Completeness of eigenvectors and measurement postulate

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MIT 8.04 Quantum Physics I, Spring 2016
View the complete course: ocw.mit.edu/8-0...
Instructor: Barton Zwiebach
License: Creative Commons BY-NC-SA
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Пікірлер: 26

@eskilandersen479 4 жыл бұрын

Genius lecturer! Amazing

@jacobvandijk6525 4 жыл бұрын

@ 2:23 It's simple because it's just the dot-product of a basisvector (= little psi) and a vector (= big psi). And that gives the component (= alpha). Simple linear algebra, presented in a difficult way. Don't forget, functions are like infinite dimensional vectors.

@hasnaamagdi4473 4 жыл бұрын

could you explain more, please? thanks

@jacobvandijk6525 4 жыл бұрын

@@hasnaamagdi4473 Take the vector V = x . i + y . j. The dot-product j . V = y (= component belonging to the basisvector j). Take a function w = f(v). For any value of v there is a value of w. In QM any value of w is a component of a vector. Because v can take on an infinite amount of values, this gives an infinite amount of components. Hope this helps a bit. If not, watch this video: kzbin.info/www/bejne/Y3nIq6mBYrGLg9U

@rajinfootonchuriquen Жыл бұрын

I don't see what's the complexification of the matter. It's standard proof of any linear algebra class. Also, it's called "inner product" when the proyection is in a infinite dimensional space, and function are not infinite dimensional vector, but the collection of all of them is infinite dimensional because composition of function is not linear, so that composition can be write it as a linear combination of basis vectors (function), which is the proof that function space are infine dimensional.

@jacobvandijk6525 Жыл бұрын

@@rajinfootonchuriquen Well, you sound smarter than I am when it comes to math. I like to reduce more or less abstract things into more or less simple things, knowing that concepts like abstract and simple are rather personal.

@rajinfootonchuriquen Жыл бұрын

@@jacobvandijk6525 I tell you because what you said was wrong, not for to be a smartass :)

@Kevinthegreaty 4 ай бұрын

What does “measurement” exactly mean here, is it any way we can use to get the value of the required quantity? Does any measurement make the wave function collapse, or to leave it the same state as measured?

@keshavmittal1077 3 жыл бұрын

when wavefunction collapse at someplace, what factors lead to certain wavefunction?

@johnschack5651 2 жыл бұрын

What is the significance of the reasoning around how a vector (or function) can be written in infinitely many ways as a superposition of vectors(or basis function)? I don't really follow this line of reasoning.. If we find eigenfunctions and organize them to be orthonormal, isn't it true then that any wavefunction can only be represented as a unique linear combination of the orthonormal basis?

@rajinfootonchuriquen Жыл бұрын

You have a canonical basis, where is easiest to write thing as a linear combination. In reality, there is infinite set of basis vector to span the space, thats why you can write it in infinitely many ways, but, given a unique set of basis, there is a unique way to write it.

@not_amanullah 2 ай бұрын

Thanks 🤍❤️

@srivishnudasu1694 4 жыл бұрын

13:45 so if wave functions are considered to be vectors, i don't need only eigen-functions to get a span of vectors right? if i know any two random wave functions i can get the span of the entire coordinate space using those two wave functions, can i not? then why only eigenfunctions?

@chensong250 4 жыл бұрын

You only need eigenvectors because you want to avoid redundancy. For instance, any vector in 3-D can be expressed by three eigenvectors, also by 4 or more vectors but 3 is the least you need. Meanwhile, two will not work so you need to take the full set of eigenvectors to span any vectors in 3-D, otherwise you are only getting a subspace of all possible states. Check the definition of linear independence and subspace in linear algebra for more insights.

@srivishnudasu1694 4 жыл бұрын

@@chensong250 agreed I was only referring to the 2D part. But are there any specific transformations so as to consider only eigenvectors instead of any vector in general ?because if we can use any general vector wouldnt using eigen vectors become redundant?

@chensong250 4 жыл бұрын

@@srivishnudasu1694 I didn't quite get your question so I will try to answer. Let's say you have for any 2-D state x = ax1+bx2 where x1, x2 is a basis (such as eigenfunctions) and a, b are the coefficients. Then any linear transformation L (for instance, a rotation) acting on x can be written as Lx = L(ax1+bx2) = aLx1+bLx2. This is powerful because if you want to know what happens to any state x, it reduces to knowing what happens to two vectors x1 and x2 instead. Now a vector space can have multiple equivalent sets of basis and eigenfunctions are just one choice of basis. It is preferred in some cases because it's easy to compute Lx1 and Lx2 which is just multiplying with the eigenvalue.

@Pranav_Bhamidipati 2 жыл бұрын

@@srivishnudasu1694 As Song Chen already said, setting the set of eigenvectors as the basis will diagonalize the Linear Transformation matrix which is a very helpful thing in complex computations.

@sangraampatwardhan1573 5 жыл бұрын

Does the 'any' really mean ANY? Does it mean that if I know ANY set of some eigenfunctions of a Hermitian operator, I can express any reasonable psi in the world in terms of those eigenfunctions?

@physicsguy877 4 жыл бұрын

Yes. Given a Hermitian operator, its eigenfunctions span the space of possible states of the system, meaning you can express any state as a linear combination of the eigenfunctions. You should be aware though that there are Hermitian operators for which the eigenvalues are continuous (like position and momentum), so then to express a state in terms of the eigenfunctions you use an integral rather than a discrete sum. The book by Shankar has a very good chapter on this.

@physicsforallpfa5866 4 жыл бұрын

@@physicsguy877 I am pretty sure that those eigenfunctions are called the basis of the Hermitian Operator and if we take required number of those basis functions with certain coefficients, then only we are able to make any wave function that is required and belongs to Hilbert Space.