1. Find the energy levels En and wave functions ?n(x) of...

1. Find the energy levels E_n and wave functions ?_n(x) of single particle, placed in an infinite potential well of size a. 2. Based on 1, find the energy levels and wave functions of a system of four distinguishable spinless particles placed in an infinite potential well of size a. 3. Use this result to infer the energy and the wave function of the ground state and the first excited state. 4. The antisymmetric wave functions for a system of N noninteracting identical particles can be written as Slater determinant: ?_a (?_1, ?_2, ... ... ?_N) = |?_n1(?_1) ?_n1(?_2) ... ?_n1(?_N)| |?_n2(?_1) ?_n2(?_2) ... ?_n2(?_N)| |... ... .... ... ... ...| |?_nN(?_1) ?_nN(?_2) ... ?_nN(?_N)| Discuss the value of ?_a in the case if any two particles occupy the same single-particle state. Conclude the Pauli Exclusion Principle for Fermions.

Transcript

00:01 Hello everyone so here if you consider a particle of mass m confined to move inside an infinitely deep asymmetric potential so here we'll get of vx is equal to this infinity just zero plus infinity at x less than zero it will be zero then equal to x less than equal to a we'll get it x greater than a so now outside that well so it will be outside well so si x is equal to zero so probability of finding the particle there is zero so inside the well so potential is zero so time independent so again scrotting the equation we know scroding the equation you know protein zel equation that is minus a scroch square by 2m del square x by del x square is equal to e si so del square by del x square will be equal to minus k square si so here k is equal to twice m k by a scroats so here the general solution of for the differential equation is so here the general solution will be i to add a sign kx plus b cost k x so where a and b are arbitrary constant so now both si n and d si by x d x are continuous so here we'll get si mu uh si zero is equal to si a so it will be zero so it will be zero so again putting as si not in equation si to x so we'll get so here putting sign not in equation shy to x so it will be 0 it would be equal to a sign plus b cross 0 so here b will be equal to 0 so so si 2 is equal to a sign kx so then si a will be equal to a sign k a so either either h0 or sign k is 0 so here it will be k a is equal n pi so k is equal to n by so k is y by a here so here n is basically so one two three so it is a natural number so again and the boundary continuous at x is equal to a so it doesn't determine the constant a so but rather that constant k n hence the possible value will be so here possible value is equal to s cross square k square by twice m so it will be equal to n square h cross square n pi square by twice m a square so in a sharp contrast to the classical case so a quantum particle in the infinite square well so we have to normalize it so we'll get zero to a size square d x so it will be equal to 0 to a, a square, sine square, kx, dx is equal to 1.

05:15 Now, so we'll get, by solving, we'll get a square is equal to twice a...

Question

Please give Ace some feedback