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Show that the symmetric and anti symmetric combinations of (819)and(820)are solutions of the two. Particle Schrödinger equation(813)of the same energy asψn(x1)ψm(x2), the unsymmetrized product(817).

Short Answer

Expert verified

It is proved that symmetric and asymmetric combination are the solution of two particles Schrodinger equation.

Step by step solution

01

Given information

The symmetric wave function isψs(x1,x2)=ψn(x1)ψn(x2)+ψn(x1)ψn(x2) .

The asymmetric wave function isψA(x1,x2)=ψn(x1)ψm(x2)ψm(x1)ψn(x2) .

02

Concept of complex number:

The expression for combination of symmetric and asymmetric wave function is given by ψSA(x1,x2)=ψn(x1)ψn(x2)±ψn(x1)ψn(x2)

03

Evaluate symmetric and asymmetric wave function

The expression for combination of symmetric and asymmetric wave function is calculated as,

Apply Schrodinger wave equation

By Schrodinger wave equation in one dimension,

Considering equation (1) and (2)

h22m2x12+2x22ψSA(x1,x2)+[U(x1)+U(x2)]ψSA(x1,x2)=EψψSA(x1,x2)

Thus, the symmetric and asymmetric combination obeys the Schrodinger equation with two particle solution.

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Most popular questions from this chapter

In its ground state, carbon's 2pelectrons interact to produce jT=0. Given Hund's rule. what does this say about the total orbital angular momentum of these electrons?

Question: What if electrons were spin32 instead of spin 12. What would be Z for the first noble gas?

Figureshows the Stern-Gerlach apparatus. It reveals that spin-12particles have just two possible spin states. Assume that when these two beams are separated inside the channel (though still near its centreline). we can choose to block one or the other for study. Now a second such apparatus is added after the first. Their channels are aligned. But the second one is rotated about the-axis by an angle ϕ from the first. Suppose we block the spin-down beam in the first apparatus, allowing only the spin-up beam into the second. There is no wave function for spin. but we can still talk of a probability amplitude, which we square to give a probability. After the first apparatus' spin-up beam passes through the second apparatus, the probability amplitude iscos(ϕ/2)2nd+sin(ϕ/2)2ndwhere the arrows indicate the two possible findings for spin in the second apparatus.

(a) What is the probability of finding the particle spin up in the second apparatus? Of finding it spin down? Argue that these probabilities make sense individually for representative values ofϕand their sum is also sensible.

(b) By contrasting this spin probability amplitude with a spatial probability amplitude. Such asψ(x)=Aete2. Argue that although the arbitrariness ofϕgives the spin cases an infinite number of solves. it is still justified to refer to it as a "two-state system," while the spatial case is an infinite-state system.

Repeat example 8.6 but assume that the upper state is the 2p12rather than the2p32

Here we consider adding two electrons to two "atoms," represented as finite wells. and investigate when the exclusion principle must be taken into account. In the accompanying figure, diagram (a) shows the four lowest-energy wave functions for a double finite well that represents atoms close together. To yield the lowest energy. the first electron added to this system must have wave function Aand is shared equally between the atoms. The second would al so have function Aand be equally shared. but it would have to be opposite spin. A third would have function B. Now consider atoms far a part diagram(b) shows, the bumps do not extend much beyond the atoms - they don't overlap-and functions Aand Bapproach equal energy, as do functions Cand D. Wave functionsAandBin diagram (b) describe essentially identical shapes in the right well. while being opposite in the left well. Because they are of equal energy. sums or differences ofandare now a valid alternative. An electron in a sum or difference would have the same energy as in either alone, so it would be just as "happy" inrole="math" localid="1659956864834" A,B,A+B, orA- B. Argue that in this spread-out situation, electrons can be put in one atom without violating the exclusion principle. no matter what states electrons occupy in the other atom.

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