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Slater Determinant: A convenient and compact way of expressing multi-particle states of anti-symmetric character for many fermions is the Slater determinant:

|ψn1x1m31ψn2x1m32ψn3x1m33···ψnNx1msNψn1x2m11ψn2x2m32ψn3x2m33···ψψn1x2msNψn3x3m31ψn2x3m12ψn3x3m33ψnNx3msN···············ψn1xNm11ψn2xNm32ψn3xNm33···ψnNxNmsN|

It is based on the fact that for N fermions there must be Ndifferent individual-particle states, or sets of quantum numbers. The ith state has spatial quantum numbers (which might be ni,i, and mfi) represented simply byni and spin quantum number msi. Were it occupied by the ith particle, the slate would beψni(xj)msi a column corresponds to a given state and a row to a given particle. For instance, the first column corresponds to individual particle state ψn(xj)ms1. Where jprogresses (through the rows) from particle 1 to particle N. The first row corresponds to particle I. which successively occupies all individual-particle states (progressing through the columns). (a) What property of determinants ensures that the multiparticle state is 0 if any two individual particle states are identical? (b) What property of determinants ensures that switching the labels on any two particles switches the sign of the multiparticle state?

Short Answer

Expert verified

(a) The property is when two columns of a determinant are identical then the determinant is considered to be zero.

(b) The property is that if two rows or two columns of a determinant are interchanged then the sign of the determinant will also be changed.

Step by step solution

01

Given data

Multi particle state is 0.

02

Concept of Determinant

A determinant is defined only for square matrix.

A=a11a12a13a21a22a23a31a32a33detA=a41deta22a23a32a33+a33deta21a22a31a32=a11a22a33-a32a23-a42a23a33-a31a23+a3a21a32-a31a22

03

Step 3(a): Find what property of determinants ensures that the multiparticle state is 0

  • A matrix is an array of number.
  • A scalar quantity that can be calculated from a square matrix of order n by n is called determinant.
  • A determinant is defined only for square matrix.
  • In case two columns of a determinant are identical then the determinant is considered to be zero.
  • It is a property of N by N matrix.
  • It assures that when two of the one-electron states have same orbital wave function and also same spin, and then wave function is identically zero.
04

Step 4(b): Find what property of determinants ensures that switching the labels on any two particles switches the sign of the multiparticle state

  • A matrix is an array of number.
  • A scalar quantity that can be calculated from a square matrix of order n by n is called determinant.
  • A determinant is defined only for square matrix.
  • When two rows or two columns of a determinant are interchanged then the sign of the determinant will also be changed.
  • It is a property of N by N matrix.
  • It assures that when given pair of coordinates are interchanged.
  • The wave function represented by the slater determinant will have changed sign.

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Most popular questions from this chapter

To investigate the claim that lowerimplies lower f energy. consider a simple case: lithium. which has twon=1electrons and alonen=2valence electron.

(a)First find the approximate orbit radius, in terms ofa0. of ann=1electron orbiting three protons. (Refer to Section 7.8.)

(b) Assuming then=1electrons shield/cancel out two of the protons in lithium's nucleus, the orbit radius of ann=2electron orbiting a net charge of just+e.

(c) Argue that lithium's valence electron should certainly have lower energy in a 25 state than in a2pstale. (Refer Figure 7.15.)

Exercise 44 gives an antisymmetric multiparticle state for two particles in a box with opposite spins. Another antisymmetric state with spins opposite and the same quantum numbers is ψn(x1)n2(x2)ψnn(x1)ψn(x2)

Refer to these states as 1 and 11. We have tended to characterize exchange symmetry as to whether the state's sign changes when we swap particle labels. but we could achieve the same result by instead swapping the particles' stares, specifically theandin equation (8-22). In this exercise. we look at swapping only parts of the state-spatial or spin.

(a) What is the exchange symmetric-symmetric (unchanged). antisymmetric (switching sign). or neither-of multiparticle states 1 and Itwith respect to swapping spatial states alone?

(b) Answer the same question. but with respect to swapping spin states/arrows alone.

(c) Show that the algebraic sum of states I and II may be written(ψn(x1)ψn'(x2)ψn'(x1)ψn(x2))(+)

Where the left arrow in any couple represents the spin of particle 1 and the right arrow that of particle?

(d) Answer the same questions as in parts (a) and (b), but for this algebraic sum.

(e) ls the sum of states I and 11 still antisymmetric if we swap the particles' total-spatial plus spin-states?


(f) if the two particles repel each other, would any of the three multiparticle states-l. II. and the sum-be preferred?

Explain.

Determine the electronic configuration for phosphorus, germanium and cesium.

Question: As the 2p levels fill from Z = 5 to 10 the element's first ionization energies tend to increase. But as the figure below clearly indicates it is slightly easier to remove an electron from Z = 8 oxygen than from Z = 7 nitrogen. What might explain this?

Question: In classical electromagnetism, the simplest magnetic dipole is a circular current loop, which behaves in a magnetic field just as an electric dipole does in an electric field. Both experience torques and thus have orientation energies -p.Eand-μ·B.(a) The designation "orientation energy" can be misleading. Of the four cases shown in Figure 8.4 in which would work have to be done to move the dipole horizontally without reorienting it? Briefly explain. (b) In the magnetic case, using B and u for the magnitudes of the field and the dipole moment, respectively, how much work would be required to move the dipole a distance dx to the left? (c) Having shown that a rate of change of the "orientation energy'' can give a force, now consider equation (8-4). Assuming that B and are general, write-μ·B.in component form. Then, noting thatis not a function of position, take the negative gradient. (d) Now referring to the specific magnetic field pictured in Figure 8.3 which term of your part (c) result can be discarded immediately? (e) Assuming thatandvary periodically at a high rate due to precession about the z-axis what else may be discarded as averaging to 0? (f) Finally, argue that what you have left reduces to equation (8-5).

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