Recent content by ProxyMelon

Well, realize that no matter what ##\sigma_x\sigma_p\geq\frac{\hbar}{2}## must hold, so your provided solution is clearly "wrong". Probably it is not even wrong, you are just using ##\hbar = 1##, could this be? Regardless, your results net you the correct value.

You are dealing with a rigid body, so you should consider where the forces are acting. In this problem it looks like the center of mass of your bar is going to stay fixed during the motion, that could help.

Griffiths has an extensive exercise about coherent states (page 129 in the second edition). The problem with Griffiths, in my opinion, is that it overuses position representation, for example in a problem of this kind you are clearly supposed to work in braket notation and not do a single...

Well, I am surely not a teacher so I can't really help you with that. In my opinion what you should do is ensure that anyone reading you paper would be able to understand what's going on clearly, even if you are not able to solve a specific problem be sure to point it out and to propose...

Well, if you wish to evaluate the dipendence on one variable you should try to do your best in order to keep the other variables fixed, clearly it would be quite hard to keep the period T fixed while you change various ropes/masses. Despite it being hard to actually perform, you could consider...

Are you talking about the period calculation?

Well, your setup seems a bit dangerous, hopefully no one gets injured :D What i think i would do is to get some kind of motor (maybe from an eletric fan) and create a rotating plate, once you have that you can measure what you need with much more precision (you can even make a wooden plate and...

I would say that the two semi-circles can be in two shapes (they are likely on the same plane), either forming a C (in this case the point in which you want to evaluate B is outside of the loop) or forming an O (in this case it is inside the loop). How whould things chance in the two setup...

It is not correct, you are never left with single bras or ket, you get, step by step (for clarity, ##\eta## is the lowering operator, it's hermitiane conjugate ##\eta^{\dagger}## is the raising operator and ##\gamma## is the eigeinvalue associated to the coherent state ##|\beta\rangle##)...

Hello fellow physics lovers, i used to lurk this forum for quite a while, now I am about to get my degree in physics, just have to tackle quantum mechanics, which I am rather confident on, and some minor exams. Hope to use this forum a lot in the future,!

Hi, a coherent state is an eigenstate of the lowering operator as you said, so you have to work with: ##\eta|\beta\rangle=\beta|\beta\rangle\Rightarrow\langle\beta|\eta^{\dagger}=\langle\beta|\beta^*## Now, I use a slightly different notation (mean value do not change as long as you keep the...