed of which ordinary circular polarized light rotates, the angular frequency of circular polarized light is too high. Angular frequency is the speed of light times the wavenumber: w=c*k, this is a very big number.^^

[ArjenDijksman]

That's an interesting question. Rotating the source (=the electrons of the source) gives unpolarized light with each photon remaining linearly polarized. I think that even if we could manage to rotate the source at the frequency of the emitted light, the individual photons would still be linearly polarized. But at that frequency the emission profile of the electrons of the source wouldn't be the same.

[mrbondisback]

excellent as always:):)