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Quantum Mechanics

Submitted by lokeshhcu on Thu, 11/07/2013 - 13:26

if you interchange two electron in quantum mechanical system.why wave function changes its sign please give me answer in physical way(what exactly alter in the system)

Quantum Mechanics

Submitted by kapoor on Sat, 11/09/2013 - 15:52

For identical particles, the state of a system does not change when their labels are interchanged. It is not possible to distinguish between two identical particles in case quantum nature is taken into account. So the STATE given by $\psi(1,2)$ is same as the STATE given by wave function after exchange of the particles $\psi(2,1)$. This leads to two possibilities the wave function must be symmetric or anti symmetric. Experiments tell us that for identical fermions antisymmetric and for identical bosons symmetric wave functions must be used. Basically, within QM it is an assumption supported by experiments. Proofs have been given in relativistic quantum field theory under very general assumptions such as relativistic invariance, causality etc.

Statistical Mechanics

Submitted by AK-47 on Sat, 11/09/2013 - 13:28

Macroscopic Variables of an open system? For open systems the system can exchange energy and particles with the environment. So energy and total number of particles are not macroscopic variables. What replaces them for description of a macroscopic system?

Statistical Mechanics

Submitted by kapoor on Sat, 11/09/2013 - 16:04

A closed system in equilibrium with a heat bath has a fixed temperature and does not have not well defined value for energy. For a given energy we can find the corresponding probability only. And we can calculate mean energy for a system in equilibrium at a given temperature. An open system can exchange energy and particles with environment and in equilibrium it is characterised by volume, temperature and chemical potential. Number of particles in system will have no meaning, only a probability can be assigned to a given number and average number of particles can be computed.

Classical and Quantum Mechanics

Submitted by AK-47 on Sat, 11/09/2013 - 16:40

A doubt sir...when we are trying to find out the cross section for Rutherford's scattering in the central force chapter. why do we need this..??? just after deriving it, what is the use..??? and then we are solving for the laboratory frame.. i didn't find the use.. is it just , we have to learn the derivation...???

Classical and Quantum Mechanics

Submitted by tanushreegope on Fri, 11/15/2013 - 19:41

why we do physics? we do to understand the phenomena occurring in nature. To understand that, we make models, do experiments and all other things. So, we need to know the derivation, how we can get the cross section for scattering and the concept used in that derivation. Thus, when you do research or deal with realistic cases not with models, you could calculate the things properly. I think, I have made myself clear.

classical mechanics

Submitted by RUDRAKSHALA KOT... on Thu, 11/28/2013 - 21:00

Canonical Transformations:: If given transformations are time dependent ,then how to check whether they are canonical are not..?if they are,then how to find generating function of transformations.

CLASSICAL MECHANICS

Submitted by RUDRAKSHALA KOT... on Sat, 11/30/2013 - 20:29

CANONICAL TRANSFORMATIONS: If transformations are time dependent how can check given transformations are canonical transformation or not? If they are canonical how to find generating function?

Canonical Transormations

Submitted by kapoor on Mon, 02/03/2014 - 11:06

Q: If transformations are time dependent how can check given transformations are canonical transformation or not? Ans: The method is same. You can check that the Poisson bracket remains unchanged for the new variables. $[ Q_i,P_j]_{PB}=\delta_{ij}, [ Q_i,Q_j]_{PB}=0 , [ P_i,P_j]_{PB}=0 $. Q: If they are canonical how to find generating function? Th generating function method does not change. You must find one of the four types of generating function which will exists? ( See Goldstien) The you compute, for example, $\sum_k p_k dq_k - \sum_k P_k dQ_k $ and express it in terms of independent variables ( Type 1, for example means in terms of q and Q, and so on). The answer should be written as a total time derivative of some function $F$. This function is just the generating function.

statistical mechanics

Submitted by lokeshhcu on Mon, 12/02/2013 - 17:07

Sir, I need lecture notes and some problems on bose-einstein condensation and ising model.

Notes on Bose Condensation and Ising Model

Submitted by kapoor on Mon, 02/03/2014 - 11:08

Please ask KPN. I do not have any notes n these topics.

Quantum Mechanics ( spin and identical particle)

Submitted by VIVEK KUMAR BHARTIYA (not verified) on Mon, 12/09/2013 - 18:09

Please tell me the method to solve following type of problem- A spin-1⁄2 particle A decays to two other particles B and C. If B and C are of spin-1⁄2 and spin-1 respectively, then a complete list of the possible values of the orbital angular momentum of the final state (i.e. B + C) is? and If there is another product C particle then how spin complete list will be ( allowed angular momentum also like the first question)?

Decay of spin 1/2 particle into two particles.

Submitted by kapoor on Mon, 02/03/2014 - 11:40

You have to know the rule for addition of angular momenta in QM. If we add two angular momenta $j_1$ and $j_2$ the resultant angular momentum $J$ can have values between $|j_1-j_2|$ and $|j_1+j_2$ in steps of 1. The same rule applies when we want to know what $j_2$ when added to $j_1$ will give $j$ as a possible value.So give $j_1$ and $j$, the allowed values of $j_2$ are from $|j_1-J|$ to $J_1+J$ in steps of 1.

Q: A spin-1⁄2 particle A decays to two other particles B and C. If B and C are of spin-1⁄2 and spin-1 respectively, then a complete list of the possible values of the orbital angular momentum of the final state (i.e. B + C) is?

Ans:

Assume the the relative angular momentum of B and C is $\ell$.

Working in the rest frame of A, the total angular momentum ( spin + orbital angular momentum) of the B and C should be 1/2. Adding spins, we will get the total spin of B and C as 1/2 or 3/2. This added to angular momentum $\ell$ should produce 1/2.

When the total spin is 1/2, the allowed values of $\ell$ with total spin 1/2 should produce angular momentum of A, i.e., 1/2:So the valus of $\ell$ are 0 and 1.
When the total spin is 3/2, the allowed values of $\ell$ with total spin 3/2 should produce angular momentum of A, i.e., 1/2: So the valus of $\ell$ are between (3/2-1/2) and (3/2+1/2) and hence 1 and 2.

Thus complete list of values of $\ell$ are 0,1,2.

Q: If there is another product C particle then how spin complete list will be
( allowed angular momentum also like the first question)?

Ans: You maen you add one more particle D ??

Basically the method is the same. Keep on adding all spins and the angular momenta. One has to then list all possible cases which will give rise to total angular momentum equal to 1/2.