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Quantum Mechanic - Evolutions of Species - Slides | PHYS 610A, Study notes of Quantum Mechanics

Material Type: Notes; Professor: Bromley; Class: QUANTUM MECHANICS; Subject: Physics; University: San Diego State University; Term: Spring 2006;

Typology: Study notes

2009/2010

Uploaded on 03/28/2010

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Lecture 10 - evolution of the species
Schr¨odinger Eqn. Soln [end of Section 4.3]
Quantum State time evolution
Free Particle [Section 5.1]
(if time) Wavepacket propagation [Section 5.1]
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Lecture 10 - evolution of the species

Schr¨

odinger Eqn. Soln [end of Section 4.3]

Quantum State time evolution

Free Particle [Section 5.1]

(if time) Wavepacket propagation [Section 5.1]

Schr¨

odinger Eqn - General soln

i ℏ

|

dtdψ

( t ) 〉 = ˆ

H | ψ ( t ) 〉 ˆ

H | E 〉 = E | E 〉

when

H

is Hermitian eigensolutions must exist!

| ψ ( t ) 〉 = ∑ | E

E

ψ

e

iEt/

a

E

(^) (0)

e

iEt/

ℏ | E

Eg. for practice, spin-

2 1

particle in magnetic field:

H

γB

0 S

z

γB

0 ℏ

Eigenstates (

E

z +

E

z −

) are the same as those of

S

z

with eigenvalues

E

z +

2 1

(^) γB

0 ℏ

and

E

z −

2 1

(^) γB

0 ℏ

.

Determinate / Stationary States

If initial state is eigenstate of Hamiltonian

ψ

E

then we calculate the state evolution simply as

| ψ ( t ) 〉 = ∑ | E

E

ψ

e

iEt/

e

iEt/

ℏ | E 〉 = | E ( t ) 〉

BUT if we do any other measurement

of this state:

P

ω

i , t

ω i | ψ ( t )

2

ω i | E ( t )

2

e

iEt/

ℏ 〈 ω i | E

2

ω

i | E

2

P

ω

i , t

ie the probability of measuring

ω

i

is constant

Engineering Quantum States

If initial state is eigenstate of

S

z

ψ

E

z +

but we place the magnetic field along y direction:

H

γB

0 S

y

γB

0 ℏ

i

i

So the state evolution via expansion in eigenvectors of

H

| ψ ( t ) 〉 = c | E

y +

d

| E

y −

c

i^1  

d

i  

After some time we then do measurements of

S

z

Free particle

Schr¨

odinger equation for free particle

V

i ℏ

d

dt

ψ ( t ) 〉 = ˆ

H | ψ ( t ) 〉 =

ˆp 2

2 m

ψ

t ) 〉

We already know eigensolns / stationary modes, ie | ψ ( t ) 〉 = | E 〉 e −

iEt/

H

E

ˆp 2

2 m | E 〉 = E | E 〉

noting that an eigenstate

ˆp | p 〉 = p | p 〉

is also an

eigenstate

ˆp 2 | p 〉 = p

ˆp | p 〉 = p 2 | p 〉

we can trial:

ˆp 2

2 m | p 〉 = E | p 〉 ⇒ ( p 2

2 m − E ) | p 〉

⇒ p = ± √ 2

mE

each eigenvalue

E

has two orthogonal eigenstates! so

α | E + 〉 + β | E − 〉 = α | p + 〉 + β | p − 〉 = | E 〉

ie. probability particle measured moving left or right

The free particle in

x

-space

Note that that proof didn’t rely on a particular space!

but for now choose position space (

ˆp

i ℏ

d

dx

ˆp 2

2 m ψ = − ℏ 2

m

d

2 ψ

dx

2

thus

d

2 ψ

dx

2 = − k 2 ψ

we’ve already found solns, to which add time bits

ψ

x

) =

Ae

ikx

Be

ikx

ψ

x, t

Ae

ikx

Be

ikx

e

iEt/

ψ

x, t

Ae

ik

( x −

ℏ k

2 m

(^) t )

Be

ik

( x

ℏ k

2 m

(^) t )

Waves have momentum

p

k

(cred to de Broglie).

so-called

plane-waves

have wavenumber

k

mE/