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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 = p^{2}/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 > <bU^{†}(t)
and,
UU^{†} = 1 = I (identity operator)
and,
<ba> > <bU^{†}(t)U(t)a> = <ba>
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 = Hψ(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 = (p^{2}/2m)ψ
= (1/2m)(ih∂/∂x)(ih∂/∂x)ψ
ih∂ψ/∂t = (h^{2}/2m)∂^{2}ψ/∂x^{2}
∂ψ(x,t)/∂t = (ih/2m)∂^{2}ψ(x,t)/∂x^{2}
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)^{2}exp(ipx/h)
.
f = (i/2mh)p^{2}f
The solution is,
f = exp(ip^{2}t/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∂/∂tψ(t) = Hψ(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> = <ψOψ>
= <dψ/dtOψ> + <dψ/dtOdψ/dt>
= i/h<ψHOψ>  i<ψOHψ>
= i/h<ψHO  OHψ>
= 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 = Hψ>
Now, operate on ψ with O.
ih∂(Oψ>)/∂t = HOψ>
LHS: ih∂(Oψ>)/∂t = Oih∂(ψ>)/∂t = OHψ>
RHS: HOψ>
∴ 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  ε^{2}HO/h^{2}  (1  iεH/h  iεO/h  ε^{2}OH/h^{2})
= 0
H and G are referred to as GROUP GENERATORS and both
are hermitian.
Consider the following examples.
Translation Symmetry

Oψ(x)> = ψ(x  ε)>
= ψ(x)  ε(∂ψ/∂x) + O(ε^{2}) ... Taylor series
Therefore,
O = I  ε(∂/∂x)
= I  iεp/h since p = ih∂/∂x
Therefore, the generator of translation is p/h. Now
H = p^{2}/2m so the commutator is:
[p^{2}/2m, p/h]
This is equal to 0 and so momentum is the conserved
quantity.
Rotation Symmetry

Oψ(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:
Lψ> = mψ>
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.
Mψ(θ)> = ψ(θ)>
If there is a reflection symmetry, the energy associated
with clockwise angular momentum states has to equal the
energy associated with anticlockwise 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 = iεL_{z}ψ
= iεyp_{x} + iεxp_{y}
Therefore,
L_{z} = xp_{y}  yp_{x}
This is the z component of L = r x p.
Likewise.
L_{x} = yp_{y}  yp_{x}
L_{y} = zp_{y}  yp_{x}
Now calculate [L_{x},L_{y}]:
[L_{x},L_{y}] = [yp_{z}  zp_{y}, zp_{x}  xp_{z}]
_{ } = ihyp_{x} + ihxp_{y}
_{ } = ih(xp_{y}  yp_{x})
_{ } = ihL_{z}
The fact that [L_{x,y,z},H] = 0 implies that angular
momentum is conserved.