EP228: Quantum Mechanics I JAN-APR 2016 Lecture 21: Ladder - - PowerPoint PPT Presentation

EP228: Quantum Mechanics I

JAN-APR 2016 Lecture 21: Ladder operators (Expectation values, Heisenberg equation, Coherent states)

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

Recap

Hamiltonian in terms of ladder operators: ˆ H = ω(ˆ a†ˆ a + 1 2) and eigenstates are |n ˆ N|n = a†a|n = n|n ˆ H|n = (n + 1/2)ω|n We have derived the operation of ladder operators on |n as follows: a|n = √n|n − 1 a†|n = √ n + 1|n + 1

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

Matrix elements of position & momentum operators

Recall position operator is ˆ x = 1 2(ˆ a + ˆ a†)

• 2

mω Similarly momentum operator is ˆ p = 1 2i (ˆ a − ˆ a†) √ 2mω Determine matrix element m|ˆ x|n and m|ˆ p|n? It turns out to be m|ˆ x|n =

• 2mω{√nδm,n−1 +

√ n + 1δm,n+1} m|ˆ p|n = i

• mω

2 { √ n + 1δm,n+1 − √nδm,n−1}

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

Determine position space wavefunctions

Recall ˆ a|0 = 0. Using this

• 2mωxo|(ˆ

x + iˆ p mω)|0 = 0 gives differential equation for ground state wavefunction ψ0(xo) = xo|0 xoxo|0 + mω ∂ ∂xo xo|0 Solving the first order differential equation, we obtain normalised wavefunction as ψ0(xo) = 1 π1/4√c e

−x2

• 2c2

where c =

• /mω

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

For the first excited state ψ1(xo) = xo|1, we solve xo|1 = xo|ˆ a†|0 = (xo − c2 d dxo )ψ0(xo) You can continue this way to determine other excited states.

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

Expectation values in the eigenstates

n|ˆ x|n = 0 Similarly, expectation value of ˆ p will be zero. What about ∆x, ∆p? For that, we need to work out ˆ x2, ˆ p2 ∆x =

• n|ˆ

x2|n similarly ∆p and verify uncertainty principle ∆x∆p ≥ /2 Explicit evaluation for n|ˆ x2|n = (n + 1/2)

mω

n|ˆ p2|n = (n + 1/2)mω

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

Heisenberg equation for x, p, a, a†

We know that for operators which do not have explicit time dependence d ˆ A dt = 1 i[ˆ A, ˆ H] Check the following dˆ p dt = −mω2ˆ x ; dˆ x dt = ˆ p m These two coupled equations becomes uncoupled for ladder operators idˆ a dt = ωˆ a ; idˆ a† dt = −ωˆ a† whose solutions are ˆ a(t) = ˆ a(0)e−iωt ; ˆ a†(t) = ˆ a†(0)eiωt We can read of ˆ x(t) using them. Similarly ˆ p(t) using them.

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

In terms of ladder operators ˆ x(t) =

• 2mω
• a†(t) + a(t)
• =
• 2mω
• a†(0)eiωt + a(0)e−iωt

=

• 2mω
• {a†(0) + a(0)} cos(ωt) + i{a†(0) − a(0)} sin(ωt)
• Rewriting RHS in terms of ˆ

x(0), ˆ p(0) ˆ x(0) =

• 2mω
• a†(0) + a(0)
• ;

ˆ p(0) = i

• mω

2

• a†(0) − a(0)
• Similar to the expected classical equation of motion

ˆ x(t) = ˆ x(0) cos(ωt) + 1 mω)ˆ p(0) sin(ωt) ˆ p(t) = ˆ p(0) cos(ωt) − mωˆ x(0) sin(ωt)

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

Direct method of evaluating ˆ x(t) in Heisenberg picture

The time evolution of any operator including position operator is ˆ x(t) = U†(t)ˆ x(0)U(t) Evaluate this for harmonic oscillator ˆ H using Baker-Hausdorff lemma ei ˆ

Ot ˆ

Ae−i ˆ

Ot = A+it[ ˆ

O, ˆ A]+ (it)2

2! [ ˆ

O, [ ˆ O, ˆ A]]++ (it)3

3! [O, [O, [O, A]]]+. . .

Expectation value of ˆ x(t) in the eigenstate |n is indeed zero. Need to find a state |α which is a superposed state to see

• scillations and classical equation.

We will look at eigenstates of ladder operator which will give such a superposed state |α

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

eigenstates of raising ladder operator

For simplicity, let’s set the parameters m, ω, to 1. Then solve eigenvalue equation ˆ a†ξ(x) = ξξ(x) First order differential equation whose solution is ξ(x) = e

x2 2 −ξx

which is not normalizable. Equivalently, any state can be expanded in {|n} basis as |ξ =

• n

cn|n ˆ a†|ξ = c0 √ 1|1 + c1 √ 2|2 . . . = ξc0|0 + ξc1|1 . . . Only trivial solution c0 = c1 = . . . = 0 for eigenstate of ˆ a†

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

eigenstates of lowering ladder operator

solve eigenvalue equation ˆ aα(x) = αα(x) First order differential equation whose solution is α(x) = e− x2

2 +αx

which is a normalizable function. Again writing the state expanded in {|n} basis: |α =

• n

dn|n ˆ a|α = d1|0 + d2 √ 1|1 . . . = α (d0|0 + d1|1 . . .) solution dn =

αn √ n!d0 determining eigenstate of ˆ

a

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

Hence eigenstate of ˆ a is |α = d0

• n

α2 √ n! |n = d0

• n

αn n! (a†)n|0 Normalisation condition α|α = 1 will fix d0 = e−|α|2/2 The operator which takes a vacuum state|0 to eigenstate of ˆ a is |α = e

−|α|2 2

eαˆ

a†|0 = e

−|α|2 2

eαˆ

a†e−α∗ˆ a|0

For two operators ˆ A and ˆ B whose commutator [ˆ A, ˆ B] = const, the following identity is obeyed:e ˆ

Ae ˆ Be−[ˆ A,ˆ B]/2 = e ˆ A+ˆ B

Using this identity |α = e(αa†−α∗ˆ

a)|0 = D(α)|0

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13

Displacement operator D(α)

|α = D(α)|0 D†(α)ˆ aD(α) = ˆ a + α |α is called coherent state reproducing classical equations of motion. α|ˆ N|α gives average number ¯ n. P(n) = |n|α|2 gives the probability of finding the state in number state|n Check out P(n) is actually Poisson distribution function.

JAN-APR 2016 () EP228: Quantum Mechanics I Lecture 21: Ladder operators (Expectation / 13