Chapter 19 - Second Quantization. 1. Identical Particles Revisited

Chapter 19 - Second Quantization 1. Identical Particles Revisited When dealing with a system that contains only a few identical particles, we were easily able to explicitly construct appropriately symmetrized state vectors. However, when the number of identical particles in the system gets very large as with electrons in a metal, superfluids or superconductors and so on, these methods become too cumbersome to use effectively. To see how to proceed in these cases, we will step back, revisit the subject of identical particles and look for an alternative way of thinking about such systems. Indistinguishability As we saw earlier, in quantum mechanics, the state of a system with identical particles can be described by a set of quantum numbers corresponding to the eigenvalues of a commuting set of single-particle operators representing single-particle observables. When we specify a state vector we designate how many particles have certain sets of quantum numbers, i.e., 1

n 1 particles have quantum number set K 1 n particles have quantum number set K and so on More complicated states are then superpositions of these states. The important fact is that, for identical or indistinguishable particles, it is impossible to state which particle has K 1, which particle has K, etc. As we saw in atoms, this indistinguishability has measurable effects on energy levels arising from particle exchange symmetries. State Vectors When we define a state vector for a system of we are working under the following assumption n identical particles any complete set of quantum numbers (observables) that can be used to describe the behavior of a single particle can also be used for n non-interacting particles of the same type. This assumption is postulated to be true even if the particles are interacting. The assumption implies that composite systems somehow retain the information related to single-particle properties. n

Mathematically it says that for each set of quantum numbers there exists an occupation number operator ˆN i such that the eigenvectors of ˆN i imply states in which a definite number, n i, of particles has the quantum numbers K i and that the eigenvalues of the ˆNi are the occupation numbers. n i We make the fundamental postulate that the set of all ˆN i forms a complete set of commuting Hermitian operators for any system of identical particles. We now construct the state vector space appropriate for the many particle system by generalizing one-particle quantum mechanics and building in indistinguishability from the start. Our postulates imply that the state vector space for the many particle system (called Fock space) has the basis vectors n 1,n,n 3,... where the notation implies that n 1 particles have quantum number set K 1 n particles have quantum number set K and so on K i 3

As we said, we assume this set is a complete orthonormal basis. In this vector space, we define 0 = 0,0,0,0,... = zero particle or vacuum state and one-particle states are of the form 0,0,0,...,0,n i = 1,0,0,... These one-particle states span the one-particle subspace of the much larger state space of the many particle system. Most of the quantum mechanics we have developed so far applies to these one-particle states. We now ask whether the postulate implies enough formalism to construct a quantum mechanics of a many-particle interacting system? To answer this question "yes", we must show that a consistent framework exists that makes predictions in agreement with experiment. The fundamental assumption that the states of an interacting system can be written in terms of states for noninteracting single particles comes from our experience with perturbation theory. It will fail to work if the noninteracting and interacting states are two inequivalent representations of the Hilbert space for the many particle system. 4

Standard perturbation theory implicitly assumes that the zero-order or unperturbed states and the exact states are equivalent representations of the Hilbert space.. Occupation Number Space Creation and Annihilation Operators Earlier we studied the â, â operators in conjunction with the harmonic oscillator problem. In that system we found that â lowered the energy of the system by 1 quantum = ω, while â raised the energy of the system by ω. In terms of the eigenvalues and eigenvectors of the number operator ˆN = â â where ˆN n = n n we found that â n = n n 1 and â n = n 1 n 1 In some sense, we can think of the â, â as annihilation/creation operators for a quantum of the energy associated with a single particle system. Another appearance of operators of this type with similar properties occurred when we studied the interaction of radiation with matter. 5

We were able to introduce photon annihilation/creation operators which k, removed/added a single photon with particular quantum numbers ( λ ). The photon operators had the same mathematical structure (commutators) as the â, â operators of the harmonic oscillator system. In the photon case, the states of the system were N k1 λ1, N k λ, N k3 λ3,... We will base our generalization on these examples. The generalization will allow us eventually to define the most general Fock space for any many particle system. Suppose that we have a potential well V( r ) with single particle energy eigenstates given by ϕ 0 ( r ) = r 0, ϕ 1 ( r ) = r 1, ϕ ( r ) = r,... We start the discussion by considering a system of n n bosons. We assume ϕ 0 ( r that all particles are in the lowest level (ground state) ) of the well. We label this state by the symbol n, where n = 0,1,, 3,... i.e., 0 is the state with no particles in the lowest level. We now introduce the operators â 0,â 0 such that â 0 n = n n 1 and â 0 n = n 1 n 1 6

By definition, these operators ϕ 0 ( r relate states of the -boson system with all n particles in ) to those states of an n ± 1 particle system with all particles in ϕ 0 ( r ). In this sense, we say that â 0 = a particle annihilation operator which removes a particle in the state ϕ 0 ( r ) from the system â 0 = a particle creation operator which adds a particle in the state ϕ 0 ( r ) to the system These operators, by construction, have the same algebra as the harmonic oscillator operators, i.e., and n = â 0,â 0 = 1, [ â 0,â ] 0 = 0 = â 0,â 0 ( â0 ) n n! 0, â 0 0 = 0 (no particles to be annihilated) This says that the state with n particles in the lowest level, n, is generated by adding n particles, in, to the vacuum state ϕ 0 ( r ) 0 Acting to the left (instead of to the right) these operators reverse their roles, i.e., n â 0 = n 1 n 1 adds particles in the state ϕ 0 ( r ) n â 0 = n n 1 removes particles in the state ϕ 0 ( r ) n 7

The operator ˆN0 = â 0â 0 measures the number of particles in a state since ˆN 0 n = â 0â 0 n = n n This is a very appealing picture, but, really, all we have done is rewrite the harmonic oscillator story using a lot of new words. We do not have any new physics yet! Before introducing the new physical ideas, we carry out this same discussion for fermions. In this case, the only allowed states are 0 =no particles in ϕ 0 ( r ) 1 =1 particle in ϕ 0 ( r ) since we cannot, according to the PEP, have two particles with the same quantum numbers (in the same state). We can still introduce annihilation/creation operators but they must have a very different algebra. We must have 8

â 0 0 = 0 there are no particles to be annihilated â 0 1 = 0 we can remove a particle when there is a particle â 0 0 = 1 we can add a particle when there are no particles â 0 1 = 0 we cannot have two particles in the same state The last relation is necessary to satisfy the PEP. We can figure out the operator algebra by using a special representation of these states and operators. We let the two states (only allowed states) be a basis and select the -dimensional representation 0 = 1 0 and 1 = 0 1 In this representation we have 0 â 0 0 = 0, 1 â 0 0 = 0, 0 â 0 1 = 1, 1 â 0 1 = 0 â 0 = 0 1 0 0 0 â 0 0 = 0, 1 â 0 0 = 1, 0 â 0 1 = 0, 1 â 0 1 = 0 â 0 = 0 0 1 0 Another way of representing these operators is via outer products,i.e., â 0 = 0 1 â 0 = 1 0 9

The operator algebra is then given by â 0 â 0 = 0 0 = projection operator on the 0 state â 0â 0 = 1 1 = projection operator on the 1 state â 0 â 0 â 0â 0 = 0 0 1 1 = Î This last relation was derived, in general, earlier and is just the sum over all projection operators. Therefore we get { â 0,â } 0 = â 0 â 0 â 0â 0 = Î The algebra involves anticommutators instead of commutators. That is the only change we need to make!! We also have { â 0,â } 0 = 0 = { â 0,â } 0 These anticommutators imply that ( ) = 0 we cannot remove two fermions from the â 0 same state (maximum of one allowed) ( â ) 0 = 0 we cannot put fermions into the same state (maximum of one allowed) 10

Summary: For one single particle level in a potential well, we can define annihilation/creation operators such that: bosons have and fermions have â 0,â 0 = Î [ â 0,â 0 ] = 0 = â 0,â 0 { â 0,â } 0 = Î { } = 0 = â 0,â 0 â 0,â 0 { } We now expand our view and consider the case where ϕ 0 ( r particles ) ϕ 1 ( r can occupy two levels of the potential well, say and ). For bosons, we write the state of a many particle system as n 0,n 1 which implies n0 particles in ϕ 0 ( r ) and n 1 particles in ϕ 1 ( r ) We now define a pair of boson annihilation/creation operators by the relations â 0 n 0,n 1 = n 0 n 0 1,n 1 â 0 n 0,n 1 = n 0 1 n 0 1,n 1 â 1 n 0,n 1 = n 1 n 0,n 1 1 â 1 n 0,n 1 = n 1 1 n 0,n 1 1 11

which imply that â 0 annihilates a particle in the state ϕ 0 ( r ) â 0 creates a particle in the state ϕ 0 ( r ) â 1 annihilates a particle in the state ϕ 1 ( r ) â 1 creates a particle in the state ϕ 1 ( r ) In the same manner as in the one level case, we must have the commutator algebra â 0,â 0 = 1 [ â 0,â 0 ] = 0 = â 0,â 0 and â 1,â 1 = 1 [ â 1,â 1 ] = 0 = â 1,â 1 For bosons, the order in which we create or annihilate particles in a state does not matter, i.e., which says â 0 â 1 n 0,n 1 = â 1 â 0 n 0,n 1 [ â 0,â 1 ] = 0 In a similar way all the other mixed commutators are also zero â 0,â 1 = 0 = â 1,â 0 = â 1,â 0 All allowed states can be constructed from the vacuum state 0,0 by using 1

Finally, ( n 0,n 1 = â1 ) n 1 n 1! ( â ) n 0 0 n 0! 0,0 ˆN 0 = â 0â 0 the number of particles in state ϕ 0 ( r ) ˆN 1 = â 1â 1 the number of particles in state ϕ 1 ( r ) and Therefore, ˆN = ˆN 0 ˆN 1 the total particle number operator ˆN 0 n 0,n 1 = n 0 n 0,n 1 ˆN 1 n 0,n 1 = n 1 n 0,n 1 ˆN n 0,n 1 = (n 0 n 1 ) n 0,n 1 Thus, for bosons we are able to just glue two single level many particle systems together to create a two-level many particle system. We are really constructing direct product states. For fermions,however, there are some extra complications that we have to deal with. We start off by following a similar procedure. For the two-level system we have only four possible fermion states, namely, 0,0, 0,1, 1,0, 1,1 We are completely free to define one set of creation/annihilation operators. 13

We define the â 1, â 1 by â 1 0,0 = 0 â 1 1,0 = 0 â 1 0,1 = 0,0 â 1 1,1 = 1,0 â 1 0,0 = 0,1 â 1 1,0 = 1,1 â 1 0,1 = 0 â 1 1,1 = 0 For â 0, â 0 we can freely ϕ 1 ( r define the operation on state with no particles in level ) â 0 0,0 = 0 â 0 1,0 = 0,0 â 0 0,0 = 1,0 â 0 1,0 = 0 We must take care, however, ϕ 1 ( r in the fermion case when a particle exists in level ),i.e., for the operations â 0 0,1 â 0 1,1 â 0 0,1 â 0 1,1 The reason we must worry about these cases is connected with our earlier discussion of a totally antisymmetric state vector for fermions,i.e., if we interchange any two identical fermions we must get a minus sign. In this formalism, how do we interchange two fermions in the state 1,1? Using only the defined relations we have... 14

Step 1: Step : Step 3: 1,1 1,0 = â 1 1,1 remove a particle from state ϕ 1 ( r ) using â 1 1,0 0,1 = â 1â 0 1,0 transfer a particle from state ϕ 0 ( r ) to ϕ 1 ( r ) by applying â 1â 0 ( â 0 followed by â 1 ) put the leftover particle back into ϕ 0 ( r ) using â 0 which gives the state relationship â 0â 1â 0 â 1 1,1 = â 0 0,1 However, all we have done is switch the particles in the original state 1,1, which means that a minus sign must appear or â 0 0,1 = 1,1 In this way, the state is completely antisymmetric under particle exchange. In a similar manner, the other relations that complete the definition of the annihilation and creation operators are Since we have â 0 1,1 = 0,1 â 0 0,1 = 0 = â 0 1,1 â 0 = ( â ) 0 15

â 0 0,1 = 1,1 â 0 â 0 0,1 = â 0 1,1 = 0,1 â 0 â 0 so that completely undoes the operation of. These definitions, which are now consistent with complete antisymmetry, correspond to the anticommutation relations { â 0,â } 0 = 1 { â 0,â 0 } = 0 { â 0,â } 0 = 0 { â 1,â } 1 = 1 â 1,â 1 { } = 0 â 0,â 1 â 0,â 1 { â 0,â } 1 = 0 { } = 0 â 1,â 1 { } = 0 { } = 0 { â 0,â } 1 = 0 We get anticommutators instead of commutators because of the complete antisymmetry under particle interchange(see argument below). The rule for constructing the allowed states is Note the order, since 0! = 1! = 1. â 0 n 0,n 1 = ( â ) n 1 1 ( â ) n 0 0 0,0 acts first and we have no factorial factors The connection between the minus sign and the anticommutators is now clear, i.e., 16

â 0 1,1 = â 0 â 1â 0 0,0 = â 1â 0 â 0 0,0 = â 1 [1 â 0â 0 ] 0,0 = â 1 0,0 = 0,1 So we could have assumed the anticommutators and derived the state operations instead of going the other way. The generalization to the many particle system where particles can occupy all of the levels is now straightforward. In generalizing, we will not only let the particles occupy all of the levels, but also have all spin orientations. Once again the states will be labeled where now n 0,n 1,n,... n i = the number of particles in ϕ i ( r) with a given spin orientation(if the particles have spin) We define creation and annihilation operators different single particle state. For boson operators we have commutation relations â i,â j = δ ij â i,â j = 0 = â i,â j â i and â i for each 17

and where n 0,n 1,n,... =... ( â ) n n! ( â 1 ) n 1 n 1! ( â 0 ) n 0 n 0! 0 = 0,0,0,0,... = the vacuum state 0 To within numerical factors, these are the same relations we found earlier for photons. Therefore, photons must be bosons!! For fermion operators we have anticommutation relations and { â i,â } j = δ ij { â i,â } j = 0 { â i,â } j = 0 n 0,n 1,n,... =...( â ) n ( â ) n 1 1 ( â ) n 0 0 0, n i = 0,1 only In both cases and An Example ˆN = ˆNi = â i a i i ˆNi, ˆN j = 0 i We now consider the complete set of plane wave states in a box using periodic boundary conditions. We have Ĥ = p op m 18

Since Ĥ, p op = 0 we have a common eigenbasis that we will label by have pop p = p p Ĥ p p = p = E p So that E = p m m The corresponding wave functions are where ϕ p ( r ) = r p = ei p = k k r V V = volume of the box p. We then The factor 1 / V normalizes the wavefunction in the box. Periodic boundary conditions are imposed by the relations ϕ p (0, y, z) = ϕ p (L x, y, z) ϕ p (x,0, z) = ϕ p (x, L y, z) ϕ p (x, y,0) = ϕ p (x, y, L z ) 19

which imply that e ik x L x e ik y L y = 1 k x = πn x L x, n x = 0, ±1, ±,... = 1 k y = πn y L y, n y = 0, ±1, ±,... We now define e ik z L z = 1 k z = πn z L z, n z = 0, ±1, ±,... â ps =the operator that creates (adds) a particle of momentum p and spin orientation s in (to) the box â ps =the operator that annihilates (removes) a particle of momentum p and spin orientation s in (from) the box The probability amplitude for finding the particle added by the box at position r ' is The operator â ps e i k r ' V = r ' ψ s ( r ) = p p e i k r V â ps adds a particle to the system in a superposition of momentum states ( ) each with amplitude â ps to 0

e i k r V This implies that the probability ψ s ( r amplitude ) r for finding the particle added to the box by at the position ' is r ' e i k r V â ps 0 = p p e i k r V r ' p = p e i k r V e i k r ' V = δ( r r ') This says that the operator ψ s ( r ) adds all the amplitude at the position r or we say ψ s ( r ) adds(creates) a particle at position r with spin orientation s. In a similar manner, the operator ( ) ei ψ s ( r ) = ψ s ( r ) p k r V â ps removes (annihilates) a particle at point. The ψ s ( r ) and ψ s( r ) are called field operators. In this new formalism, position and momentum are once again just numbers, but the wave functions are now operators. Hence the name second quantization. r 1

For bosons we have ψ s ( r ),ψ s ' ( r ') and for fermions we have [ ] = 0 = ψ s ( r ),ψ s ' ( r ') { } { ψ s ( r ),ψ s ' ( r ')} = 0 = ψ s ( r ),ψ s ' ( r ') These relations imply that for bosons adding(removing) a particle at r is an operation that commutes with adding(removing) a particle at r ' and for fermions adding(removing) a particle at r is an operation that anticommutes with adding(removing) a particle at r ' (change of sign) Finally, for bosons we get ψ s ( r ),ψ s ' ( r ') = pp ' e i k r e i k r ' V â ps,â p ' s ' e i k r e i k r ' = δ pp ' δ ss ' pp ' V and similarly for fermions we get e i k r e i k r ' { ψ s ( r ),ψ s ' ( r ')} = = pp ' pp ' V e i k r e i k r ' V { â ps,â } p ' s ' δ p p ' δ ss ' = δ( r r ')δ ss ' = δ( r r ')δ ss '

These relations imply that creating particles commutes(bosons) or anticommutes(fermions) with annihilating particles unless the two, operations occur at the same point in space. In this case, if there are no particles at ψ s ( r )ψ s ( r ) 0 r, then i.e., we cannot annihilate a particle at a point if none exists there. On the other hand ψ s ( r )ψ s ( r ) does not 0 since ψ s ( r ) adds a particle that ψ s( r ) then removes! Suppressing spin indices for simplicity, the state vector r 1, r,..., r n = 1 n! ψ ( r n )...ψ ( r )ψ ( r 1 ) 0 represents the state with one particle at, one particle at, and so on. We will use these states as a basis for the many particle, many level system. The states have the properties: (1) for bosons r, r 1,..., r n = r 1, r,..., r n r 1 r 3

() for fermions r, r 1,..., r n = r 1, r,..., r n due to the anticommutation relations which imply Since ψ ( r ) r 1, r,..., r n = ψ ( r 1 )ψ ( r ) = ψ ( r )ψ ( r 1 ) n 1 n 1! ψ ( r )ψ ( r n )...ψ ( r )ψ ( r 1 ) 0 = n 1 r 1, r,..., r n, r the commutation/anticommutation properties of the imply that the new state is automatically correctly symmetric or antisymmetric. This is one of the great advantages of the annihilation/creation operator formalism. We can show this important property this way: ψ ( r ) r 1, r,..., r n = 1 n! ψ ( r )ψ ( r n )...ψ ( r )ψ ( r 1 ) 0 ψ where = 1 n! δ( r r n ) ± ψ ( r n )ψ ( r ) ψ ( r n 1 )...ψ ( r )ψ ( r 1 ) 0 ± bosons fermions We now continue commuting ψ ( r ) with the ψ 's to the right until we have since ψ ( r ) 0 = 0...ψ ( r ) 0 = 0. If we actually carry out this process we obtain 4

ψ ( r ) r 1, r,..., r n = 1 n! [δ( r r n ) r 1, r,..., r n 1 ± δ( r r n 1 ) r 1, r,..., r n, r n... ( ±1) n 1 r,..., r n ] This says that removing r r = r a particle n or r = r at is n 1 or... or r only = r possible if 1 and if one of these conditions is true, then what remains is the correctly symmetrized combination of ( n 1 ) particle states. Because have ( ) ψ = ψ, i.e., they are Hermitian conjugate operators, we ψ ( r ) r 1, r,..., r n removes a particle ψ ( r ) r 1, r,..., r n adds a particle when acting to the right but r1, r,..., r n ψ ( r ) adds a particle r 1, r,..., r n ψ ( r when acting to the left ) removes a particle i.e., r 1, r,..., r 1 n = n! ψ ( r n )...ψ ( r )ψ ( r 1 ) 0 = 1 n! 0 ψ ( r 1 )ψ ( r )...ψ ( r n ) Note the reversal in the order of the operators. 5

According to these rules the basis states are normalized as follows: r 1 ', r ',..., r n ' ' r 1, r,..., r n where and = δ nn ' n! P ( ±1) P P[ δ( r 1 r 1 ')δ( r r ')...δ( r n r n ')] = sum over all permutations if the coordinates P ( ±1) P = What ϕ( r is 1,..., r the state n )? 1 bosons 1 fermions - even permutation 1 fermions - odd permutation φ where the particles have a wave function Since the correctly symmetrized wave function must be r 1, r,..., r n φ we must have φ = which implies that r 1, r,..., r n φ = 1 n! d 3 r1 ' d 3 r '...d 3 rn 'ϕ( r 1 ',..., r n ') r 1 ', r ',..., r n ' P ( ±1) P Pϕ( r 1,..., r n ) and thus the true wave function is properly symmetrized. 6

ϕ( r 1,..., r n ) This result is true even if is not already properly symmetrized. When it already properly symmetrized, then all n! terms are identical and r 1, r,..., r n φ = ϕ( r 1,..., r n ) We must have φ φ = 1 if ϕ( r 1,..., r n ) i.e., φ φ = is symmetrized and 1 = d 3 r1 d 3 r...d 3 rn ϕ * ( r 1,..., r n )ϕ( r 1,..., r n ) d 3 r1 d 3 r...d 3 rn ϕ *( r 1,..., r n ) r 1, r,..., r n d 3 r1 ' d 3 r '...d 3 rn 'ϕ( r 1 ',..., r n ') r 1 ', r ',..., r n ' = d 3 r1 d 3 r...d 3 rn ϕ *( r 1,..., r n )ϕ( r 1 ',..., r n ') 1 n! = d 3 r1 d 3 r...d 3 rn ϕ *( r 1,..., r n )ϕ( r 1,..., r n ) = 1 r 1, r,..., r n φ r 1, r,..., r n Now P ( ±1) P P[ δ( r 1 r 1 ')δ( r r ')...δ( r n r n ')] is the amplitude for observing particles at. It implies that φ = d 3 r1 d 3 r...d 3 rn r1, r,..., r n r1, r,..., r n φ which says that d 3 r1 d 3 r...d 3 rn r1, r,..., r n r1, r,..., r n = În 7

as it should for a complete set within the -particle subspace, i.e., it is Î only when operating on properly symmetrized n n -particle states. We then have for φ and the -particle identity operator which implies that Î = Î n ' φ = δ nn ' φ n Î n = 0 0 Î n n=0 n=1 is the identity operator when acting on properly symmetrized states for any number of particles. 3. Second Quantized Operators How do we write operators in this formalism? As a first example, let us consider the operator ρ( r ) = ψ ( r )ψ ( r ) To see what this represents physically we calculate the matrix element φ ' ρ( r ) φ n where φ and φ ' are -particle states. We obtain n 8

Now the state and therefore We thus obtain φ ' ρ( r ) φ = φ ' ψ ( r )ψ ( r ) φ = φ ' ψ ( r )Îψ ( r ) φ = φ ' ψ ( r ) 0 0 Î n ' ψ ( r ) φ n '=1 ψ ( r ) φ an (n 1) particle state and 0 0 ψ ( r ) φ = 0 Î n ' ψ ( r ) φ = 0 unless n' = n-1 n '=1 φ ' ρ( r ) φ = φ ' ψ ( r )În 1ψ ( r ) φ = d 3 r1...d 3 rn 1 φ ' ψ ( r ) = d 3 r1...d 3 rn 1 φ ' r 1,..., r n φ r 1,..., r n 1 r 1,..., r n 1, r r1,..., r n 1 ψ ( r ) φ r 1,..., r n 1, r φ Now, since the are completely symmetrized (or antisymmetrized) this can be written as φ ' ρ( r ) φ = d 3 r1...d 3 rn φ ' n = φ ' δ( r r i ) i=1 φ r 1,..., r n n δ( r r i ) i=1 r 1,..., r n φ 9

Since these two objects have all their matrix elements are identical, they must be equal. Therefore, or n i=1 ρ( r ) = δ( r r i ) ρ( r ) = a representation of the density operator in this formalism The way to think about this operator is as follows: ψ ( r )ψ ( r ) tests the density of particles at r r by attempting to remove a particle located at and then to put it back If the particles have spin, the density operator for particles at r with spin s is ψ s ( r )ψ s ( r ) and ρ( r ) = ψ s ( r )ψ s ( r ) = the total density operator ˆN = s d 3 r ρ( r ) = d 3 r ψ s ( r )ψ s ( r ) = total number operator s This agrees with our earlier result as can be seen below: ˆN = s d 3 r ψ s ( r )ψ s ( r ) = s d 3 r p e i k r V â ps p V â p ' s e i k ' r = â ps â p ' s d 3 r s pp ' e i( k ' k ) r V = â ps â p ' s δ pp ' = â ps s pp ' ps â ps 30

Now to figure out other operators. Any operator that is given by the relation f ( p op ) is easily written down in this formalism, i.e., such an operator is given by the number of times p ˆN p = â ps â ps f ( p) occurs ( ) times the value of the operator ( ) summed over ( ps ) or f ( p op ) = ps f ( p)â ps â ps Therefore, the kinetic energy operator ˆ T = ps p m â ps â ps Tˆ is given by We can rewrite this expression in a form involving the field operators that will then lead to a prescription for writing any operator in this formalism. If we invert the equations for ψ s ( r ) and ψ ( s r ) we get ψ s ( r ) e i k r â ps = d 3 r V â ps = k r e i d 3 r ψ s ( r ) V Physically, the first equation implies that to create a particle with momentum p we create particles r ei k at different points with relative amplitudes r / V 31

and so on for the second equation, which is exactly what we have been assuming all along! Using these equations we get T ˆ = 1 m = m 1 V 1 V ps ps ( ) d 3 r d 3 r ' pe ik r d 3 r ( pe i k r ' )ψ s ( r )ψ s ( r ') d 3 r ' ( r e i k r ) ( r ' e i k r ' )ψ s ( r )ψ s ( r ') We now integrate by parts, assuming that ψ s ( r ) and ψ s ( r ') 0 r so that "surface" terms 0. We get T ˆ = 1 m V d 3 r d 3 r 'e ik r e i k r ' ( r ψ s ( r )) ( r ' ψ s ( r ')) Using 1 V and doing the ps r ' ˆ T = m ps e i k r e i k r ' = δ( r r ') integration we get d 3 r ψ ( r ) ψ ( r ) This is very similar to the expectation value of the kinetic energy operator for a single particle, i.e., as ˆ T = m d 3 r ϕ *( r ) ϕ( r ) 3

Similarly, ϕ *( r )ϕ( r the density operator resembles the probability density ) for finding a single particle with wave function ϕ at the point r. This leads to the general concept of second quantization where one-particle wavefunctions appear to have become operators that create/annihilate particles single-particle expectation values appear to have become operators for physical observables position and momentum appear to have become ordinary numbers We can now exploit this similarity to write down other operators, for example, j( r ) = 1 im ψ ( r )( ψ ( r )) ( ψ ( r ))ψ ( r ) = the current density operator S( r ) = 1 ψ s ( r ) σ ss ' ψ s ' ( r ) where σ = ( ˆσ x, ˆσ y, ˆσ z ) ss ' = the operator for the density of spin at point r 33

An Example - A Gas of N Non-Interacting Spin 1/ Fermions We assume that the system is in its ground state. As we discussed earlier, the ground state φ 0 corresponds to all momentum states (energy levels) being filled starting from the lowest level. Since we have a finite number of particles, there will be a maximum momentum value, = the Fermi momentum. p F In the new formalism, we describe the ground state via a set of occupation numbers n p = n p = φ 0 â p â p φ 0 Now N = = 1 p pf 0 p p F n ps = 1 ps p p F Converting the sum to an integral we have N = V p F d 3 p = (π) 3 where we have used the properties 0 p F 3 3π 3 V 34

which imply and thus We then get where spacing between p x values is π L x spacing between p y values is π L y spacing between p z values is π L z p x p L x π V (π) 3 dp x p F = ( 3π n) 1/3 n = N V d 3 p = average particle density The expectation value of the density operator is Now ρ( r ) = φ 0 ψ s ( r )ψ s ( r ) φ 0 s φ 0 = sp p ' e i k r e i k ' r V φ 0 â ps ( r )â p ' s ( r ) φ 0 = δ pp ' n ps â ps ( r )â p ' s ( r ) φ 0 35

since if we remove a particle of momentum p from the ground state, we can only get the ground state back if we add a particle of the same momentum p' Therefore,. Where we have used ˆn ps = â ps ( r )â ps ( r ) ρ( r ) = 1 V ps n ps = n = constant The density of the non-interacting fermion gas is uniform. A useful physical quantity is defined by which is We have G s ( r r ') = φ 0 ψ s ( r )ψ s ( r ) φ 0 the amplitude for removing(annihilating) a particle with spin s at r ' from the ground state and then returning to the ground state by replacing(creating) a particle with spin s at G s ( r r ') = = 1 V pp ' e i k r e i k ' r pp ' V e i k r e i k ' r V φ 0 r â ps ( r )â p ' s ( r ) φ 0 δ p p ' n ps Changing to an integral we obtain = 1 V p e i k ( r r ') n ps 36

where and G s ( r r ') = = p F d 3 p e i k ( r p r 1 F 1 ') = p dp dµ (π) 3 4π 3 0 1 4π 3 p = 3n p F dp e ik r r ' µ e ik r r ' µ 0 sin x x cos x x = p F n = 1 3π x 3 r r ' ( ) As a function of x this looks like (for n=1000) 3 p F 3 0 1 e ik r r ' µ 37

The single-particle correlation function oscillates with a characteristic 1 / k F r = r period under an envelope which falls to zero. We have for ', G s (0) = n / = r r density of particles with spin orientation s. For small ' we have G s ( r r ') = 3n 3n 1 3 x3 1 30 x5 x 3 sin x x cos x x 3 3n = n 1 x 10 x x3 6 x5 10 = 1 1 10 p F r r ' x x 1 x4 4 x 3 G s ( r r ') is called the one-particle density matrix. 4. Pair Correlation Function As we have seen earlier, in a system of fermions, there is a tendency, due to the PEP, for particles with the same spin to avoid each other. This says that the amplitude for being close together must be small. How do we calculate the relative probability of finding a particle at if we know there is a particle at? r ' r 38

One way is as follows: (1) Remove (mathematically) a particle with spin s at r from the system. We are then left in the N-1 particle state φ '( r,s) = ψ s ( r ) φ 0 () Calculate the density distribution of the particles with spin s in the new state This density is φ '( r,s) ψ s ' ( r ')ψ s ' ( r ') φ '( r,s) where g ss ' ( r r ') = φ 0 ψ s ( r )ψ s ' ( r ')ψ s ' ( r ')ψ s ( r ) φ 0 = n g ss ' ( r r ') the pair correlation function. An equivalent way of asking the same question is the following: (1) Remove(annihilate) a particle from () Remove(annihilate) a particle from r using ψ ( s r ) r ' using ψ ( s ' r ') The relative amplitude for ending up in some N- particle state φ i " is φ i " ψ s ' ( r ')ψ s ( r ) φ 0 39

(3) Sum over a complete set of N- particle states This gives the total probability of removing two particles and ending up in any N- particle state. We get total probability of removing two partcles= φ i '' ψ s ' ( r ')ψ s ( r ) φ 0 i = φ 0 ψ s ( r )ψ s ' ( r ') φ i '' φ i '' ψ s ' ( r ')ψ s ( r ) φ 0 i = φ 0 ψ s ( r )ψ s ' ( r ')ψ s ' ( r ')ψ s ( r ) φ 0 = n g ss ' ( r r ') We can evaluate g ss ' ( r r ') by shifting to the creation/annihilation operator formalism. We get Now n g ss ' ( r r ') = φ 0 pp ' q q ' e i( p p ') r / e i( q q ') r '/ V φ 0 â ps ( r )â qs ' ( r ')â q ' s ' ( r ')â p ' s ( r ) φ 0 = 0 â ps â qs ' â q ' s ' â p ' s φ 0 unless we put back particles with the same spin and same momentum that we remove, i.e. if s s', then p' = p and q ' = q This implies that φ 0 â ps â qs ' â q ' s ' â p ' s φ 0 = φ 0 â ps â ps â qs ' â qs ' φ 0 = n ps n qs ' We then get 40

or n g ss ' ( r r ') = 1 V g ss ' ( r r ') = 1 for s s' n ps n qs ' = n s n s ' = n pq This implies r r that the relative probability for finding particles at and ' for different spins is independent of r r '. This is the same result as one obtains in a classical non-interacting gas. The PEP does not influence particles of different(opposite in this case) spins. On the other hand, if the spins are the same, s=s, then we have two possibilities, namely, p' = p and q ' = q or p' = q and q ' = p p' = q ' φ 0 â ps Note that if, then â q ' s â p ' s φ 0 = 0 since â qs â q ' s â p ' s ( â ) p ' s = 0 (for fermions) Therefore, we have φ 0 â ps â qs â q ' s â p ' s φ 0 = δ pp ' δ qq ' φ 0 â ps â qs â qs â ps φ 0 δ pq ' δ qp ' φ 0 â ps â qs â ps â qs φ 0 We must use the superposition of both (indistinguishable) possibilities. 41

This becomes φ 0 â ps â qs â q ' s â p ' s φ 0 = (δ pp ' δ qq ' δ pq ' δ qp ' ) φ 0 â ps â ps â qs â qs φ 0 = (δ p p ' δ q q ' δ p q ' δ q p ' )n ps n qs where we have used the anticommutation relations for fermion operators q p { â } ps,â qs = 0 = { â } ps,â qs q = p... = 0 We finally obtain n g ss ' ( r r ') = 1 V = n pq 1 e i( p q ') ( r r ')/ [ G s ( r r ')] n ps n qs where G s ( r r ') is the single particle density function. This then becomes g ss ' ( r r ') = 1 9 ( sin x x cos x) x 6 where x = p F r r ' As a function of x this looks like 4

This result implies a substantial reduction in the probability for finding two fermions of the same spin at distances less than / p F. The PEP causes large correlations in the motion of the particles with the same spin. It seems like fermions of the same spin repel each other at short distances. This effective repulsion is due to the exchange symmetry (PEP) of the wave function and not from any real additional potentials. 43

r r ' g ss g ss ' = 1 At large. as we might have guessed since at large separations the PEP should have no effect and spin effects should drop out. What happens if we consider a system of non-interacting bosons instead? Suppose the system is in the state φ = n p0,n p1,... The density in this state is φ ψ ( r )ψ ( r ) φ = 1 V p n p = n The calculation of the pair correlation function is the same as for fermions up to this point. n g( r r e i( p p ') r / e i( q q ') r '/ ') = φ V â â φ 0 pâ qâ q ' p ' 0 p p ' q q ' In this case, φ 0 â pâ qâ q ' â p ' φ 0 0 only if p = p', q = q ' or p = q ', q = p' p = q These two cases are not distinct if. Therefore, in the same manner as before, we have 44

φ 0 â pâ qâ q ' â p ' φ 0 ( ) δ pp ' δ qq ' φ 0 â pâ qâ q â p φ 0 δ pq ' δ qp ' φ 0 â pâ qâ p â q φ 0 = 1 δ p q ( )n p n q δ pq δ pp ' δ qq ' n p (n p 1) = ( 1 δ ) pq δ pp ' δ qq ' δ pq ' δ qp ' δ p q δ p p ' δ q q ' φ 0 â pâ pâ p â p φ 0 where we have used â p,â p = 1 n g( r r ') = φ ψ ( r )ψ ( r ')ψ ( r ')ψ ( r ) φ = n 1 V p in the last term. Therefore, we obtain n p e i k ( r r ') This differs from the fermion result in p 1 n V p (n p 1) (1) the sign of the second term ( instead of -), which is due to different exchange symmetry properties. () the existence of the third term, which is due to the fact that we can have many bosons in the same state. If all the particles are in only one state, n g( r r ') = φ ψ ( r )ψ ( r ')ψ ( r ')ψ ( r ) φ = n n 1 N(N 1) N(N 1) = V V The pair distribution function is position independent. p 0, then we have 45

The above result implies that the relative amplitude for removing the first particle is N/V, while the amplitude for removing a second particle is (N-1)/V since there are only N-1 remaining after removing the first particle. Now suppose that n p = smoothly varying distribution = Ce -α ( This could be a beam of particles of momentum centered at Gaussian spread about. p 0 p- p 0 ) p 0 with a If we let V large, with N/V=n fixed, then the third term is of order 1/V smaller than the first two terms and we can neglect it. Converting the remaining terms to integrals we have n g( r r ') = φ ψ ( r )ψ ( r ')ψ ( r ')ψ ( r ) φ = n = n 1 e - ( d 3 p (π) n 3 p e i r- r') A plot of this result looks like: α k ( r r ') 46

The e - ( r- r') α term is due to exchange symmetry. In this case, exchange symmetry increases the probability that two bosons will be found at small separations. In fact, the probability of finding two bosons on top of each other is equal to two times the probability of finding two bosons with a large separation Hanbury-Brown and Twiss Experiment Do bosons really tend to clump together? 47

At this point we will discuss the Hanbury-Brown and Twiss(HBT) experiment that shows that they do tend to clump. The HBT experiment measures the probability of observing photons simultaneously at different points in a beam of incoherent light. Incoherent light, as we saw earlier, can be described in terms of occupation numbers of photon states. The apparatus looks like: The half-silvered mirror splits the beam into two identical beams. The amplitude for a photon to be transmitted/reflected by the mirror is 1 / (the probability is 1/). HBT measured the following I 1 (t) = light intensity at detector 1 at time t I (t τ ) = light intensity at detector at a later time t τ and then they averaged the quantity I 1 (t)i (t τ ) over t, keeping τ fixed. 48

This is the same as measuring the relative probability of observing two photons separated by a distance cτ in the beams ( c = speed of light). g( r r The experimental result is just ') for r r ' = cτ. This confirmed the theory or so one thought at that time. We ask the following question: Is this experiment a verification of the quantum mechanical theory for identical bosons? The answer might, in fact, be no! Can we understand the result completely using classical electromagnetic theory and wave superposition? The experiment might imply that the boson properties of the photon follow from the superposition principle obeyed by classical fields and we might not need the quantum mechanical concept of a photon in this experiment. Let us see how. Consider the setup below: 49

We have two sources of photons, A and B. A emits coherent light with amplitude α and wave number k. B emits coherent light with amplitude β and wave number k '. We assume that the relative phase of the two coherent beams is random and that they have the same polarization. (A 1) αe ikr 1 The amplitude for light and the amplitude for light (B 1) βe ik ' r 1 '. Therefore, the superposition principle implies that a 1 = total amplitude at detector 1 and that =αe ikr 1 βeik ' r 1 ' I 1 = intensity at detector 1 = αe ikr 1 βeik ' r 1 ' = α β Re( α * βe i(k ' r 1 ' kr 1 ) ) Therefore, Similarly, and that Therefore, I 1 = I 1 averaged over the random phase = α β a = total amplitude at detector =αe ikr ' βe ik ' r I = intensity at detector = αe ikr ' ( ) βe ik ' r = α β Re α * βe i(k ' r kr ') I = I averaged over the random phase = α β 50

The product of the averaged intensities is clearly independent of the separation distance between the two detectors. The product of the intensities, however, behaves very differently, however, i.e., I 1 I = a 1 a = α e ik(r 1 r ) β e ik '(r 1 'r ') αβ e ikr 1 eik ' r ' e ik ' r 1 ' e ikr I 1 I ( ) α and β Multiplying out and averaging over the random phases of eliminates the terms proportional to αβ α, αβ β, etc and we get I 1 I = α 4 β 4 α β e ikr 1 eik ' r ' For well-collimated beams e ik ' r 1 ' e ikr = I 1 I α β cos[ k '(r 1 ' r ') k(r 1 r )] r 1 ' r ' r 1 r [ ] I 1 I = I 1 I α β cos (k ' k)(r 1 r ), which gives Therefore, the correlated intensities have a term that depends on the detector separation. This term is a maximum when the detectors are at the same point. If we average over all the different k and k ' present in the beam (using a Gaussian distribution) we get the same form as the quantum result. 51

This seems to imply that the "photon bunching" effect seen in the HBT experiment is a consequence of the superposition principle applied to light from noisy sources. The quantum mechanical interpretation of the classical result is 1st term = amplitude for both photons to come from A nd term = amplitude for both photons to come from B 3rd term = amplitude for one photon to come from A and the other to come from B There are two ways to do this (1) A 1 B () A B 1 These two ways are indistinguishable and the interference between them gives the cosine term. So, we corroborate superposition and bunching, but we do not seem to need the quantum concept of a photon to do it. When one studies this problem in more detail one can prove that a photon with quantum properties must exist. 5

5. The Hamiltonian Finally, we write the Hamiltonian in second quantized form. Suppose the particles interact via a two-particle potential V( r r '). The interaction energy operator then becomes ν = 1 ss ' d 3 r d 3 r 'V( r r ')ψ s ( r )ψ s ' ( r ')ψ s ' ( r ')ψ s ( r ) The order of the operators in this expression is very important. This form for ν can be confirmed by comparing its matrix elements to the matrix elements in the standard formalism. ν We interpret in this way: (1) it tries to remove particles from r and r ' () if successful it counts V( r r ') and then replaces the particles (3) it replaces the last particle removed first (4) it sums over all pairs (the factor 1/ avoids double counting of pairs) of points and r ' The second quantized Hamiltonian for particles of mass such a pairwise interaction potential is then H = d 3 r m ψ ( r ) s ψ s ( r ) 1 d 3 r d 3 r 'V( r r ')ψ s ( r )ψ s ' ( r ')ψ s ' ( r ')ψ s ( r ) s ss ' r m with We now calculate the ground state energy of our gas of spin = 1/ fermions. 53

We will treat the interaction potential as a perturbation. To lowest(zeroth) order we have E (0) = T = p m n = p ps = 3 5 p F p m N = 3 5 E F p m = V p F p 0 E (1) d 3 p (π) 3 The first order energy correction is the expectation value of ν in the unperturbed ground state. We get m If we let E (1) = 1 d 3 r d 3 r 'V( r r ') φ0 ψ s ( r )ψ s ' ( r ')ψ s ' ( r ')ψ s ( r ) φ 0 ss ' = 1 d 3 r d 3 n r 'V( r r ') ss ' g ss ' ( r r ') = 1 d 3 r d 3 r 'V( r r ') n G s ( r r ') s v 0 = d 3 r V( r ) then we have 1 d 3 r d 3 r 'V( r r ')n = Nnv 0 This is the average interaction of a uniform density of particles with itself (no correlations). It is called the direct or Hartree energy. 54

The second term represents the exchange energy E ex = 1 d 3 r d 3 r 'V( r r ') Gs ( r r ') s This term takes account of the tendency of particles of the same spin to stay apart. The effects of the short-range part of V( r r ') are overcorrected in the direct energy and fixed up in exchange energy. We have E ex N = 9n 4 and to first order An Example d 3 r E 0 = 3 5 sin p F r 6 V(r) p r F cos p r F p F r p F m nv 0 E ex N Consider a gas of electrons of average density n interacting via a Coulomb potential V( r r ') = e r r ' The conduction electrons in a metal form such a gas. We note that in a real physical system of this type, we never have an isolated electron gas. There always exists enough positive charges to make the overall system electrically neutral. 55

To first approximation in a metal or a plasma we can replace the positive ions by a uniform background of positive charge of density ne. The electrostatic self-energy of this background 1 d 3 rd 3 r ' e n r r ' plus the average electrostatic interaction between the positive background and the electrons e n d 3 rd 3 r ' r r ' exactly cancels the Hartree energy as it must because the electrostatic energy of a neutral system can only be proportional to the volume for a large system(not a higher power of the volume!). Therefore, the net interaction energy of the electron gas (to first order) is E ex E ex N = 9πne dx ( sin x x cos x) p F x 5 0 For a typical electron gas, this energy is written in terms of a parameter r s = d a 0 = average interparticle spacing Bohr radius 56

Now, we also have Therefore, and a 0 = me n 4π 3 d 3 = 1 r s = 9π 4 1/3 me p F E =.1 0.916 r s r s e a 0 where the first term is the kinetic energy and the second term is the exchange energy. Finally, if we write formalism we get v = 1 V where V k = pp ' ν qq ' ss ' in the creation/annihilation operator V p ' p d 3 re i k r V( r ) δ p q, p ' q ' â p ' s â q ' s ' â qs ' â ps This is a sum over scattering processes. Feynman developed a clever diagrammatic method of thinking about such terms. 57

(1) The delta-function represents conservation of energy and momentum in the scattering process () The process itself is represented by the diagram We shall return to this idea later and develop a method of thinking about the universe as expressed by these diagrams. Bogoliubov Transformation, Quasiparticles and Superfluidity Let us consider a weakly interacting Bose gas. From our earlier discussions we have, in general, V p ' p δ p q, p ' q ' â â ââ p ' = 1 q ' q p so that v = 1 V p p ' qq ' Ĥ = ω â kâ k k 1 k V k p q V kp q V q âk q â p q â p âk V q âk q âk q â p â q where we have used the delta function (which corresponds to momentum conservation). This corresponds to the process 58

k p ( k q) ( p q) or we have momentum transfer of bosons. q between the two interacting At low temperatures, a Bose-Einstein condensation takes place in the k = 0 mode, i.e., the k = 0 mode is macroscopically occupied or N 0 = Φ 0 aa Φ 0 N N N 0 = # excited particles << N 0 This means that we can neglect the interaction of excited particles with one another and restrict our attention to the interaction of the excited particles with the condensed particles. This gives Ĥ = ω â kâ k k 1 V V â 0â0 0â 0 â 0 k 1 V (V 0 Vk )â 0â 0 â â k k 1 V k 0 k 0 V â k (â k k â 0 â 0 â 0â 0â kâ k )... The effect of â 0 and â 0 on the state with N 0 particles in the condensate is â 0..., N 0,... = N 0..., N 0 1,... â 0..., N 0,... = N 0 1..., N 0 1,... 59

Since N 0 is a very large number ( 10 3 ), both of these relations correspond to multiplication by N 0. It is physically clear that the removal or addition of one particle from the condensate will make no difference to the physical properties of the system. In comparison to N 0, the effect of the commutator â 0,â 0 = 1 is negligible. This says that, in this case, the operators â 0 and â 0 can be approximated by a number. We then have Ĥ = k â m â k k 1 V N 0V 0 N 0 V k 0 We can write N = N 0 âk â k k 0 N 0 â k (V 0 Vk )â N 0 k V which says that (carrying out the algebra) 1 V N 0V 0 = 1 V N V 0 N V V kâ k â V 0 k V k 0 The Hamiltonian then becomes Ĥ = k 0 k â m â k k N V V â kâ k k k 0 k 0 N V V 0 N V k, k ' 0 k 0 k 0 â â k kâ â k ' k ' V â k (â k k â kâ k ) V â k (â k k â kâ k ) up to terms with 4 creation/annihilation operators, which are of order n' where 60

n' = N N 0 V = density of particles not in condensate We make the approximation of neglecting these anharmonic terms, which is good for n' << n. The remaining Hamiltonian is a quadratic form which we need to diagonalize to solve the problem. We use the so-called Bogoliubov transformation. We assume that âk = u k ˆα k v ˆα k k â k = uk ˆα k vk ˆα k with real coefficients. We then require that the operators satisfy Bose commutation relations ˆα k, ˆα k ' = ˆα k, ˆα k ' = 0 ˆα This requires that ˆα k, ˆα k ' = δ k k ' u k v k = 1 The inverse transformations are ˆα k = ukâ k v kâ k ˆα k = u kâ v k kâ k 61

We also have that â â k k = u k â â k k α k The Hamiltonian becomes Ĥ = 1 V N V 0 α k v α k k α k uk v(α k k α k α k α k ) = u k α k α k v k α k α k u k v(α k k α k α k α k ) â kâ k = u k α k α k v α k k α k uk v(α k k α k α k α k ) k 0 N V k m nv k u k α k α k v α k k α k uk v(α k k α k α k α k ) Vk k 0 (u k v k )(α k α k α k α k ) u k v(α k k α k α k α k In order for the non-diagonal terms to vanish, we must have k m nv k u k v N k V V(u k k v k ) = 0 This equation together with u k v k = 1 uk determine and vk. If we define ω k = k m nv k ( nvk ) = then we get ω k k m nv k u k = ω k v k = 1/ ω k k m nv k ω k k m ) is sufficient to nk V k m 1/ 6

and u k v k = nv k ω k v k = ( nvk ) ω k ω k k m nv k Finally, the Hamiltonian becomes where Ĥ = 1 V N V 0 1 1 V N V 0 1 k 0 ω k α k α k k m nv ω k k ω k α k k 0 k 0 α k k m nv ω k k = ground state energy E 0 k 0 = sum of oscillators or excitations The excitations (oscillators) or quanta that are created by the are called quasiparticles. α k All the excited states correspond to different numbers of noninteracting bosons where each boson has the excitation energy Ek = ω k = k m nv k ( nvk ) 1/ = k m nk V k m Quasiparticles appear in all kinds of physical systems at all energy scales. 1/ 63

The ground state of the system quasiparticles are excited, α k 0 0 = 0 for all k is fixed by the condition that no The number of particles outside the condensate(ground state) is given by N ' = 0 â â k k 0 = 0 ˆα k ˆα k 0 = vk For small where k k 0 k 0 we have E k = ω k c s k c s = nv 0 m k 0 = speed of sound in fluid This says that the long wavelength excitations have a linear dispersion relation ( E versus k ). For large k we have Ek = ω k = k m nv k This corresponds to the dispersion relation for free particles whose energy is shifted by a mean potential nvk. We can now understand qualitatively how superfluidity comes about. 64

Consider a small particle(call it a cluster) which could be several dirt atoms or a piece of a surrounding wall moving through a quantum liquid with the dispersion relations just described for the quasiparticle excitations, i.e., the quasiparticle energy rises linearly with k at low k and quadratically at large k. The cluster can lose energy (experience friction) only by causing excitations in the fluid. For T>0, there are already excitations present in the fluid at which the cluster may scatter and thus lose energy, but at T=0, q this is not the case. Let the initial momentum of the cluster be and the momentum of the excitations be k. In a scattering event of the cluster with the fluid, energy and momentum are conserved. q ' q = q ' k q m = q' m E(k) where is the momentum of the cluster after the scattering. The elementary excitations of the fluid (quasiparticles) are given by Ek = ω k = k m nk V k m Consider the energy conservation equation. We have 1/ 65

or q m = ( q k) m 0 = m E(k) q k k m E(k) α This says that (let be the angle between q and k ) cosα = 1 E(k) v k 1 k v m E(k) where v = q = initial velocity of the cluster m Now for the quasiparticle excitations E(k) k > c s therefore, for the excitation(emission) of a quasiparticle the cluster velocity must be larger than c s, i.e., v > c s. This follows from the above relation as k 0 or the angle α becomes imaginary! v < v critical = c s A cluster moving with (in this model) cannot lose energy to the fluid. Thus, there is no friction and one has superfluidity. For liquid helium v critical << c s and the physics is even more dramatic. 66