Recent content by Sammywu

Checked with a book, I found my answer to E3.1) B) will basically lead to "looking for the set of maximum commutable observables". That is not what Eye asked any way. Eye's question already assumed there is a Hilbert space and we found multiple "generalized kets" for one eigenvalues of a...

E 3.1) B). 5). In 3), we actually have used the two property of "tensor product": ( | x_1 > + | x_2 > ) \otimes | y > = | x_1 , y > + | x_2 , y > | x > \otimes ( | y_1 > + | y_2> ) = | x , y_1 > + | x , y_2 > The last property \alpha | x, y > = | \alpha x> \otimes | y>...

E 3.1) B) 3) In looking what we already have, | \psi_x > = | x> < x | \psi_x > , P(y) dy = < \psi | y> < y | \psi > dy , | \psi_y > = | y> <y | \psi_y > , P(x) dy = < \psi | x> < x | \psi > dx and our believe that | \psi > = | x, y > < x,y | \psi > and P(x,y) dx...

E 3.1) a) Even though my knowledge of "tensor product" of vector spaces might not be enough, I did give a thought on how to handle this exercise. 1). How do we know the spectrum of Q is degenerate? Unless, we have another set of eigenbasis and self-adjoint operator, said E, and we...

< A > = \sum_{S{p(a)}} a P(a) \ + \ \int_{S{c(a)}} a P(a) da = ( By E 3.4 and E 3.5 , we derive below: ) \sum_n \sum_{k=1}^{g(n)} a < a_n^k | \rho | a_n^k > \ + \ \int_{S{c(a)}} a < a | \rho | a > da = ( Using the definition of eigenkets as, A a_n^k = a a_n^k A | a...

E 3.6 ) A). If the probability of an obeservable showing a is 1, of course we will expect it always shows a. If the observable has multiple possible outcomes, we really can not say which one it will definiitely show, we basically make a math. average of them and saying this is its average...

E 3.5) By taking the generlized EQ. you posted at #110, < \psi | L | \varphi > = \int < \psi | L | a> < a | \varphi > da or < a \prime | L \varphi > = \int < a \prime | L | a> < a | \varphi > da , this is pretty easy. Taking < \psi | I | \varphi > = \int < \psi | a> < a |...

Just try to summarize what I learn here before continue on E 3.5) 1) Let me start with \sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a | da = I , so \int_{S_{c(a)}} | a > < a | da = I - \sum_{S_{P(a)}} P_{a_{n}} . 2). Take derivative of a to it at the continuous part...

E 3.5) By P3, P(a) = | < a | \psi > |^2 da = < a | \psi > \overline{< a | \psi >} da . < a | \rho | a > = < a | ( \psi > < \psi | a > ) = < a | \psi > < \psi | a > We already know what is < a | \psi > , but unclear about what < \psi | a > could be. By comparing the two...

E 3.2) \forall \psi \in H \ , < \psi | \ ( \sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a | da ) \ | \psi > = < \psi | 1 | \psi > = < \psi | \psi > = 1 < \psi | \ ( \ \sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a |da \ ) \ | \psi > = < \psi | \ ( \...

E 3.4) is very similar to E 3.3). (a) Using P_{a_{n}} | \psi > = \sum_{k=1}^{g(n)} | a_n^k > < a_n^k | \psi > , we get P ( a_n ) = < \psi | P_{a_{n}} | \psi > = < \psi | ( \sum_{k=1}^{g(n)} | a_n^k > < a_n^k | \psi > ) = \sum_{k=1}^{g(n)} < \psi | a_n^k > < a_n^k | \psi > =...

I will do E 3.3) first. (a) Using P_{a_{n}} | \psi > = | a_n > < a_n | \psi > for nondegerate eigenvalue a_n because the projector will map | \psi > to its subcomponent of | a_n > , we get P ( a_n ) = < \psi | P_{a_{n}} | \psi > = < \psi ( | a_n > < a_n | \psi > ) = < \psi |...

E 3.1) The one I can think of that mixes discrete and continuous spetrum seems to be the simple one in that the discrete eigenvalues are basically the degenerate case of [ q , q + \triangle q ] . By basically taking a function that maps the underlining continuous interval to discrete...

Eye, My previous post regarding the object \psi_q seems to make sense, but I just got into troube when trying to verify that with the inner products or norms. So, I guess it's not working any way. You can just disregard that and just move on. To me, I really appreciate what you...

Eye, I roughly got you, still reading it, glad that you clarify many points here . I definitely agree that you can just bypass #98-102. About my \psi_q and |q>, I tried some more in clarifing what's going on: I). Using an self-adjoint operator as example, starting from a discrete...