Recent content by yogeshbua

Caution: The applications come in after mastering the basics. And indeed, you probably can, and certainly may, ask doubts. Just message me the link to the corresponding thread... Cheers,

I'd recommend 1) Intro to Quan Mech by David J Griffiths 2) Principles of Q M by R Shankar 3) Modern Quan Mech by J J Sakurai (The latter only to be read after the first two) for Quan Phy. For modern Physics, no idea! There's so much that any particular author will be good at something and...

I followed (all) the arguments you made in the quoted post. It does not seem confusing. However, I should add that this third operator, A, need not commute with O! (Consider diagonal matrix 1 1 2 for O (In a particular basis). Diagonal matrix 1 2 3 for T. And A such that it has it's eigen...

Danke.

Which book is this?

You may try out Introduction to Electrodynamics by David J Griffiths...

We should ask someone the conditions for this implication. That is, that of, 'physically distinguishable states' \Rightarrow 'there exists some observable for which the two states have different eigenvalues' It would be worth knowing... Sorry, but tell me, are all of us students, or do...

Here's something we can think about. I was convinced by Jensa that different 'physical states' would mean ones that can be 'observed' distinctly and hence, states, I quote (recollected version) 'that can be physically distinguished from one another'. Suppose |1,1> and |1,2> are the eigen...

Yes. We can 'find' (have in existence, though not 'observe') states that are neither 'different' nor the 'same'. But we cannot 'find' 'physical states' (and 'find' for 'physical states' means 'observe') that are neither 'different' nor the 'same'!

Yes... Thank you. I had not taken that perspective. Edit 1: I agree with you completely, given the interpretation of 'physical states' as you have. Thank you...

Agreed... Thank you, again. Do I need to mark the question as solved/ whatever? How? Cheers.

I hope that the next time you see an old post opened for discussion again, you'll not be surprised. Here's why: Phy Forums keeps old threads? Why? For people in the future to come and see if their doubts are solved. If a future person has to add to the discussion, (s)he should do so, right...

But you see, the wiki article talks about, to quote, "In physics two or more different physical states are said to be degenerate if they are all at the same energy level" If we say that, to quote, "When we talk about two different "physical states" we mean physically distinguishable states"...

So may we say that 'We assume the eigenvalues are nondegenerate'. Or does non-degeneracy follow from the assumption that the eigen-ket set is a basis? (It should not, for one can have an eigen-basis which includes degenerate eigen-values; at least for finite dimensional spaces. Here, it's an...

That is, use that p\Delta p=m\Delta E \Rightarrow \Delta p\le\sqrt{m\Delta E} \Rightarrow \Delta x\ge \frac{\hbar}{2\sqrt{m\Delta E}}