...and given that so many have applauded this dismissive post, I am done. If your goal was to scare off someone who wants to learn, mission accomplished.
This is a kinder, gentler form of exactly the thing I am talking about. You, of course, have zero evidence that my question was ill-considered. It may seem that way from your perspective, but it is patently unfair to imply anything of the sort on my part.
My question was genuine, I got flamed...
What do you think could have been different? See post 3.
I have been back on this site primarily to exchange some PMs. If you guys want this to be a place where people who are not as knowledgeable as most of you are feel welcome, well, I think you could be a tad more charitable. Let me be...
Thanks to those who answered courteously and with respect. However, I have decided to leave this forum so I will not be reading any more replies. Thanks again.
I just learned that if two linear operators do not commute, this means when we use operators to characterize observables in quantum mechanics, the corresponding observables cannot both be definite at the same time. This seems hard to believe to me since I have a strong intuition, perhaps...
Great. Got it, very clear explanation. However, do you not agree that the speaker misleads us just a little when he says "B-alpha has to be the same eigenvector as alpha since they both have eigenvalue lambda". Shouldn't he have said something like this after introducing the nondegenerate case...
Since this is a "general lounge" area, I will assume this rather personal story is not out of line. I am 64 years old (relevance of my age should become clear shortly) and graduated from an Ivy League University in 1980 with a degree in Electrical Engineering. I also scored over 700 on my math...
Here again with another question about the Quantum Sense video series. Thanks for all the useful feedback to my last question. My question concerns a very short chunk of about 20 seconds beginning at 4:25 of this link:
At around 4:34, he says "B-alpha has to be the same eigenvector as alpha...
Thanks to both who answered. I have not had time yet to go through your answers in detail, but I can think of one immediate follow-on. You both obviously think an expression of the form |x> <y| is legitimate - operator on right, ket on left. I was thinking that since the form <a| is used to...
I suspect it will help if you know about my background: I did some linear algebra in university but never used it and am now in my mid 60s. I am interested in understanding the mathematics of quantum physics. I have read a number of layman's texts on quantum mechanics, but they all gloss over...
I hear you, but I doubt that math is the issue - clarity of explanation, or my own basic ability to follow the details of an explanation, is this issue, I believe. There is little math in Carroll's text (perhaps none?) and it is definitely not the math that is tripping me up. Either the author...
I got about 90% through and then gave up. Let me be clear: I have heard his lectures and he certainly seems to be trying his utmost to make this exotic and abstract subject accessible to regular shmoes like me (I have an undergrad engineering degree and have read a fair bit of the material out...
Sean Carroll (in a video) claims that regions of empty space (vacuum) that are near each other must be highly entangled. He appears to argue that if they were not, there would be "a lot of energy contained there" which - my conclusion - would not be consistent with these regions being low energy...