Lecture 4 | The Theoretical Minimum

January 30, 2012 - In this course, world renowned physicist, Leonard Susskind, dives into the fundamentals of classical mechanics and quantum physics. He discovers the link between the two branches of physics and ultimately shows how quantum mechanics grew out of the classical structure. In this lecture, he continues his discussion on the vectors and operators that define the language of quantum physics.
Stanford University:
www.stanford.edu/
Stanford Continuing Studies:
continuingstudies.stanford.edu/
Stanford University Channel on KZbin:
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[@andrewcriscione]

These will be viewed as the modern Feynman Lectures, thanks Stanford!

[@smajidy]

0:00:00 to 0:23:09 - Review of postulates 0:23:10 to 0:56:40 - Evolutions in time 0:56:41 to 1:11:55 - Expectation value 1:11:56 to 1:37:34 - Time evolution of expectation values 1:37:35 to 1:47:22 - Conservation of average energy

[@elliotmacrae-sadek6752]

Thanks beast

[@petergreen5337]

Thank you

[@Lucas-ss5xi]

Lecture 4 - Time and Change 0:00:00 - Review of 3.2 The Principles 0:23:10 - 4.1 A Classical Reminder & 4.3 Determinism in Quantum Mechanics 0:26:45 - 4.2 Unitarity 0:30:48 - 4.4 A Closer Look at U(t): (U(t)†U(t)=I) 0:39:07 - 4.5 The Hamiltonian 0:49:31 - Generalized time dependent Schrödinger Equation 0:56:43 - 4.7 Expectation Values (Average Values) 1:11:56 - 4.9 Connections to Classical Mechanics 1:37:35 - 4.10 Conservation of Energy 1:26:40 to 1:30:00 - 4.6 What Ever Happened to ħ? the names of these chapters are taken from the following book: Leonard Susskind, Art Friedman - Quantum Mechanics: The Theoretical Minimum (2014)

[@lotusenmarid4715]

thanks! this is indeed really useful to those who have the book.

[@rshxrma]

I'M DONE. I'm really jealous of him eating those sweet delicious-looking whatever those are!

[@parikshitkulkarni3551]

same!!!!! everytime i see him eating those it makes me want to eat something too!!!!!

[@stevejones4275]

44:00, he keeps dropping these bombshells! Hamiltonian and energy are the same. Before he said that 'normal to' and 'orthogonal' are the same.

[@wagsman9999]

These are fantastic!

[@capitanmission]

2nd law of thermodynamics correctly restated by Lenny: Entropy probably never decrease :D

[@supern0is349]

this is awesome, thanks a lot!!!!

[@amirhosseinkzm9278]

Thank you very much.

[@sandeepchandappillai9814]

Thank You !

[@user-lm4vi4dw2r]

If you have two orthogonal vectors corresponding to the same eigenvalue, say 1, how could you possibly distinguish between those two states? The only measurement you’ll see is “1”! But you’re supposed to be able to distinguish between orthogonal vectors?

[@enisten]

That's a really good question. Ask Google Bard and ChatGPT: "How can you distinguish experimentally between two eigenvectors with the same eigenvector in quantum mechanics?" I just did. And they answered there are some effects like the Zeeman effect and Stark effect that allow you to distinguish between them experimentally. So the claim that orthogonal states can always be distinguished seems to be true in general.

[@danfara6126]

Thus, is the misterious "i" in Schröndinger equation being explained by saying U=I+a H with H hermitian and U unitary and "a" a complex number necessarily "a" has to be purely imaginary?

[@greenfloatingtoad]

It has to be purely imaginary to connect infinitesimal generators of unitary operators to observables. The infinitesimal generators of unitary operators are anti-hermetian. Observables are hermetian. A factor of i takes you from one to the other

[@gershunistepan]

Entropy never deacreases except when it does xD

[@TheVivek199]

1:23:49 Learning a bit of french certainly helped there

[@MarcusMChase]

Also useful for Poisson contemporaries: Lagrange (the barn) and Laplace (the place)

[@enisten]

Can someone explain why the probability of observing a degenerate eigenvalue is the same regardless of which set of orthogonal eigenvectors we choose to use? Depending on this choice, we will get a different set of inner products with the state vector. So, how come the total probability of observing that degenerate eigenvalue remains unchanged? PS. I understand it must be invariant because the sum of the probabilities of remaining eigenvalues doesn't change. But I'm looking for a direct and intuitive explanation.

[@chimaru8943]

For his explanation of orthogonal space vectors, if you are "prepared" with an "up" or "right" and measure along the z axis and measure "-1" wouldn't that tell you that it can't be up and thus has to be left?

[@attilakun7850]

It's not left though. You measure again along the x axis and you'll get a random answer.

[@annamallett879]

You are correct that if you measure along the z axis and get -1 then you know the spin was originally 'prepared right', as it can't have been prepared up. (Likewise, if you measure along the x axis and get -1 then you know the spin was prepared up, as it can't have been prepared right.) The problem is that *even if* the spin is prepared right, there's only a 50% probability that the result of a measurement along the z axis will be -1, and if the result is +1 then you still can't be sure of the original preparation. For state vectors to be orthogonal, there has to be a measurement you can make that will tell them apart under _every_ possible outcome. For example, if you have a spin that is prepared either 'up' or 'down' and you make a measurement along the z axis, you are guaranteed to get the result +1 or -1 respectively. No matter what the result of the measurement, you know with 100% certainty how the spin was prepared. There is no equivalent unambiguous measurement you can make for a spin that you know has been prepared either 'up' or 'right', so those state vectors are not orthogonal.

[@chimaru8943]

Anna Mallett I see,, thank you !!

[@zphuo]

@26:00 Although QM is full of randomness, it is still reversibility & Information conservation. Why? @52:00 Generalized time dependent Schrodiger Equation.

[@Spielix]

All the randomness you are talking about comes from Observations and therefore in the Copenhagen interpretation from a collapse of the wave function. The operator corresponding to such a collapse cannot be unitary and your reversibility and Information Conservation go overboard. This is the reason why people who think about the interpretation of QM tend to like the "Many Worlds"-interpretation instead. Most physicists just take the "shut up and calculate"-ansatz and don't think about this at all. One could argue that the biggest reason for the Copenhagen Interpretation to be the only one given in many books is that most physicists learned it as students and then gave up about thinking about the interpretation because you don't need an interpretation to calculate some theoretical results and compare them with the corresponding experiment. In the end this is pretty much a philosophical matter and therefore shunned in modern science.

[@zphuo]

I learned so much from you. Thanks very much.

[@Spielix]

@polka You are contradicting yourself. The Copenhagen interpretation explicitly states WF collapse. What you are describing pretty much sounds like the Many Histories interpretation, which is modified Many Worlds. I never said that any of these is truth or sth like that. No one knows.

[@Spielix]

@polka If the WF of a photon would look like a delta before hitting a screen, we wouldn't observe the outcome of the two slit experiment as it is.

[@addis11100]

@21.47 he is crying for oreo or quantum entaligment?

[@sabrewolf479]

9:37 What's the odds of forgetting #4?

[@Dragonlordz7]

From 1:06:00, when he starts writing the probability of lambda, I'm getting confused with the notations there. What is lambda(i)? is he taking it as an eigen vector and and another lambda (i) as its corresponding eigen value and what is lambda, where he's put inside the expected or average value? When he's trying to find the probability of lambda (i) at 1:07:20, what is lambda(i) there?

[@capitanmission]

i guess it is the eigenvalue representing the output of an experiment.

[@capitanmission]

A measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured, the eigenvalue this eigenstate belongs to being equal to the result of the measurement. - P.A.M. Dirac (1958) in The Principles of Quantum Mechanics p. 36

[@roberthayter157]

He should be writing eigenvector |A(λ)> but he is cheating with the notation a bit and simply writing |λ> to represent the eigenvector. He mentioned (in the previous lecture? I can't remember) that he would do this.

[@Gwunderi25]

"It's pure pluginology" he sais at min. 1:09:25 - so nothing simpler than QM maths : )))

[@guitar_jero]

Wouldn't it be better if we defined the Poisson brackets as: {A,B} = lim h->0 (2πi/h)*[B,A] ??? So, it's a 0/0 limit and therefore the left side can exist and be unique.

[@jinchoi8296]

Hyero Nimus Then it would make quantum mechanics continuous, which has no difference with classical mechanics

[@Spielix]

Yes, he generally is using the equal sign very freely, at least in these "Theoretical Minimum" lectures. Mathematicians would have their problems with this at many points. Best read his equal signs in those situations as "corresponds to" or sth.

[@jeffreylanz719]

8:35 haha looks like a child's drawing of a dinosaur

[@raghavendrakaushik4871]

At 14:47 shouldn't the probability be as discussed in the previous lectures? It is not just square, it is square of length of the component.

[@cubeik-jakubvlcek698]

I'm confused about this too. But I think the definition from the previous video should be correct, since you don't get a real number from squaring a complex one.

[@cubeik-jakubvlcek698]

But maybe it can actually be the same thing since multiplication is considered to be an inner product that consists of vector A and its conjugate.

[@netrapture]

at 1:11:12 of the previous lecture (Lecture 3) he mentions and writes that = ||^2. The right hand side is not the square of a complex number, it's the square of the length of what's inside (the vector projection of A onto the eigenvector associated with Lambda). In any complex vector space length(z)^2 = |z|^2 = z*z just like for complex numbers.

[@12388696]

at 1:10, isn't the lambda i used twice - once by L hit ket lambda i and another time ket lambda i was written again?

[@hasanshirazi9535]

When L hits |λi> it gives eigenvalue λi and eigenvector |λi>.

[@morgellonbetancor1453]

DE TEORIA A LA PRACTICA SI EN CASO QUE SE DIESE PRESTO,SALUDOS

[@johnboro64]

I have absolutely no idea what he’s talking about, but I know it’s in English. It’s fascinating listening to you someone so clever, the best way I’ve found to sleep ever, I love it

[@guestguest9603]

33:30 I'm not able to prove that the inner product remains constant by simply expanding into basis. All the proofs I found online already assume U is unitary, but we can't assume that because he later derives that from the above proof. Anyone got any suggestions?

[@brendawilliams8062]

Do you know if any math exist that is technical on Hawkins radiation available on yt.?

[@MrAngryCucaracha]

We have two vectors, v1 and v2. We have two normal bases, n1 and n2. The vectors can be expressed as v1=a1*n1+b1*n2, v2=a2*n1+b2*n2, the dot product is a1*a2+b1*b2 (i ignore complex conjugation for simplicity), because n1*n1=n2*n2=1 and n1*n2=0. Now we apply the transform that preserves angles, because it is linear we get v1=a1*U*n1+b1*U*n2. Same for v2. Now we multiply the vectors and get the same as before, dot product is a1*a2+b1*b2, because U*n1*n1*U=U*n2*n2*U=1 (the module of all vectors was said to be 1), and U*n1*n2*U=0 because it preserves orthogonality. Since all vectors can be separated in their orthogonal components, if you preserve the angles between orthogonal components you preserve all angles and so all dot products. I recommend the linear algebra classes by gilbert strand.

[@enisten]

So, preservation of orthogonality is not enough to prove preservation of inner products. You have to also assume preservation of lengths (i.e. probabilities in quantum mechanics). When you assume a linear transformation preserves BOTH orthogonality AND lengths, then you get that it actually preserves all inner products, of which 0 (with orthogonal vectors) and 1 (with itself) are just special cases. Am I right?@@MrAngryCucaracha

[@MrAngryCucaracha]

@@enisten yes, if a transformation rotated the vectors and doubled one of then, the new inner product would be the double of the old one. But since all vectors are stated to be of length one, all transformations must preserve the length.

[@sansha2687]

00:18:00 , 1:04:50

[@shivaagarwal4093]

The time developement operator is acting on state Psi(0) and getting us Psi(t). But linear operators aren't supposed to change the eigenvectors themselves. Is this the problem of notations or something else?

[@danielstone8775]

SHIVA AGARWAL Acting on psi(0) can be expanded as a superposition of the wave function and eigenvectors. Psi(t) would be an eigenvector in this expansion corresponding to the eigenvalue the apparatus records after measurement. But I dont know if this is how it works when the operator corresponds to time. I guess its the same.

[@shivaagarwal4093]

Daniel Stone I still require clarification on this one.

[@shivaagarwal4093]

Since psi(0) is transformed to psi (t) only after the operator acts on it... Am I correct that operators don't change the state vector? This is the confusion.

[@danielstone8775]

SHIVA AGARWAL Yeah I know. I am a newcomer to quantum physics. This is "easy." Everything becomes hard when you get to entanglement and quantization.

[@danielstone8775]

SHIVA AGARWAL Operators are mathematical tools to determine what are the possible states in which a system may be left after measuring the observable corresponding to the operator (also the values which the apparatus may record). You are right: The notation is confusing since there is no indication that the measurment has been made.

[@artkirakosyan2633]

Fourth lecture trying to figure out if there is any relation between spin up or down. I mean does the vectors have even anything in common? Can somebody answer ?

[@yesegg3596]

Spin up and down are considered orthogonal, they are linearly independently and cannot be written in function of each other

[@deinauge7894]

and of course the relation has been made clear from the start of lecture 1: 'up' in one reference frame is 'down' in a mirrored reference frame

[@xinzeng-iq7zv]

conditional probablities for beginners

[@mikikaboom9084]

1:04:03 T... ten?

[@dancrepeau1]

30:20 "I've given up trying ... I actually never tried." 😂

[@vishwasshankar3929]

38:08 😅 nice wordplay

[@mohammadnooralam5968]

58:00 what name profesor says

[@charlesb3300]

Murray Gell-Mann i believe

[@jimdogma9890]

In response to one comment in this thead, information has essentially ALWAYS been free in this country at least. It is just always been a matter of your industriousness to find it, which most people don't have or are too lazy to have. I published my first article in cog neuroscience in 1995 using only the local library, card catalogs, and interlibrary loans. The internet is, yes, great, but it did not make learning possible, it just made it easier.

[@lina31415]

the "law" of large numbers is literally a theorem, meaning there is a proof for it..

[@samistheman32]

A mathematical proof, you can't do it physically

[@enisten]

Which mathematical theorem can be proved physically?@@samistheman32

[@muzafarganie4339]

What the hell he is doing, inserting only equations and symbols.

[@ryanlafferty5948]

I bet the good poor people you knew didn't know any physics.

[@teflondon4210]

21:20 :DDD

[@FarFromEquilibrium]

nobody owes you a damn thing. be grateful that this exists and you have access to it, and remember that being poor does not make you a good person any more than being rich makes you a bad one! Some of the worst people I've ever known were poor and some of the most decent ones were well off, and vise versa. Of course i've known a lot of good poor people and slimeball rich people too, most of us have.

[@hasanshirazi9535]

@polka Its also easy to be an AH if you are rich.

[@bzqp2]

U owe me 5 bucks

[@brendawilliams8062]

Well, monetary gain may not be the only objective. It’s just nice that those who made enough of it in the Professor’s positions shared. I don’t know that I wouldn’t have been stingy with greedy. You never know if you never will know.

[@ghenkhoash2440]

Last week I bought a second hand iphone 6.

[@ghenkhoash2440]

@@caleboh9321 $100