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3.1 Schrödinger Equation
Postulate 6: Schrödinger Equation
The time evolution of a quantum system is determined by the Hamiltonian
or total energy operator H(t) through the Schrödinger equation
i
d
 (t )  H (t )  (t )
dt
= Operator corresponding to total energy
Derived from classical Hamiltonian
•What kind of terms do we expect it could contain?
•How is H different from E?
Let’s see what this equation tells us about our states if H is timeindependent
 is a functional: traditionally a map from a vector space of
functions usually to real numbers. In other words, it is a
function that takes functions as its argument and returns a real
number (or it can be written in matrix form as an operator)
•As an operator corresponding to an observable, what properties do
we expect it to have?
•If we diagnonalize it, what will the eigenvalues correspond to?
Let’s write the eigenvalue equation for this operator (assume H is timeindependent for now)
H Ei  Ei Ei
Ei : energy eigenvalue
Combine the above equation with the Schrodinger equation to
determine what we can say about the eigenvectors (eigenstates)
H is an observable  Hermitian operator
 Eigenvectors form a complete basis.
   ci Ei
i

 (t )   ci (t ) Ei
i
3.1 Schrödinger Equation
Postulate 6: Schrödinger Equation
The time evolution of a quantum system is determined by the Hamiltonian
or total energy operator H(t) through the Schrödinger equation
i
d
 (t )  H (t )  (t )
dt
= Operator representing total energy
Argued Friday: If H is time-independent, and we write this in the
continuous functional form: ih(dψ(t)/dt) = Hψ(t)
The time derivative returns itself, with an i out front
So we can write ψ(t) = eiαtn, where n is time-independent
So in Dirac notation we may expect: |ψ(t)> = |eiαtn>
 is a functional: a map from a vector space of functions to numbers. In
other words, it is a function that takes functions as its argument and
returns a number – often used in the calculus of variations to find the
minimizing function – here that would be finding the state that
minimizes energy
As an operator corresponding to an observable, we know it’s
Hermitian, so it must contain a complete basis of eigenkets and
eigenvalues which correspond to energies of our system
Let’s write the eigenvalue equation for this operator (assume H is timeindependent for now)
H Ei  Ei Ei
Ei : energy eigenvalue
Here |Ei> is our |n>, we’ve taken out the time-dependence
**NOTE H is an operator, Ei are real, scalar,
eigenvalues – the Hamiltonian is NOT the energy!
Let’s put time back into the states – if the full eigenstates |ψ(t)>
had no time-dependence then d/dt would yield zero, but the time
dependence doesn’t “live” in the eigenstate of the hamiltonian
since H has no time dependence. AND, the eigenvectors must
satisfy the completeness relationship, so we must be able to write:
   ci Ei
i

 (t )   ci (t ) Ei
i
Shove    ci Ei
i
equation

 (t )   ci (t ) Ei
into the Schrödinger
i
Use orthonormality to simplify
Verify what our understanding of continuous functions told us
Simplify
Get general form of |ψ(t)> for all time-independent Hamiltonians:
|ψ(t)> = ΣiCoe-iwit|Ei>
that’s an wi in the exponential! (Yucky microsoft… )
What will happen if our system starts in one
particular energy eigenstate at t=0, say |E2>, then we
watch it for some time t – how do we write the state
at time t?
•How will the probability of finding the state with
energy E2 (or any energy) after time t differ from at
time t=0?
•How will the probabilities associated with any
observations of this state after time t differ from at
t=0?
What name might you give such an initial “pure”
state to explain how it behaves with time?
If our initial state at t=0 is a|E1> + b|E2>, what will be
our state after some time t?
•How will the probability of finding the state with
energy E2 (or any energy) after time t differ from at
time t=0?
•How will the probabilities associated with any
observations of this state after time t differ from at
t=0?
Reminders: found that the energy eigenstates are
“stationary states”, and that if H is time independent,
we can expand any state as: |ψ(t)> = ΣiCie-iEit/ħ|Ei>
Refresher calculation: our initial state at t=0 is a|E1>
+ b|E2>, what will be our state after some time t, and
what is the probability of finding the state with energy
E2 (or any energy) at any time t?
Your turn: limit to 2-level system for simplicity
•How will the probabilities associated with any
observations of this state after time t differ from at
t=0?
Energy eigenstates
• Regardless of our initial energy eigenstate, the
probability of observing the energy of the state does not
change with time (that’s because our Hamiltonian,
which represents total energy, is time independent!)
• (This is true for any observable that commutes with H
as well – since commuting observables share
eigenstates)
• However, if we are considering a general observable,
our probability will oscillate with time!
• angular frequency of the “time evolution” = Bohr
Frequency ω12 = E2-E1/ħ
Measurement of some other observable A – say position, momentum, or spin!!
(1) If [A, H ] = 0, then observations won’t change with time – “stationary states”
since A and H must share eigenstates
(2) If [A, H ]  0, then observations will oscillate with time
Look at example of time evolution of square well states using the
PhET simulation (http://phet.colorado.edu)
Explain what you observe in terms of our new understanding of
“time evolution” of quantum states
Recipe for solving a standard time-dependent quantum mechanics problem
with a time-independent Hamiltonian
Given Hamiltonian H and an initial state |y(0), what is the probability that an
is measured at time t?
• Diagonalize H (find eigenvalues Ei and eigenvectors |Ei).
• Write |y(0) in terms of energy eigenstates |Ei.
• Multiply each eigenstate coefficient by e
 (0)   ci Ei
i

i
 (t )   ci e
Ei
t

i
i
• Calculate probability P(an )  an  (t )
Ei
t

to get |y(t).
Ei   ci e iit Ei
i
2
3.2 Spin Precession
Find our measurements depend on energy differences, so our Hamiltonian needs to
only include terms that will involve energy differences in the two possible spin
states
•Only the dipole potential energy does this: recall that U = -μ·B
•Choice for zero point of potential energy is arbitrary if we only care about
energy differences
•No kinetic term is needed here, K=0 below
Hamiltonian of a spin-1/2 system in a uniform magnetic field (The electron g-factor is
a bit more than two, and has been measured to twelve decimal places: 2.0023193043617)

B


 
e  
H  K  U    B 
S B
me c

q 
where   g
S
2me c
e 

S
me c
3.2.1 Magnetic Field in z-direction
The uniform magnetic field is directed along the z-axis.

B  B0 z , so
eB0
e 
 eB0
H    B 
S  B0 z  
S z  0 S z , 0 
me c
me c
me c
o  1 0 


or : H 
2  0  1
This is a time independent hamiltonian!!
H , S z   0
H and Sz share common eigenstates. Therefore,
what would we expect about measurement of Sz?
We can just “write down” the eigenstates since they must
be the same as for H – since Sz and H commute
Eigenvalue equations
for the Hamiltonian
Eigenvalues and
Eigenvectors
 0
H   0 S z  
  E 
2
 0
H   0 S z   
  E 
2
 0
E 
, E  
2
 0
E  
, E  
2
Example (1) The initial state is spin up along z-axis:
 (t )  e
i
E
t

 e
i
0 t
2
 (0)  

The initial state is an energy eigenstate.
 The time evolved state simply has a phase factor in front.
 therefore there is no physical change of the state with time
The probability for measuring the spin to be up along the z-axis
P()    (t )
2
 e
i
0 t
2
2

1
time independent
|+ and |- are stationary states, if we start purely in one of them, we stay there!.
Example (2) The most general initial state:
 (0)   n  cos

 (t )  e
i

2
E
t

i
  e sin
cos

2

2
 e
i

E
t
i

e sin

2

In matrix formalism
 

 cos 
2 
 (0)  
 ei sin  
2


 i E t
   i 20t
 
 e
cos 
e
cos
2   
2 
 (t )   E
 i  t i
   i 20t i

e sin   e e sin 
e
2 

2
Only this term changes – it is
rotated to a new angle phi, but
theta has stayed the same!
 e
This overall phase out
front does not effect
measurements
i
0 t
2



cos


2


 ei 0t   sin  
2

White board activity
• Find the probability finding the general
state in spin up in z, in x, or in y (row 1, 2
and 3):



0 t 
cos

i
2
2 

|ψ(t)> 
e

 ei 0t   sin  
2

Probability for measuring the spin projection along the z-axis:
2
P()    (t )
2
 1 0e
i
0 t
2



2
 t
cos


i 0

2
2

  e 2 cos
 cos

2
2
 ei ( 0 t ) sin 
2

• Time independent since the Sz eigenstates are also the energy eigenstates, and are
therefore stationary states.
• Consistent with the fact we found the polar angle q to be constant.
Probability for measuring the spin projection along the x-axis:
2



2
cos


i
2
1
1


2
  cos  ei ( 0 t ) sin
1 1e 2 
Px ()  x   (t ) 
2
2
2
2
 ei ( 0 t ) sin  
2

1




 cos 2  cos sin ei ( 0 t )  e i ( 0 t )  sin 2 
2
2
2
2
2
1
use : sin( 2 )  2 sin  cos 
 1  sin  cos(  0 t )
2
0 t


• Time dependent since the Sx eigenstates are not stationary states ([Sz,Sx] is not zero).
White board activity
•
•
•
•
Find <Sx>
<Sy>

And <Sz>
t
i 0 
2 
e
For the general state |ψ(t)>



cos

2

 ei 0t   sin  
2

Expectation Value of Spin Angular Momentum:
Sz
 
 
    P()     P()   (t ) S z  (t )
 2
 2
e
i
0 t
2


 cos
2



  cos
2
2



 t
cos

i 0 
1
0





2

e 2 
e i ( 0 t ) sin  
2  2  0  1
 ei ( 0 t ) sin  
2




cos

 
 i (   0 t )
2

e
sin  
2    ei ( 0 t ) sin  


2

 2

2
 cos  sin   cos 
2
2
2 2
S x   (t ) S x  (t )



 t
 t
cos

i 0 
i 0 
0
1







i
(



t
)
2
0

e 2 
 e 2  cos
e
sin  
2
2  2 1 0
 e i ( 0 t ) sin  

2


 i (   0 t )
sin 
e




2
  cos
e i ( 0 t ) sin  

2
2
2

cos

2




 i (   0 t )  i (   0 t ) 
 cos sin e
e
 sin  cos(  0 t )
2
2
2
2


S y   (t ) S y  (t )



0 t
0 t 
cos

i
i
0

i







2

e 2 
 e 2  cos
e i ( 0 t ) sin  
2
2 2i 0 
 e i ( 0 t ) sin  

2



i (   0 t )
sin 
  ie




2
  cos
e i ( 0 t ) sin  

2
2
2

i cos

2






i (   0 t )
 i (   0 t )
 cos sin  ie
 ie
 sin  sin(   0 t )
2
2
2
2



S (t )  S x xˆ  S y yˆ  S z zˆ


xˆ sin  cos(  0 t )  yˆ sin  sin(   0 t )  zˆ cos    nˆ( ,  0 t )
2
2
B0 zˆ
Spin Precession
Larmor frequency:
frequency of precession
eB0
0 
me c


S (t )  nˆ ( ,   0t )
2

S (0)
w0 t
q q

 nˆ ( ,  )
2
f
f +w0 t

S (t )  S x xˆ  S y yˆ  S z zˆ
x


xˆ sin  cos(  0 t )  yˆ sin  sin(   0 t )  zˆ cos     nˆ( ,   0 t )
2
2
y

S (t )  S x xˆ  S y yˆ  S z zˆ


xˆ sin  cos(  0 t )  yˆ sin  sin(   0 t )  zˆ cos    nˆ( ,  0 t )
2
2
What is <S(t)>:
• if we start in a |+>x state?
• If we start in a |+>y state?
• If we start in a |+> state?
What does this mean?
Classical expectations:
• Assume the magnetic moment is
aligned with its angular momentum
• γ is the gyromagnetic ratio,
proportional to q/2m
• Torque will be perpendicular to the
moment and to B! It will change
the direction of the angular
momentum – just like precession
of a top
• If γ>0, precession is clockwise. In
the spin-1/2 case for electron, it is
negative (spin and magnetic
moment are anti-parallel) so the
precession is counterclockwise
• This precession shows our system
has angular momentum!!
  B
  L
dL d (  /  )


 B
dt
dt
d
   B
dt
3.2.2 Magnetic Field in general direction

B  B0 zˆ  B1 xˆ

B

B0
q
B1
tan  
B0

B1
Larmor frequencies:
eB0
eB1
0 
, 1 
me c
me c
Hamiltonian:

e  
e
H    B 
SB 
( S x B1  S z B0 )  0 S z  1S x
me c
me c

In matrix representation
0  1 0  1  0 1    0

 

  
H 
2  0  1 2  1 0  2  1
1 

 0 
This will not have the same eigenvalues and eigenvectors as the last case!
We must diagonalize it!!
Characteristic equation:

0  
2

1
2
I  H  0

2
1





 

2
 0   0     0      1   0

2
 2
 2 
 0  
2
What does this
2
2

   
    0    1   0    
02  12
2
2  2 
become if B is
along z, so Bx
= 0?
2
cos  
NOW DO A TRICK:
Rewrite the Hamiltonian
  0
H  
2  1
B1 1
tan  

B0 0
0


2
2
1  
2
2  0  1
 
0  1 
1
 0  2

 2 2
0
1

w0 q
sin  
w1
1


2
2
0  1 

0


2
2 
0  1 
0
02  12
1
02  12
cos  
Now we simply get:
2
2   cos 
H  0  1 
2  sin 
sin  

 cos  
w0 q
sin  
w1
0
02  12
1
02  12
Let n be the unit vector in the direction of the magnetic field:



n  cos z  sin x
 
S n  S  n  S x sin   S z cos 
 0 1
 1 0 
  cos 
 sin   
 cos   
S n  
2 1 0
2  0  1
2  sin 
sin  

 cos  
H  02  12 S n
Now we’re using our field direction to define the coordinate system:
This makes sense – we’re shifting our perspective to a new axis z’ which has
been rotated from z by an angle of theta
We already know how to deal with this system, because we know the
eigenstates corresponding to Sn:

2
2


E







cos


sin


0
1
Eigenstates:
n
2
2
2



2
2
E





 n  sin   cos 

0
1
2
2
2
We want to see if it is possible for a system that starts in the state |+> to end up in the
state |-> (or vice versa) in this magnetic field that is aligned along z and x Let’s start
in |+>:
Initial state
 ( 0)     
 
nn
 cos
Time evolved state
2
 (t )  e


n   

nn
n

E
t

cos

2



 n  sin
i

n

2
nn
nn


 


z
n
n
 n e
i
E
t

sin

2

n
B
Probability of a spin flip

 (0)  
P(  )    (t )

  e

e
e
i
E
i  t

E
t

E
i  t

 sin 2

2
cos
cos


2

2
 n e
  n e
2


cos sin  e
2
2
cos 2

2
1 e
i
n
n
 cos
 sin
E
i  t


2
  cos

sin  n 
2

sin
sin
( E  E ) 2
i
t

2
  sin

E
i  t

E
t



2


2

2
2


B0
q

B1
z
2


B
2
n

x
B

2
 (t )
  cos 
2
2
1
( E  E ) 

 sin 2  1  cos 
t
2



 ( E  E ) 
 E 
 sin  sin  
t   sin 2  sin 2 
t
 2

 2 
2

z
Probability oscillates
with frequency
dependent on delta(E)
 E 
P(  )  sin  sin 
t
 2 
2
z
B
2
P (   )
Note if B = Bz, θ is zero, and no spin flip occurs – which
is what we would expect. That system just precesses
around the z-axis!
However, if B = Bx, sin(θ)=1, and we get maximum
probability for the flip
 E 
P(  )  sin 2  sin 2 
t
 2 
sin 2 
The probability is always
less than one if the angle is
between 0 and 90, so the
flip is never 100% certain
to happen
2
E
t
1

2
2
0  1 and, sin  
Since E  
we can rewrite the probability:
2
2
2
0  1
Rabi Formula
 2 2
12
0
1
2
P(  )  2
sin

0  12
2


t


This is generalizable to any 2state system with nondiagonal H – the probability
of being in a non-eigenstate
will oscillate between two
values
Understanding “spin space”
• You know 3-d space isn’t the same as our
spin space
– Evidence: the “length” of the spin vector is
longer than we can ever measure – more or less
meaning we can’t align real space perfectly
parallel to spin space
– We must be careful when interpreting any
physical picture of spin! (they are all
“unsatisfactory” and can be misleading if this
fact is forgotten)
In words of QM theorist at Los
Alamos:
• Quantum objects are like what classical things look to someone
with bad eyesight: a quantum spin half object is like a cylinder
with one red end and one green end looked at by someone with
such bad eyesight that it looks more like a fuzzy sphere. So, the red
side basically points out a hemisphere towards which it points. The
abstract vector space is a space in which every classical
hemisphere (or smaller solid angle for larger spin value) is a point.
You can naively associate the central direction as the classical
vector, as long as you remember the irreducible fuzzyness.
• the superposition principle that lies behind the construction of the
state vector space is the most difficult bit in QM. The hemisphere
picture works somewhat but is not very clean… I know that
almost everybody involved in understanding why quantum
computation is faster than classical computation for some problems
would like to understand this. It is also at the heart of most
controversies about interpretation of quantum mechanics. So, if
textbooks don't have a good picture, its because the author doesn't.
Some possibly interesting papers
• “Delirium Quantum: Or where I will take quantum
mechanics if it will let me” by Christopher Fuchs at
Bell Labs, 2009
• http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1968
v1.pdf
• “Nonlinear Quantum Mechanics, the Superposition
Principle, and the Quantum Measurement Problem”
by Kinjalk Lochan and T.P. Singh atTata Institute,
2009
• http://arxiv.org/PS_cache/arxiv/pdf/0912/0912.2845
v2.pdf
Extra practice to help with lab
• Write down the projection operator needed for
combining the two ports from the S-G along X
• Calculate the probabilities you’d get out of the |+>
and out of the |-> ports on the last S-G device
• Consider the spin-1 case, and write down the
projection operator for combining multiple ports as
needed for lab 3
Neutrino Oscillations
• What are neutrinos?
• http://www.particleadventure.org/index.html
• Tau neutrino discovered at Fermilab in 2000 – strong
evidence for 3 “generations” of standard model
• 1998 was first report of evidence for neutrino
oscillation (at Super-K in Japan)
• “missing solar neutrinos” in 2001 – realized to be
due to oscillations
– Proton mass is ~1GeV
– Electron mass ~1MeV
– Neutrino mass <1eV!
Oscillations:
Produced viaNeutrino
the weak interaction:
• |Ve> and |Vμ> are eigenstates of the hamiltonian for
the weak interaction – in weak processes they will be
stationary states
However, in “free space” (i.e., when they are produced in
the sun and travel to the earth) their energy is
described by the Hamiltonian for their relativistic
energy
• Eigenstates for this hamiltonian are NOT the same – IF
these two neutrinos don’t have the same mass - these
two hamiltonians won’t commute!
• Write the “free space” eigenstates as |V1> and |V2>
“mixing”
• 2-level system (if we consider only electron and
muon neutrino, tau is less common) – expand it
using our spin ½ knowledge:
– |Ve> = cos(θ/2)|V1> + sin(θ/2)|V2>
– |Vμ> = sin(θ/2)|V1> - cos(θ/2)|V2>
• Call θ/2 the “mixing angle”
• If we start with an electron neutrino |Ve> write
how our state would look with time. (think
carefully: What eigenvalues are you using in your
exponent – the ones from the weak hamiltonian or
the relativistic hamiltonian?)
Simplify the relativistic energy
eigenvalues
2
2 2
• Ei = √[(pc) + (mic ) ]
• (note the energies would be the same if they
had the same mass)
• Simplify using the fact that mc2 << pc
• Write the probability for finding this neutrino
that started as an electron neutrino to later be a
muon neutrino
• Simplify E1-E2, and use t = L/c, where L is the
distance from the sun to the earth
“Super-K” (Kamiokande)
• http://en.wikipedia.org/wiki/Super-Kamiokande
Why are neutrinos massive?
• In Standard Model, fermions have mass because of
interactions with Higgs field (Higgs boson), but
this can’t explain neutrino mass
• Mass of neutrino is at least 500,000 times smaller
than the mass of an electron, so any correction to
standard model wouldn’t explain why the mass is
SO small
• This remains unexplained
• http://en.wikipedia.org/wiki/Neutrino_oscillations
Extending
our
knowledge
Cabibbo–Kobayashi–Maskawa matrix
•
• specifies the mismatch of quantum states of quarks
when they propagate freely and when they take part in
the weak interactions
• IF this matrix were diagonal, there would be no
“mixing”
• Necessary for understanding Charge Parity violation
(important research topic)
• (Nobel prize for 2008)
• http://en.wikipedia.org/wiki/Cabibbo%E2%80%93Ko
bayashi%E2%80%93Maskawa_matrix