Quantum Mechanics 1 Lecture 12: Time Dependent Phenomena

Quantum Mechanics 1 Lecture 12:

Time Dependent Phenomena

Massimiliano Grazzini

University of Zurich

Outline

The interaction picture

Time dependent perturbative expansion

The Fermi golden rule

The driven harmonic oscillator

- time ordered product and time evolution

- the S-matrix and coherent states

- periodic perturbation

The Interaction Picture

In Lecture 3 we have seen that one can define two pictures The Schrödinger picture

Time dependence is carried by states while operators are time

independent ^{| ψ}

^{(t)⟩ A}

The Heisenberg picture

Time dependence is carried by operators and the states are time

independent ^A

^{| ψ}

^⟩

⟨A⟩

(t) = ⟨ψ

(t) | A

| ψ

(t)⟩ = ⟨ψ

(t

) | U

(t, t

)A

U(t, t

) | ψ

(t

)⟩

Moving from one picture to another the matrix elements are invariant since

| ψ

(t)⟩ = U(t, t

) | ψ

(t

)⟩

with U(t, t

) = exp{− i

ℏ H(t − t

)}

A

(t)

In perturbation theory it is convenient to use a third picture, called interaction picture

This picture is in some sense intermediate between the Schrödinger and Heisenberg pictures

H = H

+ V(t)

Let us suppose that

where the perturbation _V(t) may depend on time In the interaction picture we define

| ψ

(t)⟩ ≡ exp { i

ℏ H

t } | ψ

(t)⟩

The operators are connected to those in the Schrödinger picture through

A

(t) ≡ exp { i

ℏ H

t } A

exp { − i

ℏ H

t }

Therefore their time dependence is given by the free (unperturbed) Hamiltonian _H

(1)

We of course have

⟨ψ

(t) | A

| ψ

(t)⟩ = ⟨ψ

(t)exp{− i

ℏ H

t}exp{ i

ℏ H

t} | A

| exp{− i

ℏ H

t}exp{ i

ℏ H

t}ψ

(t)⟩ = ⟨ψ

(t) | A

(t) | ψ

(t)⟩

We can now study the state evolution in the interaction picture

iℏ ∂ ∂t | ψ

(t)⟩ = iℏ exp { i

ℏ H

t } ( i

ℏ H

) | ψ

(t)⟩ + exp { i

ℏ H

t } ( H

+ V(t) ) | ψ

(t)⟩

= exp { i

ℏ H

t } V(t) exp { − i

ℏ H

t } exp { i

ℏ H

t } | ψ

(t)⟩ = V

(t) | ψ

(t)⟩

Therefore the time evolution of states in this picture is entirely given by

which is the perturbation potential in the interaction picture ^V

^(t)

V

(t)

We note that if the potential in the interaction picture depends on time even if _V(t) is time independent ^[H

^{, V(t)] ≠ 0}

We now define the evolution operator in the interaction picture

| ψ

(t)⟩ ≡ U

(t, t

) | ψ

(t

)⟩

A

(t)

Since

| ψ

(t)⟩ = U(t, t

) | ψ

(t

)⟩

we have

| ψ

(t)⟩ = exp { i

ℏ H

t } | ψ

(t)⟩ = exp { i

ℏ H

t } U(t, t

) | ψ

(t

)⟩

= exp { i

ℏ H

t } U(t, t

)exp { − i

ℏ H

t

} | ψ

(t

)⟩

and we can identify

and _{| ψ}

I

(t)⟩ ≡ U

(t, t

) | ψ

(t

)⟩

U

(t, t

) = exp { i

ℏ H

t } U(t, t

)exp { − i

ℏ H

t

}

iℏ ∂ ∂t U

(t, t

) = V

(t)U

(t, t

)

where is generally different than the time at which we have defined the interaction picture in Eq. ^t

₍₁₎ ^{t = 0}

(2)

The evolution operator _U

_{(t, t}

₎ fulfils the equation

Note that, since in general _[V

_{(t), V}

_{(t′ )] ≠ 0} we have

U

(t, t

) ≠ exp { − i ℏ ∫

t

t₀

V

(t)dt′ }

The equation can be solved in iterative form ₍₂₎ Indeed if _V

_{(t) = 0} we have

iℏ ∂ ∂t U

(t, t

) = V

(t)U

(t, t

) = 0

and thus

U

(t, t

) = 1

If we now plug this zero-order solution on the right hand side of we find ₍₂₎

iℏ ∂ ∂t U

(t, t

) = V

(t)U

(t, t

) = V

(t, t

)

and thus

U

(t, t

) = 1 − i ℏ ∫

t

t₀

dt′ V

(t′ )

This means that equation can be solved in iterative form ₍₂₎

U

(t, t

) = 1 − i ℏ ∫

t

t₀

dt′ V

(t′ ) + ( − i ℏ )

2

∫

t

t₀

dt′ ∫

t′

t₀

dt′ ′ V

(t′ )V

(t′ ′ ) + . . .

The solution can also be written by using the time-ordered product

T{O

(t

)O

(t

) . . . O

(t

)} = O

(t

)O

(t

) . . . O

(t

)θ(t

− t

)θ(t

− t

) . . . θ(t

− t

)

U

(t, t

) = 1 − i ℏ ∫

t

t₀

V

(t′ )dt′ + ( − i ℏ )

2

1 2! ∫

t

t₀

T{V

(t′ )V

(t′ ′ )}dt′ dt′ ′ + . . . ≡ T exp { − i ℏ ∫

t

t₀

V

(t′ )dt′ }

we can write + permutations

T{O

(t

)O

(t

)} = O

(t

)O

(t

)θ(t

− t

) + O

(t

)O

(t

)θ(t

− t

)

(3)

The solution in Eq. is known as Dyson series ₍₃₎

If we now plug this first-order solution on the right hand side of we find ₍₂₎

U

(t, t

) = 1 − i ℏ ∫

t

t₀

dt′ V

(t′ ) + ( − i ℏ )

2

∫

t

t₀

dt′ ∫

t′

t₀

dt′ ′ V

(t′ )V

(t′ ′ )

Time dependent perturbations

We now discuss the prototype of problem that one has to face with time dependent perturbations

We have a system which is in an eigenstate of the Hamiltonian and it is subject to a perturbation ^{| ψ}

n(0)

⟩ H

V(t)

The states and could be associated to different energies, or, in case in which there is degeneracy, could have the same energy ^{| ψ}

n(0)

⟩ | ψ

⟩

This is the typical situation of scattering problems The transition amplitude at the time is _t

a

(t) = ⟨ψ

(t) | U(t, t

) | ψ

(t

)⟩

a

(t) = ⟨ψ

(t) | U

(t, t

) | ψ

(t

)⟩

and it can be rewritten by using the interaction picture as

The perturbation is weak, so it will not change dramatically the energy levels

but it can induce a transition to another level _{| ψ}

_⟩

a

(t) = ⟨ψ

| U

(t, t

) | ψ

⟩

and use the expansion of _U

_{(t, t}

₎ in Eq. to find ₍₃₎

a

(t) = δ

+ ( − i

ℏ ) ∫

t

t₀

dt′ ⟨ψ

| V

(t′ ) | ψ

⟩

This result can be interpreted as a perturbative expansion for _a

_(t)

a

(t) = ∑

i=0

a

(t) + ( − i

ℏ )

2

∫

t

t₀

dt′ ∫

t′

t₀

dt′ ′ ⟨ψ

| V

(t′ )V

(t′ ′ ) || ψ

⟩ + . . . .

⟨ψ

(t) | = exp { i

ℏ E

t } ⟨ψ

| exp { − i

ℏ H

t } = ⟨ψ

|

but

and analogously

| ψ

(t

)⟩ = exp{ i

ℏ H

t

} | ψ

(t

)⟩ = exp{ i

ℏ E

t

}exp{− i

ℏ E

t

} | ψ

⟩ = | ψ

⟩

we can write

The driven harmonic oscillator

We consider the Hamiltonian _{H = H}

_{+ V(t)}

where and is a time-dependent potential

corresponding to a classical force ^H

^{= p}

2

/2m + 1/2mω

x

V(t) = − F(t)x F(t)

We suppose that the system in the fundamental state at and we want to find the probability to be in the -state at _{n t → + ∞} ^{t → − ∞}

x = ℏ

2ωm ( a + a

) = x

2 ( a + a

)

We first need to evaluate the potential in the interaction picture

We have _V

_{(t) = exp {} ⁱ

ℏ H

t } V(t)exp { − i

ℏ H

t }

we can show that

a

(t) = exp { i

ℏ H

t } a exp { − i

ℏ H

t } = ae

a

(t) = exp { i

ℏ H

t } a

exp { − i

ℏ H

t } = a

e

By using _H

0

= ℏω ( a

a + 1 2 ) [a, a

] = 1

This result could also be shown by using the equation of motion for an operator in the Heisenberg picture

d

dt O

(t) = − i

ℏ [O

, H

] + ∂O

(t)

∂t a

(t) = exp { i

ℏ H

t } a exp { − i

ℏ H

t }

indeed from

we have

a ·

(t) = i

ℏ H

a

(t) − i

ℏ a

(t)H

= − i

ℏ [a

(t), H

] = − iωa

(t)

and thus _a

I

(t) = ae

and analogously

a

(t) = a

e

and [a, H

] = ℏω[a, a

a] = ℏωa

We have therefore

V

(t) = − F(t) x

2 ( a

(t) + a

(t) ) = − F(t) x

2 ( ae

+ a

e

)

And we find for the first-order transition probability from the fundamental state to a state _n

= 1 2ℏωm ∫

+∞

−∞

dt F(t) ( e

⟨n | a | 0⟩ + e

⟨n | a

| 0⟩ )

2

P

(t) = 1 ℏ

∫

+∞

−∞

dt⟨n | V

(t) | 0⟩

2

= 1 2ℏωm | F(ω) ˜ |

δ

At first perturbative order we can have transitions only to the first excited state

Fourier

transform of _{f (t)}

We now want to solve the problem exactly

S ≡ U

(+ ∞, − ∞) = T exp { − i ℏ ∫

+∞

−∞

V

(t′ )dt′ } = T exp { − i ℏ ∫

+∞

−∞

f (t) ( ae

+ a

e

) dt }

To do so we need to find a closed form for the time-ordered exponential

f (t)

This can be done by observing that the commutator _{[a, a}

_{] = 1}

The time ordering can be shown to produce just an irrelevant phase factor

S = T exp { − i ℏ ∫

+∞

−∞

f (t) ( ae

+ a

e

) dt } = e

exp { − i ℏ ∫

+∞

−∞

f (t) ( ae

+ a

e

) dt }

= e

exp { β(ω)a

− β*(ω)a } with _{β(ω) =} ⁱ

2m ωℏ F(ω) ˜

We can now use the Baker-Campbell-Hausdorff formula e ^A+B = e ^A e ^B e ⁻

^[A,B]

which is valid if _{[A, [A,} B]] = [B, [A, B]] = 0 as in our case with

S = exp { β(ω)a

} exp { − β*(ω)a } exp { − 1

2 | β(ω) |

}

Neglecting the irrelevant overall phase we find _Φ

A = β(ω)a

B = − β*(ω)a

The matrix generates coherent states with _{S β = β(ω)}

S | 0⟩ = e

exp { β(ω)a

} | 0⟩

= e

∑

n=0

β

(ω)

n ! | n⟩ = | β(ω)⟩

This means that

Recall coherent states (Lecture 5)

| β⟩ = e

∑

n=0

| n⟩ β

n !

- Eigenstates of the annihilation operator _{a | β⟩ = β | β⟩} with _{β ∈ ℂ} - They have the property of having minimal uncertainty

- They are characterised by a complex number _β

- The fundamental state is a particular coherent state with

(indeed _{a | 0⟩ = 0} ) ^{β = 0}

Their explicit expression is

- They keep minimal uncertainty under time evolution

The transition probability to the -state is given by _n

| ⟨n | S | 0⟩ |

= e

| β |

n !

The result in Eq. is nothing but a Poisson distribution ₍₄₎

More generally one can show that the effect of the driving term in our harmonic oscillator is to put the oscillator into a coherent state if it was originally in a

coherent state

P

(n) = e

λ

n !

Poisson distribution: the discrete probability distribution of a given number of independent events occurring in a fixed interval of time or space with a known average _λ

For example: the number of particle emitted by a radioactive source in a given time interval

At the first perturbative order we recover the previous result

| ⟨n | S | 0⟩ |

∼ | β(ω) |

δ

(4)

When we start from the fundamental state the expected energy transfer is thus

ΔE = ℏω⟨n⟩ = ℏω | β(ω) |

= | F(ω) ˜ |

/2m

in agreement with what one would obtain with a classical calculation

⟨n⟩ = e

∑

n=0

n λ

n ! = λe

∑

n=1

λ

(n − 1)! = λ

The average number of “events” is

Where in our case and is related to the Fourier transform of the driving term

λ = | β(ω) |

β(ω) = i

2m ωℏ F(ω) ˜

∞

∑

P

(n) = e

∑

n=0

λ

n ! = 1

The distribution is normalised as

The Fermi golden rule

Let us now consider the case in which we have a perturbation which is active just from the time _{t = 0} , but is for the rest independent on time

V(t) = Vθ(t)

This allows us to discuss the case in which the system is prepared in an

initial state in a region in which the potential is negligible, let interact with the potential, and then measured in a region where the potential is still

negligible: this is the typical situation in scattering processes

V

(t) = exp { i

ℏ H

t } V(t)exp { − i

ℏ H

t } a

(t) = δ

+ ( − i

ℏ ) ∫

t

0

dt′ ⟨ψ

| V

(t′ ) | ψ

⟩

The transition amplitude at the first perturbative order is

with

a

(t) = δ

+ ( − i

ℏ ) ∫

t

0

dt′ exp{ i

ℏ (E

− E

)t′ }V

We obtain

with _V

_{= ⟨ψ}

_{| V | ψ}

_⟩ The integral gives (we assume _{m ≠ n} )

a

(t) = − 1

E

− E

( exp{i(E

− E

)t / ℏ} − 1 ) V

The transition probability per unit time is thus

1

t P

= 1 t | a

(t) |

= | V

|

t(E

− E

)

exp { i(E

− E

)t

ℏ } − 1

2

But we have

exp { i(E

− E

)t

ℏ } − 1

2

= exp{ i(E

− E

)t

2ℏ } ( exp{ i(E

− E

)t

2ℏ } − exp{− i(E

− E

)t 2ℏ } )

2

= 4 sin

[ (E

− E

)t

2ℏ ]

(5)

The transition probability per unit time is thus

1

t P

= 4 | V

|

t (E

− E

)

sin

[ (E

− E

)t

2ℏ ]

We now consider the limit of large , in which the perturbation acts for a very large

time ^t

The function _{f (x) = sin}

^(xt)

tx

For large is strongly peaked in the origin but its integral is constant* ^t

∫

+∞

−∞

dx sin

(xt)

tx

= π

It provides a representation of the distribution _δ

δ(x) = 1

π lim

t→∞

sin

(xt)

tx

*How to compute _{I = ∫}

?

−∞

dx sin

(xt) tx

First of all we can change variable and show that the integral is independent

on _t ^{x → x / t}

∫

+∞

−∞

dx sin

(xt)

tx

= ∫

+∞

−∞

dx sin

(x)

x

= 2 ∫

+∞

0

dx sin

(x) x

We define _{F(r) = 2 ∫}

such that

0

dxe

sin

(x)

x

I = F(0)

We have _{F′ (r) = − 2 ∫}

and

0

dxe

sin

x

x F′ ′ (r) = 2 ∫

∞

0

dxe

sin

x

can now be computed integrating by parts and we find

F′ ′ (r)

F′ ′ (r) = 2 ∫

∞

0

dxe

sin

x = 4

r (r

+ 4)

Integrating twice with respect to we find _{r F(r) = c}

_{+ c}

_{r − 2 arctan(r /2) + r ln} ^r

4 + r

By using the constraint _lim we can fix and and we find

r→∞

F(r) = 0 c

= π c

= 0

I = F(0) = π

We now go back to the probability per unit time and set _{x =} ^E

^{− E}

2ℏ 1

t P

= 4 | V

|

t (E

− E

)

sin

[ (E

− E

)t

2ℏ ] = | V

|

ℏ

tx

sin

(tx)

In the large limit we find _t

1

t P

⟶ | V

|

ℏ

πδ ( E

− E

2ℏ ) = 2π | V

|

ℏ δ(E

− E

)

This formula (and its extension to the case of periodic perturbation to be discussed later) was derived by Pauli in 1928 and because of the number of different applications was called by Fermi “golden rule”

It is particularly useful in the case of the continuous spectrum, in which it

expresses the conservation of energy

Periodic perturbation

We now consider a sinusoidally varying time-dependent potential

V(t) = θ(t) ( 𝒱e

+ 𝒱

e

)

a

(t) = ( − i

ℏ ) ∫

t

0

dt′ e

( 𝒱

e

+ 𝒱

e

)

At first order we find (assuming _{m ≠ n} )

= − 1 ℏ [

e

− 1

ω

+ ω 𝒱

+ e

− 1

ω

− ω 𝒱

]

where we have set _ω

_{= (E}

_{− E}

_{)/ ℏ}

This formula is completely analogous to Eq. ₍₅₎

a

(t) = − 1

E

− E

( exp{i(E

− E

)t / ℏ} − 1 ) V

with the replacement _ω

_{→ ω}

± ω

Therefore we have for the probability per unit time

1 t P

= 2π

δ(E

− E

+ ℏω) 2π

δ(E

− E

− ℏω)

We see that there is no energy conservation of the quantum system alone:

in the first case (stimulated emission) the energy is given by the

system to the potential ^ℏω

in the second case (absorption) the energy is taken by the system from the potential and the system ends up in an excited state ^ℏω

E _n

E

ℏω ℏω

stimulated emission absorption

This formalism finds a natural applications to the interactions of atomic electrons with the classical radiation field

Note that

| 𝒱

|

= | 𝒱

|

E

E _n