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 = − + mω2x2. (1) 2m 2
This yields eigenstates φn(x) with energies En = ~ω(n + 1/2). We saw that we can recast this Hamiltonian in terms of “ladder” operators,
† H = ~ω(ˆa aˆ + 1/2), (2) wherea ˆ† is a “raising” operator anda ˆ is a “lowering” operator. When acting on a real-space basis of harmonic † oscillator eigenstates,a ˆ φn = φn+1, andaφ ˆ 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 + mω2x2. (3) i 2m 2 i
In general, our wave function will have to be written as ψ(x1, x2). Our Schrodinger equation is
Hψ(x1, x2) = Eψ(x1, x2) X p2 1 − i + mω2x2 ψ(x , x ) = Eψ(x , x ). (4) 2m 2 i 1 2 1 2 i=1,2
Can we solve this by writing the separable form ψ(x1, x2) = ψ1(x1)ψ2(x2)? Let’s try.
p2 p2 1 1 − 1 − 2 + mω2x2 + mω2x2 ψ (x )ψ (x ) = Eψ (x )ψ (x ) 2m 2m 2 1 2 2 1 1 2 2 1 1 2 2 p2 p2 1 1 − ψ (x ) 1 ψ (x ) − ψ (x ) 2 ψ (x ) + ψ (x ) mω2x2ψ (x ) + ψ (x ) mω2x2ψ (x ) = Eψ (x )ψ (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 − ψ1(x1) − ψ2(x2) + mω x1 + mω x2 = E = Ep1 + Ep2, (5) ψ1(x1) 2m ψ2(x2) 2m 2 2 so this is a separable differential equation, and ψ1(x1) and ψ2(x2) are harmonic oscillator eigenstates with energies Ep1 and Ep2, respectively!
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,
ψbosons(x1, x2) = ψbosons(x2, x1) (6)
ψfermions(x1, x2) = −ψfermions(x2, x1). 2
We can construct these wave functions as 1 ψbosons(x1, x2) = √ (ψ1(x1)ψ2(x2) + ψ1(x2)ψ2(x1)) 2 1 ψfermions(x1, x2) = √ (ψ1(x1)ψ2(x2) − ψ1(x2)ψ2(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
ψbosons,0(x1, x2) = φ0(x1)φ0(x2), (8) which is already symmetrized. The energy of this state is Z Z ∗ E0 = dx1 dx2ψbosons,0(x1, x2)Hψbosons,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) ω ω = ~ + ~ . (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 ψfermions,0(x1, x2) = √ (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 ψfermions,0(x1, x2) = √ (φ0(x1)φ1(x2) − φ0(x2)φ1(x1)) . (11) 2 The energy of this state is Z Z ∗ E0 = dx1 dx2ψfermions,0(x1, x2)Hψfermions,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 ω 3 ω 3 ω ω = 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 ω 3 ω 3 ω ω × ~ 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 ω 3 ω 3 ω ω = ~ + ~ + ~ + ~ 2 2 2 2 2 ω 3 ω = ~ + ~ . (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 ψ(x1, x2) = φ0(x1)φ1(x2), and their interaction energy would be Z Z ∗ Eint = dx1 dx2ψ (x1, x2)Hintψ(x1, x2) Z Z ∗ ∗ = dx1 dx2φ0(x1)φ1(x2)gδ(x1 − x2)φ0(x1)φ1(x2) Z 2 2 = g dx|φ0(x)| |φ1(x)| g = √ , (14) 8πa p where a = ~/mω. Now, suppose we have two indistinguishable bosons in these states. The symmetrized state would be 1 ψbosons(x1, x2) = √ (φ0(x1)φ1(x2) + φ0(x2)φ1(x1)) , (15) 2 so the interaction energy would be Z Z ∗ Eint = dx1 dx2ψbosons(x1, x2)Hintψbosons(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 dx|φ0(x)| |φ1(x)| g = √ . (16) 2πa It’s twice as big! Now, suppose we have two indistinguishable fermions in these states. The antisymmetrized state would be 1 ψfermions(x1, x2) = √ (φ0(x1)φ1(x2) − φ0(x2)φ1(x1)) , (17) 2 so the interaction energy would be Z Z ∗ Eint = dx1 dx2ψfermions(x1, x2)Hintψfermions(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
ˆ ˆ† [bn, bm] = δmn ˆ ˆ ˆ† ˆ† [bn, bm] = [bn, bm] = 0. (19) For fermions, the operators obey
ˆ ˆ† {fn, fm} = δmn ˆ ˆ ˆ† ˆ† {fn, fm} = {fn, fm} = 0. (20) Let’s say that these guys act on Fock, or number states,
|N0,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)
ˆ p bn|N1,N2,...,Nn,...i = Nn|N1,N2,...,Nn − 1,...i ˆ† p bn|N1,N2,...,Nn,...i = Nn + 1|N1,N2,...,Nn + 1,...i = Nn|N1,N2,...,Nn,...i. (22) We can define the number operator
ˆ† ˆ nˆn = bnbn. (23) When the number operator operates on a Fock state, it gives
ˆ† p nˆn|N1,N2,...,Nn,...i = bn Nn|N1,N2,...,Nn − 1,...i = Nn|N1,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 ˆ† ˆ Hbosons = ~ωnbnbn. (25) n where ~ωn = hn|H|ni = ~ω(n + 1/2). For the example above, the relevant Fock state is |1, 1i. Let’s calculate the single-particle energy,
E = h1, 1|Hbosons|1, 1i X ˆ† ˆ = ~ωnh1, 1| bnbn |1, 1i n ˆ†ˆ ˆ†ˆ = h1, 1| ~ω0b0b0 + ~ω1b1b1 |1, 1i
= ~ω0 + ~ω1 ω 3 ω = ~ + ~ . (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. Then, we can simply put them back and calculate the overlap between their wave functions,
1 X H = V ˆb†ˆb†ˆb ˆb . (27) int 2 klmn k l n m klmn 5
R R ∗ ∗ where hkl|V |mni = dx1 dx2φk(x2)φl (x1)V (x1 − x2)φm(x1)φn(x2). For our state |1, 1i, we have
Eint = h1, 1|Hint|1, 1i 1 X = V h1, 1|ˆb†ˆb†ˆb ˆb |1, 1i 2 klmn k l n m klmn 1 = h1, 1|V ˆb†ˆb†ˆb ˆb + V ˆb†ˆb†ˆb ˆb + V ˆb†ˆb†ˆb ˆb + V ˆb†ˆb†ˆb ˆb |1, 1i 2 1212 1 2 2 1 2121 2 1 1 2 1221 1 2 1 2 2112 2 1 2 1 g = 2V1212 = √ . (28) 2πa This is the same result as before! Now, let’s do the same thing for fermions. We find
Eint = h1, 1|Hint|1, 1i 1 X = V h1, 1|fˆ†fˆ†fˆ fˆ |1, 1i 2 klmn k l n m klmn 1 = h1, 1|V fˆ†fˆ†fˆ fˆ + V fˆ†fˆ†fˆ fˆ + V fˆ†fˆ†fˆ fˆ + V fˆ†fˆ†fˆ fˆ |1, 1i 2 1212 1 2 2 1 2121 2 1 1 2 1221 1 2 1 2 2112 2 1 2 1 1 = h1, 1|V fˆ†fˆ†fˆ fˆ + V fˆ†fˆ†fˆ fˆ − V fˆ†fˆ†fˆ fˆ − V fˆ†fˆ†fˆ fˆ |1, 1i 2 1212 1 2 2 1 2121 2 1 1 2 1221 1 2 2 1 2112 2 1 1 1 = 0. (29)