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Why Ladder Operators Are Useful Why ladder operators are useful Ryan M. Wilson (Dated: February 13, 2015) I. HARMONIC OSCILLATOR Let's consider a 1D harmonic oscillator with the Hamiltonian p2 1 H = − + m!2x2: (1) 2m 2 This yields eigenstates φn(x) with energies En = ~!(n + 1=2). We saw that we can recast this Hamiltonian in terms of \ladder" operators, y H = ~!(^a a^ + 1=2); (2) wherea ^y is a \raising" operator anda ^ is a \lowering" operator. When acting on a real-space basis of harmonic y oscillator eigenstates,a ^ φn = φn+1, andaφ ^ n = φn−1. II. TWO PARTICLES What if we have two particles in a 1D harmonic oscillator? Then our Hamiltonian is H = H1 + H2, where p2 1 H = − i + m!2x2: (3) i 2m 2 i In general, our wave function will have to be written as (x1; x2). Our Schrodinger equation is H (x1; x2) = E (x1; x2) X p2 1 − i + m!2x2 (x ; x ) = E (x ; x ): (4) 2m 2 i 1 2 1 2 i=1;2 Can we solve this by writing the separable form (x1; x2) = 1(x1) 2(x2)? Let's try. p2 p2 1 1 − 1 − 2 + m!2x2 + m!2x2 (x ) (x ) = E (x ) (x ) 2m 2m 2 1 2 2 1 1 2 2 1 1 2 2 p2 p2 1 1 − (x ) 1 (x ) − (x ) 2 (x ) + (x ) m!2x2 (x ) + (x ) m!2x2 (x ) = E (x ) (x ) 2 2 2m 1 1 1 1 2m 2 2 2 2 2 1 1 1 1 1 2 2 2 2 1 1 2 2 2 2 1 p1 1 p2 1 2 2 1 2 2 − 1(x1) − 2(x2) + m! x1 + m! x2 = E = Ep1 + Ep2; (5) 1(x1) 2m 2(x2) 2m 2 2 so this is a separable differential equation, and 1(x1) and 2(x2) are harmonic oscillator eigenstates with energies Ep1 and Ep2, respectively! A. Bosons and fermions Now, when we have more than one quantum object, we have to worry about whether we have distinguishable or indistinguishable particles. If they're distinguishable, we can proceed as before. If they're indistinguishable, things become more complicated. If the particles are indistinguishable bosons, their wave function must be symmetrized. If the particles are indistinguishable fermions, their wave function mush be anti-symmetrized. So, bosons(x1; x2) = bosons(x2; x1) (6) fermions(x1; x2) = − fermions(x2; x1): 2 We can construct these wave functions as 1 bosons(x1; x2) = p ( 1(x1) 2(x2) + 1(x2) 2(x1)) 2 1 fermions(x1; x2) = p ( 1(x1) 2(x2) − 1(x2) 2(x1)) : (7) 2 You can check that these wave functions have the properties we need. Also, you can check that these wave functions maintain a normalization of unity. Now, what's the ground state of two bosons in a harmonic trap? This would be two harmonic oscillator states with n = 0 for both particles, or bosons;0(x1; x2) = φ0(x1)φ0(x2); (8) which is already symmetrized. The energy of this state is Z Z ∗ E0 = dx1 dx2 bosons;0(x1; x2)H bosons;0(x1; x2) Z Z ∗ ∗ = dx1 dx2φ0(x1)φ0(x2)(H1 + H2)φ0(x1)φ0(x2) Z Z ∗ ∗ = dx1φ0(x1)H1φ0(x1) + dx2φ0(x2)H2φ0(x2) ! ! = ~ + ~ : (9) 2 2 What's the ground state of two fermions in a harmonic trap? This would be two harmonic oscillator states with n = 0 for both particles (that is antisymmetrized), or 1 fermions;0(x1; x2) = p (φ0(x1)φ0(x2) − φ0(x2)φ0(x1)) = 0: (10) 2 Wait! This state doesn't exist! This is a reflection of the Pauli exclusion principle - two indistinguishable fermions can't occupy states with the same quantum numbers! So, we have to add one quanta of energy to one of the fermions to get 1 fermions;0(x1; x2) = p (φ0(x1)φ1(x2) − φ0(x2)φ1(x1)) : (11) 2 The energy of this state is Z Z ∗ E0 = dx1 dx2 fermions;0(x1; x2)H fermions;0(x1; x2) 1 Z Z = dx dx (φ∗(x )φ∗(x ) − φ∗(x )φ∗(x )) (H + H )(φ (x )φ (x ) − φ (x )φ (x )) 2 1 2 0 1 1 2 0 2 1 1 1 2 0 1 1 2 0 2 1 1 1 Z Z ! 3 ! 3 ! ! = dx dx (φ∗(x )φ∗(x ) − φ∗(x )φ∗(x )) ~ φ (x )φ (x ) − φ (x ) ~ φ (x ) + φ (x ) ~ φ (x ) − ~ φ (x )φ (x ) 2 1 2 0 1 1 2 0 2 1 1 2 0 1 1 2 0 2 2 1 1 0 1 2 1 2 2 0 2 1 1 1 Z Z = dx dx (φ∗(x )φ∗(x ) − φ∗(x )φ∗(x )) 2 1 2 0 1 1 2 0 2 1 1 ! 3 ! 3 ! ! × ~ φ (x )φ (x ) − φ (x ) ~ φ (x ) + φ (x ) ~ φ (x ) − ~ φ (x )φ (x ) 2 0 1 1 2 0 2 2 1 1 0 1 2 1 2 2 0 2 1 1 1 ! 3 ! 3 ! ! = ~ + ~ + ~ + ~ 2 2 2 2 2 ! 3 ! = ~ + ~ : (12) 2 2 So, I've said that symmetrization/antisymmetrization is very important, but the energies we obtain for proper boson and fermionic two-body wave functions are just those of two distinguishable, boring particles. Now, what if we have interactions? 3 B. Interactions A two-body Hamiltonian with a simple interaction would look like H = H1 + H2 + Hint = H1 + H2 + gδ(x1 − x2): (13) Let's just focus on the interaction part for now. Suppose we have two distinguishable particles in the ground and first-excited state of a harmonic oscillator. Their wave function would look like (x1; x2) = φ0(x1)φ1(x2), and their interaction energy would be Z Z ∗ Eint = dx1 dx2 (x1; x2)Hint (x1; x2) Z Z ∗ ∗ = dx1 dx2φ0(x1)φ1(x2)gδ(x1 − x2)φ0(x1)φ1(x2) Z 2 2 = g dxjφ0(x)j jφ1(x)j g = p ; (14) 8πa p where a = ~=m!. Now, suppose we have two indistinguishable bosons in these states. The symmetrized state would be 1 bosons(x1; x2) = p (φ0(x1)φ1(x2) + φ0(x2)φ1(x1)) ; (15) 2 so the interaction energy would be Z Z ∗ Eint = dx1 dx2 bosons(x1; x2)Hint bosons(x1; x2) g Z Z = dx dx (φ∗(x )φ∗(x ) + φ∗(x )φ∗(x )) δ(x − x )(φ (x )φ (x ) + φ (x )φ (x )) 2 1 2 0 1 1 2 0 2 1 1 1 2 0 1 1 2 0 2 1 1 g Z = dx (φ∗(x)φ∗(x) + φ∗(x)φ∗(x)) (φ (x)φ (x) + φ (x)φ (x)) 2 0 1 0 1 0 1 0 1 Z 2 2 = 2g dxjφ0(x)j jφ1(x)j g = p : (16) 2πa It's twice as big! Now, suppose we have two indistinguishable fermions in these states. The antisymmetrized state would be 1 fermions(x1; x2) = p (φ0(x1)φ1(x2) − φ0(x2)φ1(x1)) ; (17) 2 so the interaction energy would be Z Z ∗ Eint = dx1 dx2 fermions(x1; x2)Hint fermions(x1; x2) g Z Z = dx dx (φ∗(x )φ∗(x ) − φ∗(x )φ∗(x )) δ(x − x )(φ (x )φ (x ) − φ (x )φ (x )) 2 1 2 0 1 1 2 0 2 1 1 1 2 0 1 1 2 0 2 1 1 g Z = dx (φ∗(x)φ∗(x) − φ∗(x)φ∗(x)) (φ (x)φ (x) − φ (x)φ (x)) 2 0 1 0 1 0 1 0 1 = 0: (18) The interaction energy is zero! This is because the fermions don't really overlap in space! Crazy! 4 C. Ladder operator reproduction We can reproduce this using ladder operators with the appropriate commutation relations. For bosons, the operators obey ^ ^y [bn; bm] = δmn ^ ^ ^y ^y [bn; bm] = [bn; bm] = 0: (19) For fermions, the operators obey ^ ^y ffn; fmg = δmn ^ ^ ^y ^y ffn; fmg = ffn; fmg = 0: (20) Let's say that these guys act on Fock, or number states, jN0;N1;:::i (21) where N0 is the number of particles in the h.o. state n = 0, N1 is the number of particles in the h.o. state n = 1, etc. When they operate on these states, they produce (we're using boson operators without loss of generality) ^ p bnjN1;N2;:::;Nn;:::i = NnjN1;N2;:::;Nn − 1;:::i ^y p bnjN1;N2;:::;Nn;:::i = Nn + 1jN1;N2;:::;Nn + 1;:::i = NnjN1;N2;:::;Nn;:::i: (22) We can define the number operator ^y ^ n^n = bnbn: (23) When the number operator operates on a Fock state, it gives ^y p n^njN1;N2;:::;Nn;:::i = bn NnjN1;N2;:::;Nn − 1;:::i = NnjN1;N2;:::;Nn;:::i; (24) so the number operator just counts the number of atoms in state n. Let's write the Hamiltonian for bosons in a harmonic oscillator, X ^y ^ Hbosons = ~!nbnbn: (25) n where ~!n = hnjHjni = ~!(n + 1=2). For the example above, the relevant Fock state is j1; 1i. Let's calculate the single-particle energy, E = h1; 1jHbosonsj1; 1i X ^y ^ = ~!nh1; 1j bnbn j1; 1i n ^y^ ^y^ = h1; 1j ~!0b0b0 + ~!1b1b1 j1; 1i = ~!0 + ~!1 ! 3 ! = ~ + ~ : (26) 2 2 This is the same result as above, and it's the same for fermions. Now, what about interactions? For interactions to take place, there needs to be at least two particles present in the system. Thus, if we start out our interaction operator by \annihilating" two particles, we will ensure that we have at least two in our system.
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