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Last modified: January 27, 2018

Time Evolution
--------------

Recall from classical mechanics that the Hamiltonian 
describes how x and p change with time.  Thus, it 
describes flow with time through phase space (the time 
evolution of phase space).

H = p2/2m + U(x)

Hamilton's equations are:
        .
∂H/∂p = x
 
         .
∂H/∂x = -p

Consider some arbitrary function F(x,p)
                    .          .
dF(x,p)/dt = (∂F/∂p)p + (∂F/∂x)x

           = -(∂F/∂p)(∂H/∂x) + (∂F/∂x)(∂H/∂p)

           = {F,H} where {} are Poisson brackets

Therefore, the Hamiltonian is the generator of time 
evolution.

Return to QM:  The basic requirement is that the 
inner product between states does not change with 
time.  This means that whatever the operator is it 
must be unitary (U).  Therefore,
       
|a> -> U(t)|a>

and,
                 
<b| -> <b|U†(t)

and,

UU† = 1 = I (identity operator)

and,
          
<b|a> -> <b|U†(t)U(t)|a> = <b|a>

Therefore, the states are unchanged. 

Let's apply an operator to move a state forward in 
time by an infinetisimally small amount, ε.
          
|ψ(t + ε)> = U(ε)|ψ(t)>
         
When ε = 0, U(0) = 1
        
Consider U(ε) =  1 - (iεH/h)  

Therefore,

U(ε)†U(ε) = (1 - iεH/h)(1 + iεH†/h) = 1 

(iε/h)(H† - H) = 0

Thus,

H† = H 

|ψ(t + ε)> = (1 - iεH/h)|ψ(t)>

Therefore,

|ψ(t + ε)> - |ψ(t)> = (-iεH/h)|ψ(t)>  

Divide both side by ε.  The LHS becomes a differential 
and we get:

ih∂|ψ(t)>/∂t = H|ψ(t)> 

This is the Time dependent Schrodinger equation that 
describes how a state function changes with time.  

Solution to the TDSE
--------------------

ih∂ψ/∂t = Hψ = (p2/2m)ψ

        = (1/2m)(-ih∂/∂x)(-ih∂/∂x)ψ

ih∂ψ/∂t = -(h2/2m)∂2ψ/∂x2

∂ψ(x,t)/∂t = (ih/2m)∂2ψ(x,t)/∂x2

Solutions to the equation:

The eigenvectors of the momentum operator are: exp(ipx/h)

Let ψ(t) = f(t)exp(ipx/h)

Plug this into SE:
.   .
ψ = fexp(ipx/h) = (ih/2m)f(ip/h)2exp(ipx/h)
.
f = (-i/2mh)p2f

The solution is,

f = exp(-ip2t/2mh)

  = exp(-iEt/h) ... The QUANTUM PROPAGATOR 

f is the operator that describes the process of a physical 
system in time and must be a unitary operator.  We can say

f*f = ff* = I

This leads to:

ψ(t) = ψ(x)exp(-iEt/h)

     = exp(ipx/h)exp(-iEt/h)

     = exp(ipx/h)exp(-iωt)   where E = hω

     = ih∂/∂t|ψ(t) = H|ψ(t)>

     = |ψ(t)> = exp(-iHt/h)|ψ(0)>


We can also obtain the quantum propagator by evolving 
a state using n incremental time periods, ε.  Therefore,

ε = t/n

and we can write:

f = (1 - itH/hn)n 

  = exp(-itH/h)

Therefore, we can write:

|ψ(t + ε)> = exp(-itH/h)|ψ(t)>

Expectation Value of the Time Derivative of an Operator
-------------------------------------------------------

 .
<O> = <ψ|O|ψ>

        = <dψ/dt|O|ψ> + <dψ/dt|O|dψ/dt>

        = -i/h<ψ|HO|ψ> - i<ψ|OH|ψ> 

        = -i/h<ψ|HO - OH|ψ> 

        = -i/h<ψ|[H,O]|ψ>  

 .
<O> = -i/h[O,H]

This is a very important relationship that we will
use frequently.

Symmetry Operations
-------------------

Let O be a unitary operator corresponding to a 
rotation or a translation.  Let U corresponds to 
a time translation.  

OO† = (1 - iεO)(1 + iεO†)

 = 1 + iε(O† - O)

For this to be equal to 1 O† - 0 which means 
that O must be Hermitian.

Recall the TDSE:

ih∂|ψ>/∂t = H|ψ> 

Now, operate on ψ with O.

ih∂(O|ψ>)/∂t = HO|ψ>

LHS:  ih∂(O|ψ>)/∂t = Oih∂(|ψ>)/∂t = OH|ψ>

RHS:  HO|ψ>

∴ OH = HO 

If O commutes with H (i.e. [O,H] = 0) then there is 
a symmetry.  Therefore, to find symmetries we look 
for operators that commute with the time evolution 
operator (the Hamiltonian).  This is the quantum 
mechanical equivalent of NOTHER'S THEOREM in classical 
mechanics which states that if there is a symmetry 
there is an associated conserved quantity.  In this 
case the conserved quantity is the energy. 

A symmetry is a unitary operation that commutes with 
the Hamiltonian and preserves the relationship between
states.  If the states have a particular relationship, 
that relationhip will be maintained.  For example, if 
the states are orthogonal they will remain orthogonal.

We can replace U with I - iεH and O with I - iεO to 
get the commutator:

(I - iεH/h)(I - iεO/h) - (I - iεO/h)(I - iεH/h) 

= I - iεO/h - iεH/h - ε2HO/h2 - (1 - iεH/h - iεO/h - ε2OH/h2)

= 0

H and G are referred to as GROUP GENERATORS and both 
are hermitian.

Consider the following examples.

Translation Symmetry
--------------------

O|ψ(x)> = |ψ(x - ε)>

        = ψ(x) - ε(∂ψ/∂x) + O(ε2) ... Taylor series

Therefore,

O = I - ε(∂/∂x)

  = I - iεp/h  since p = -ih∂/∂x

Therefore, the generator of translation is p/h.  Now 
H = p2/2m so the commutator is:

[p2/2m, p/h]

This is equal to 0 and so momentum is the conserved 
quantity.

Rotation Symmetry
-----------------




O|ψ(θ)> = |ψ(θ - ε)>

        = ψ(θ) - ε(∂ψ/∂θ)

Therefore,

O = I - ε∂ψ/∂θ

By analogy with the momentum generator, we can define 
the angular momentum generator, L, as:

L = -ih(∂/∂θ)

Therefore,

O = I - iεL/h

The eigenvalue equation is:

L|ψ> = m|ψ>

-ih∂ψ(θ)/∂θ = mψ(θ)

Therefore, for angular momentum the eigenvectors and
eigenvalues are:

Eigenvector:  ψ(θ) = exp(-imθ/h)  Eigenvalue: m

Likewise, for the opposite direction we get:

Eigenvector:  ψ(θ) = exp(imθ/h)  Eigenvalue: -m

The energies corresponding to m and -m are the same 
but the states are different therefore the energy 
levels are at first glance appear to be DEGENERATE.  
However, this degeneracy can be easily broken by the 
presence of a magnetic field.  Therefore, rotation 
symmetry by itself is not enough to tell you that 
the levels are truly degenerate.  In order to prove 
this we need to add another symmetry  - reflection 
symmetry.

Reflection Symmetry
-------------------

Reflection symmetry is a discrete symmetry.

         ^
         |
         |--->
         |
------------------------
         |
         |--->
         |
         v

Consider the angular momentun again.



Consider the reflection operator, M, which relects about
the x axis.

M|ψ(θ)> = |ψ(-θ)>

If there is a reflection symmetry, the energy associated 
with clockwise angular momentum states has to equal the 
energy associated with anti-clockwise angular momentum 
states.  

It can also be shown that if any 2 symmetries don't 
commute with each other it implies degeneracy.  We will
show this for [M,L].  Consider,

[M,L]exp(imθ/h) = (ML - LM)exp(imθ/h)

                = MLexp(imθ/h) - LMmxp(imθ/h)

                = Mmexp(imθ/h) - Lexp(-imθ/h) 

                = mexp(-imθ/h) + mmexp(-imθ/h)

                ≠ 0

Closed Symmetry Groups
---------------------

Consider 2 hermitian operators A, B.  Assume:

[A,H] = 0  

[B,H] = 0

[A,B] = iC

If C is indeed a new operator.  It is easy to show 
that this new operator also commutes with H.  

[C,H] = [[A,B],H] - [H,[A,B]]

      = (AB - BA)H - H(AB - BA)

      = ABH - BAH - HAB + HBA

      = 0

In this case, the process of commutation is 'closed' 
in that any 2 generators produce the third generator 
in the group.  As an example, consider the following 
2D rotation:

      
                      
δx = -εy
δy = εx

δψ = (∂ψ/∂x)δx + (∂ψ/∂y)δy

   = -ε(∂ψ/∂x)y + ε(∂ψ/∂y)x  = iεLzψ

   = -iεypx + iεxpy

Therefore,

Lz = xpy - ypx

This is the z component of L = r x p.

Likewise.

Lx = ypy - ypx

Ly = zpy - ypx

Now calculate [Lx,Ly]:

[Lx,Ly] = [ypz - zpy, zpx - xpz]

        = -ihypx + ihxpy

        = ih(xpy - ypx)

        = ihLz

The fact that [Lx,y,z,H] = 0 implies that angular 
momentum is conserved.

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