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Time Dependent Schrödinger Equation
Up until now we have been talking about “stationary states” which don’t change withtime, but recall that the state function
Ψ^ is also a function of time,
Ψ(q^ ,... qn^1
, t) in
m ,^
f^
f^
f^ m ,
(q^ ,^1
qn,^ )^
general. When will we care about time dependence?
Most of the time!
For
examples,spectroscopy
⇒^ molecule interacting with electromagnetic radiation and implies transitions between statesand implies transitions between states.chemical reactions imply transitions between at least two states.The time evolution of a state function is given by
Postulate V
, which says that the
g^
y^
y
state function
Ψ^ satisfies the simple appearing differential equation:
(x, t)
ˆ^ (x, t) = i
∂Ψ t
Ψ^
H
and^
is the^ Hamiltonian operator. So we must solve this equation. We will do so by means of
Separation of Variables.
This is a very important
technique, one that we shall use again and again. Essentially no problem is solvable
t∂
ˆH^ q^
g^
g^
y^ p
unless we can separate variables. Note carefully the logic in what follows:First,^ assume that
Ψ (x, t) is a the product form
Ψ(x, t) =
ϕ(x) f(t)
We then try to solve the Schrödinger equation with such a separable form
If we
We then try to solve the Schrödinger equation with such a separable form.
If we
succeed in finding a general solution with this restrictive assumption, then wehave justified our separability assumption, and all is well.
(x) f(t)
f(t)^
f(t)
ˆ^
d
ϕ∂
∂
So, making this assumption
,^ the Schrödinger equation becomes^ (x) f(t)
f(t)^
f(t)
(^ (x) f(t)) = i
=^ (x) i
(x) i
t^
t
d^ dt
ϕ ϕ
ϕ
ϕ
∂^
∂^
=
∂^
∂
=^
=^
=
H So, with this assumption, the right side of the above equation simplifies asabove.,Now, if in addition,
contains no explicit time dependence, then f(t) is also a t^ t^ ith
t t^ th
ti^
i di^ t d b
d f(t)^
b^ b^
ht
ˆH^
constant with respect to the actions indicated by
, and f(t) can be brought
through the
operator. (What does it mean for
to have no time dependence?)
In any event, we obtain
H
ˆH^
ˆH
y^
, ˆ
f(t)
f(t)^
(x)^ =
(x) i
t ϕ^
ϕ^
∂ = ∂
H^
t ∂
an equation which we need to solve.
ˆ^ (x) = E^
(a)
ϕ H (x) ϕ
andi df(t) E =^
(b) = f(t)^ dt f(t)^ dt
df(t)^
iE Look at (b)^ df(t)dt^
iE = −f(t)^ = Integrate this equation to obtain
f^ t^
i Et e ( )^ =
−^ =
for the time dependence of
Ψ(x,t). (^ )
The spatial dependence comes from solving equation (a), rewritten as
ˆ^
ˆ^
(x) = E
(x)
ϕ^
ϕ
H
a relation that we recognize as the total energy eigenvalue equation. This is exactlythe equation that we solved earlier to obtain the energy states of the particle in abox!^ This total energy eigenvalue equation is best known as thebox!^ This total energy eigenvalue equation is best known as the
Time Independent Schrödinger Equation The existence of a
product form
solution enabled the one differential p
equation in two variables to be written as two separate differential equations, eachhaving only one variable. This is useful because we can kill any differential equationin one variable with only computing and no thinking.
This concept is very important and will occur again and again.
After assuming a product form solution^ ψ(x,t) =
ψ(x) (t), the TDSE becomes ψ(x,t)
ψ(x)
(t), the TDSE becomes^2
2
1
1
i^
V^
E
∂ϕ
∂ ψ −=
+^
=
=
= If the potential energy function V in the Schrödinger
2
2
i^
V^
E
t^
m^
x
ϕ ∂
ψ ∂ =^
+^
=
= If the potential energy function V in the SchrödingerEquation is a function of time, as well as x, [V = V(x,t)]would separation of variables still work; that is, would
p^
;^
,
there still be solutions to the SE of the form
Ψ(x,t) =
ψ(x)
(t)?
A) Yes, always
B) No, never
C) Depends on the functional dependence of VC) Depends on the functional dependence of Von x and t
TIME DEPENDENCE of
Now we have seen how to separate variables and obtain asolution to the above equation when the state function couldbe written as a product
Ψ(q t) =
ψ(q)f(t). We saw that
be written as a product,
Ψ(q,t)
ψ(q)f(t). We saw that
Ψ(q,t) was
NOT
a function of time when
ψ(q) was an
eigenfunction of
ˆ H
eigenfunction
of^
.
We demonstrate a more general solution, but one still
H m^
m^
g^
,
restricted to the case that does not contain explicit timedependence
This restriction allows us to view the propagation dependence.
This restriction allows us to view the propagation of a^ wave packet
and make additional connection with our
i^ i i
d^
di^
W^ d
l^ i^
f
intuitive understanding
.^ We demonstrate a solution for a two
term form, but the general result will be obvious.
1
2
t^
t
-iE^
-iE
ˆ^
( , )
exp^
exp
h^
h
1
2
s^
1
2
x t^
(x)^
+^
(x)
E^
E
ψ
ψ
Ψ^
= H^
h^
h
and the right side as the following:
s^
1
1
2
2
1
2
(x, t)^
t^
t
iE^
iE^
iE^
iE
i^
= i^
(x) exp
+ i^
(x) exp
ψ
ψ
∂^
−^
−^
⎡^
⎤^
⎡^
Ψ^
⎢^
⎥^
⎢^
∂^
⎣^
⎦^
⎣^
=^
=^
=^
=^
=^
1
2
( )^
p^
( )^
p
t
ψ
ψ
⎢^
⎥^
⎢^
∂^
⎣^
⎦^
⎣^
=^
=^
=^
=^ E
(x)
iE t
(x)^
iE t
ψ^
ψ
exp^
exp
−^
−
=^
=
( )^
( )
ψ^
ψ p^
p
=^
=
Since the two sides are identical, this sum of products formis a solution to the time dependent Schrödinger equation
whenever
has no time dependence. Note that now
Ψ(x, t) hass
time dependence.
ˆ H
(^
)^
(^
)
1 2
1 2
i E^ E^
t^
i E^ E^
t
−^
(^ )^
(^ )^
(^
)^
(^
)
(^
) 1 2
1 2
*^
i E^ E^
t^
i E^ E^
t
s^ s^
x^
x^
e^
e E^
E^ t
ψ^
ψ^
ψ ψ
ψ ψ −
Ψ Ψ
=^
+^
+^
⎛^
⎞ =^
(^ )^
(^ )^
(^
) 1
cos 1 2
E^
E^ t
x^
x ψ^
ψ^
ψ ψ^
− ⎛^
⎞
=^
+^
+^
⎜^
⎟ ⎝^
⎠
Use Mathcad to see some other examples of acceptable state functions for theparticle in a box. Let the box extend from 0 to
π.^ Consider a state function
which is the sum of the first 100 eigenfunctions.Set ranges for x and nSet ranges for x and n.n=1, 2,... 100
0 < x <
π
Define the eigenfunctions.
f(^ Are the f(x n) orthogonal? Normalized? Check one numerically! ),x n 2 .sin(^ π^
.)n x Are the f(
x,n) orthogonal? Normalized? Check one numerically!
=d
x
2 f( ),x 4
YES!Now define our 100 term function, g(x).
0 Is it normalized?
g(^ )x^
100 .0.1 = 1n
f(^ ),x n = 1n
and what is the probability density?
(^2022) g( )x^0
0.^
0
x
What is the average kinetic energy?
How does it propagate in time?
How does this relate to the so-called
δ^ function??
Also, if
we set
E=1, then E^1
(^2) = n, and we easily findn
〈E〉^ as
,^1
,^ n
y^
〈^ 〉
〈E〉^ =
〈E〉 How would this state function evolve in time?
Lets look at an example.
Consider the evolution of a localized particle on the left side of a box.First we construct a state function localized on the left sideFirst, we construct a state function localized on the left sideof the box. We could put a Gaussian or some such functionon the left side of the box, with whatever width we choose.`A simpler way to get a feeling for this is to employ the sumA simpler way to get a feeling for this is to employ the sumof the first 10 particle in a box eigenfunctions as our initial state.The box extends from x = 0 to x =1.
We construct the time dependent
state function with the first 10
eigenfunctions having equal weight.
f^
f^
g^ f^
g^ q^
g
Thus
Ψs
iE t
x t^
j^ x e
j
sin
=^ ∑
πb g^
j=^1
The individual components are all in phase at t = 0.This will give a peak near the left side of the box at t=0.This will give a peak near the left side of the box at t 0.What do the individual components look like?We then propagate the function in time, using the expression
p^ p g
g^
p
above.
The box extends from 0 to 1W^
t^
t th^
ti^
d^
d^ t t t
^ We construct the time dependent statefunction with all of the first 10eigenfunctions having equal weighteigenfunctions having equal weight ^ They are all in phase at t = 0. This will givea peak near the left side of the box at t=0a peak near the left side of the box at t=0. ^ What do the individual components looklike?like? ^ We then propagate the function in time,using the expression from notes.using the expression from notes.
Individual components of
Ψ
t = 0.
0.120.12^ 0.1 0.08 .0.06 0. f(^ ),x^ t^ f(^ ),x^ t
0 0.
0.^
0.6^ 0.
1
0.04 0.02 00
1
0
x
