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

Recap Hamiltonian in terms of ladder operators: a + 1 ˆ a † ˆ H = � ω (ˆ 2) and eigenstates are | n � ˆ a † a | n � = n | n � N | n � = ˆ H | n � = ( n + 1 / 2) � ω | n � We have derived the operation of ladder operators on | n � as follows: √ n | n − 1 � a | n � = √ a † | n � = n + 1 | n + 1 � Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

Matrix elements of position & momentum operators Recall position operator is � x = 1 2 � a † ) ˆ 2(ˆ a + ˆ m ω Similarly momentum operator is √ p = 1 a † ) ˆ 2 i (ˆ a − ˆ 2 � m ω Determine matrix element � m | ˆ x | n � and � m | ˆ p | n � ? It turns out to be � √ 2 m ω {√ n δ m , n − 1 + � � m | ˆ x | n � = n + 1 δ m , n +1 } � √ n + 1 δ m , n +1 − √ n δ m , n − 1 } � m ω � m | ˆ p | n � = i { 2 Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

Determine position space wavefunctions Recall ˆ a | 0 � = 0. Using this � x + i ˆ � p 2 m ω � x o | (ˆ m ω ) | 0 � = 0 gives differential equation for ground state wavefunction ψ 0 ( x o ) = � x o | 0 � ∂ x o � x o | 0 � + � � x o | 0 � m ω ∂ x o Solving the first order differential equation, we obtain normalised wavefunction as − x 2 1 o ψ 0 ( x o ) = π 1 / 4 √ c e 2 c 2 � where c = � / m ω Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

For the first excited state ψ 1 ( x o ) = � x o | 1 � , we solve a † | 0 � = ( x o − c 2 d � x o | 1 � = � x o | ˆ ) ψ 0 ( x o ) dx o You can continue this way to determine other excited states. Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

Expectation values in the eigenstates � n | ˆ x | n � = 0 Similarly, expectation value of ˆ p will be zero. What about ∆ x , ∆ p ? x 2 � , � ˆ p 2 � For that, we need to work out � ˆ � x 2 | n � ∆ x = � n | ˆ similarly ∆ p and verify uncertainty principle ∆ x ∆ p ≥ � / 2 x 2 | n � = ( n + 1 / 2) � Explicit evaluation for � n | ˆ m ω p 2 | n � = ( n + 1 / 2) � m ω � n | ˆ Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

Heisenberg equation for x , p , a , a † We know that for operators which do not have explicit time dependence d ˆ dt = 1 A i � [ˆ A , ˆ H ] Check the following d ˆ p d ˆ dt = ˆ x p dt = − m ω 2 ˆ x ; m These two coupled equations becomes uncoupled for ladder operators a † i � d ˆ a i � d ˆ a † dt = � ω ˆ a ; dt = − � ω ˆ whose solutions are a (0) e − i ω t ; ˆ a † ( t ) = ˆ a † (0) e i ω t a ( t ) = ˆ ˆ We can read of ˆ x ( t ) using them. Similarly ˆ p ( t ) using them. Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

In terms of ladder operators � � � � a † ( t ) + a ( t ) x ( t ) ˆ = 2 m ω � � � a † (0) e i ω t + a (0) e − i ω t � = 2 m ω � � � � { a † (0) + a (0) } cos( ω t ) + i { a † (0) − a (0) } sin( ω t ) = 2 m ω Rewriting RHS in terms of ˆ x (0) , ˆ p (0) � � m � ω � � � � � a † (0) + a (0) a † (0) − a (0) x (0) = ˆ ; ˆ p (0) = i 2 m ω 2 Similar to the expected classical equation of motion 1 x ( t ) ˆ = x (0) cos( ω t ) + ˆ m ω )ˆ p (0) sin( ω t ) p ( t ) ˆ = ˆ p (0) cos( ω t ) − m ω ˆ x (0) sin( ω t ) Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

Direct method of evaluating ˆ x ( t ) in Heisenberg picture The time evolution of any operator including position operator is x ( t ) = U † ( t )ˆ ˆ x (0) U ( t ) Evaluate this for harmonic oscillator ˆ H using Baker-Hausdorff lemma Ot ˆ Ot = A + it [ ˆ A ]+ ( it ) 2 A ]]++ ( it ) 3 e i ˆ Ae − i ˆ O , ˆ 2! [ ˆ O , [ ˆ O , ˆ 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 oscillations and classical equation. We will look at eigenstates of ladder operator which will give such a superposed state | α � Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

eigenstates of raising ladder operator For simplicity, let’s set the parameters m , ω, � to 1. Then solve eigenvalue equation a † ξ ( x ) = ξξ ( x ) ˆ First order differential equation whose solution is x 2 2 − ξ x ξ ( x ) = e which is not normalizable. Equivalently, any state can be expanded in {| n �} basis as � | ξ � = c n | n � n √ √ a † | ξ � = c 0 ˆ 1 | 1 � + c 1 2 | 2 � . . . = ξ c 0 | 0 � + ξ c 1 | 1 � . . . a † Only trivial solution c 0 = c 1 = . . . = 0 for eigenstate of ˆ Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

eigenstates of lowering ladder operator solve eigenvalue equation ˆ a α ( x ) = αα ( x ) First order differential equation whose solution is α ( x ) = e − x 2 2 + α x which is a normalizable function. Again writing the state expanded in {| n �} basis: � | α � = d n | n � n √ ˆ a | α � = d 1 | 0 � + d 2 1 | 1 � . . . = α ( d 0 | 0 � + d 1 | 1 � . . . ) α n solution d n = n ! d 0 determining eigenstate of ˆ √ a Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13

Hence eigenstate of ˆ a is α 2 α n � � n ! ( a † ) n | 0 � | α � = d 0 √ | n � = d 0 n ! n n Normalisation condition � α | α � = 1 will fix d 0 = e −| α | 2 / 2 The operator which takes a vacuum state | 0 � to eigenstate of ˆ a is −| α | 2 −| α | 2 a † | 0 � = e a † e − α ∗ ˆ e α ˆ e α ˆ a | 0 � | α � = e 2 2 For two operators ˆ A and ˆ B whose commutator [ˆ A , ˆ B ] = const , the following identity is obeyed: e ˆ A e ˆ B e − [ˆ A , ˆ B ] / 2 = e ˆ A +ˆ B Using this identity | α � = e ( α a † − α ∗ ˆ a ) | 0 � = D ( α ) | 0 � Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 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. Lecture 21: Ladder operators (Expectation JAN-APR 2016 () EP228: Quantum Mechanics I / 13