Lecture 15 Solving the time dependent Schrdinger equation Lecture 15 Solving the time dependent...
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Transcript of Lecture 15 Solving the time dependent Schrdinger equation Lecture 15 Solving the time dependent...
Lecture 15Lecture 15
Solving the time dependent SchrSolving the time dependent Schrödinger dinger equationequation
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t
uiVuu
m
22
2
Poisson’s equation
t
uiVuu
m
22
2
2
2
22 1
t
u
cu
Introduction to PDEsIntroduction to PDEs
In many physical situations we encounter quantities which depend on two or more variables, for example the displacement of a string varies with space and time: y(x, t). Handing such functions mathematically involves partial differentiation and partial differential equations (PDEs).
t
u
hu
22 1
02 u
0
2
uAs (4) in regions
containing mass, charge, sources of heat, etc.
Electromagnetism, gravitation,
hydrodynamics, heat flow.
Laplace’s equation
Heat flow, chemical diffusion, etc.
Diffusion equation
Quantum mechanicsSchrödinger’s
equation
Elastic waves, sound waves, electromagnetic
waves, etc.Wave equation
x
Introduction to PDEsIntroduction to PDEs
A wave equation is an example of a partial differential equation
Think of a photo of waves (i.e. time is fixed)
If you make either x or t constant, then you return to the expected SHM case
2 2
2 2 2
( , ) 1 ( , )y x t y x t
x c t
You all know from last year that a possible solution to the above is
)cos( tkxAy
and that this can be demonstrated by substituting this into the wave equation
Or a movie of the sea in which you focus on a specific spot (i.e. x is fixed)
)cos( constkxAy
)cos( consttAy
y
ty
Step 4: Boundary conditions could then be applied to find A and B
Step 2: The auxiliary is then and so roots are
Step 1: Let the trial solution be So andmtx e mxmedt
dx mt xmemdt
xd mt 222
2
Unstable equilibrium
xxm 22 m
Step 3: General solution for real roots is m
)()( 2
2
2
txdt
txd
tt BeAetx )(
Introduction to PDEsIntroduction to PDEs
Thing to notice is that x(t) only tends towards x=0 in one direction of t, increasing exponentially in the other
tAetx )(
t
Step 1: Let the trial solution be So and
Introduction to PDEsIntroduction to PDEs
Harmonic oscillator
mtx e mxmedt
dx mt xmemdt
xd mt 222
2
Step 2: The auxiliary is then and so roots are xxm 22 im
Step 3: General solution for complex is
where = 0 and = so
)cossin( tDtCex t
tDtCx cossin
im
Thing to notice is that x(t) passes through the equilibrium position (x=0) more than once !!!!
Step 4: Boundary conditions could then be applied to find C and D
)()( 2
02
2
txdt
txd
tCx sin
Introduction to PDEsIntroduction to PDEs
Half-range Fourier sine series
1
sin)(n
n d
xnbxf
d
n dxd
xnxf
db
0sin)(
2
where
3)(x
xf 9
1
9)(
xxf
4
d
12
d
d
d
d
n dxd
xnx
ddx
d
xnx
db
4
4
0sin
99
12sin
3
2
A guitarist plucks a string of length d such that it is displaced from the equilibrium position as shown. This shape can then be represented by the half range sine (or cosine) series.
d
1
sin)(n
n d
xnbxf
2 2
2 2 2
( , ) 1 ( , )y x t y x t
x c t
Find the solution to the wave equation to predict the displacement of the guitar string at any later time t
The One-Dimensional Wave Equation
A guitarist plucks a string of length L such that it is displaced from the equilibrium position as shown at t = 0 and then released.
Let’s go thorugh the steps to solve the PDE for our specific case …..
2 2
2 2 2
( , ) 1 ( , )y x t y x t
x c t
)()(),( tTxXtxy
Ndt
tTd
tTcdx
xXd
xX
2
2
22
2 )(
)(
1)(
)(
1
SUMMARY of the procedure used to solve PDEs
9. The Fourier series can be used to find the particular solution at all times.
1. We have an equation with supplied boundary conditions
2. We look for a solution of the form
3. We find that the variables ‘separate’
4. We use the boundary conditions to deduce the polarity of N. e.g.
5. We use the boundary conditions further to find allowed values of k and hence X(x).
6. We find the corresponding solution of the equation for T(t).
7. We hence write down the special solutions.
8. By the principle of superposition, the general solution is the sum of all special solutions..
2kN
L
xnBxX nn
sin)( kxBkxAxX sincos)( so
kctDkctCtT sincos)(
nn L
ctnEtT cos)(
nnn L
ctn
L
xnBtxY
cossin),(
1
cossin),(n
nn L
ctn
L
xnBtxy
L
ct
L
x
L
ct
L
x
L
ct
L
x
L
ct
L
xdtxy
7cos
7sin
49
15cos
5sin
25
13cos
3sin9
1cossin
8),(
2
www.falstad.com/mathphysics.html
But any superposition such as also satisfies the
TDSE, and thus represents a possible state of the system.
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
L
x
Lx
sin
2)0,(1
L
x
Lx
2sin
2)0,(2
)0,()0,()0,( 2211 xCxCx
1),(2
dxtx
12
2
2
1 CC
The TDSE is a linear equation, so any superposition of solutions is also a solution. For example, consider two different energy eigenstates, with energies E1 and E2. Their complete normalised
wavefunctions at t = 0 are:
When we superpose, the resulting wavefunction is no longer normalised.
Recall that all wavefunctions must obey the normalization condition:
However it can be shown that the normalisation condition is fulfilled so long as:
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
t
txitxtxV
x
tx
m
),(
),(),(),(
2 2
22
t
txi
x
tx
m
),(),(
2 2
22
Consider the time dependent Schrödinger equation in 1 dimensional space:
Within a quantum well in a region of zero potential, V(x,t) = 0, this simplifies to:
Let’s solve the TDSE subject to boundary conditions (0, t) = (L, t) = 0 (as for the infinite potential well) For all real values of time tand for the condition that the particle exists in a superposition of eigenstates given below at t = 0 .
Question
L
x
L
x
L
x
Lx
3sin3
12sin3
1sin3
12)0,(
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
What does the wavefunction look like?
L
x
L
x
L
x
Lx
3sin3
12sin3
1sin3
12)0,(
-1.5
-1
-0.5
0
0.5
1
1.5
displacement from x = 0 to x = L
tota
l am
plit
ud
e
-1.5
-1
-0.5
0
0.5
1
1.5
displacement from x = 0 to x = L
tota
l am
plit
ud
e
-1.5
-1
-0.5
0
0.5
1
1.5
displacement from x = 0 to x = L
tota
l am
plit
ud
e
-1.5
-1
-0.5
0
0.5
1
1.5
displacement from x = 0 to x = L
tota
l am
plit
ud
e
n = 3
n = 2
n = 1
Superposition at t = 0
If we measure the energy of the state Ψ(x,t) described above we will measure either E1 or E2 or E3 each with the probability of 1/3.
These curves arent normalised – figs intended just to show shape
Step 1: Separation of the Variables
Our boundary conditions are true at special values of x, for all values of time, so we look for solutions of the form (x, t) = X(x)T(t). Substitute this into the Schrödinger equation:
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
t
txi
x
tx
m
),(),(
2 2
22
In a region of zero potential, V(x,t) = 0, so :
dt
tdTxXitT
dx
xXd
m
)()()(
)(
2 2
22
dt
dT
T
i
dx
Xd
Xm
2
22 1
2
Separating variables:
Step 2: Rearrange the equation
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
Step 3: Equate to a constant
Now we have separated the variables. The above equation can only be true for all x, t if both sides are equal to a constant. It is conventional (see PHY202!) to call the constant E.
Edx
Xd
Xm
2
22 1
2
X
mE
dx
Xd22
2 2
Edt
dT
T
i
T
iE
dt
dT
So we have:
which rearranges to (i)
which rearranges to (ii)
dt
dT
T
i
dx
Xd
Xm
2
22 1
2
where giving
For X(x)Our boundary conditions are (0, t) =(L, t) = 0, which means X(0) = X(L) = 0. So clearly we need E > 0, so that equation (i) has the form of the harmonic oscillator equation. It is simpler to write (i) as:
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
Step 4: Decide based on situation if E is positive or negative
We have ordinary differential equations for X(x) and T(t) which we can solve but the polarity of E affects the solution …..
Xkdx
Xd 22
2
22 2
mE
k kxBkxAxX sincos)(
XmE
dx
Xd22
2 2
If where then applying boundary conditions
gives
For X(x)
Our boundary conditions are (0, t) =(L, t) = 0, which means X(0) = X(L) = 0.
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
22 2
mE
k kxBkxAxX sincos)(
Step 5: Solve for the boundary conditions for X(x)
X(0) = 0 gives A = 0 ; we must have B ≠ 0 so X(L) = 0 requires ,
i.e. so L
xnBxX nn
sin)( for n = 1, 2, 3, ….
0sin kL
L
nkn
Step 6: Solve for the boundary conditions for T(t)
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
iEteTT 0Equation (ii) has solution as it’s only a 1st order ODE T
iE
dt
dT
A couple of slides back we decided that in order to have LHO style solutions for X(x) we must have E > 0. So here we must also take E > 0.
dtiE
T
dT
dt
iE
T
dT
ciEt
T
ln tiE
ceeT
Proof of statement above
tiE
eTT
0Replace the constant with T0
Step 7: Write down the special solution for (x, t)
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
tiEnnnn
neL
xnBtTxXtx
sin)()(),( 2
22222
22 mL
n
m
kE nn
where
(These are the energy eigenvalues of the system.)
Question asks for the solutions of the TDSE for real values of time
sincos ie i
tEnwhere
Real values are thereforetE
L
xnBtTxXtx nnnnn cossin)()(),(
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
Step 8: Constructing the general solution for (x, t)
We have special solutions:
The general solution of our equation is the sum of all special solutions:
11
cossin),(),(n
nn
nn
tE
L
xnBtxtx
(In general therefore a particle will be in a superposition of eigenstates.)
tE
L
xnBtTxXtx nnnnn cossin)()(),(
-1.5
-1
-0.5
0
0.5
1
1.5
displacement from x = 0 to x = L
tota
l a
mp
litu
de
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
tE
L
xtE
L
xtE
L
x
Ltx 321 cos
3sin3
1cos
2sin3
1cossin
3
12),(
2
22
2 2
4
mLE
2
22
3 2
9
mLE
If we know the state of the system at t = 0, we can find the state at any later time.
Since we said that
Then we can say:
and
Step 10: Finding the full solution for all times
L
x
L
x
L
x
Lx
3sin3
12sin3
1sin3
12)0,(
Superposition at t = 0
11
cossin),(),(n
nn
nn
tE
L
xnBtxtx
where2
22
1 2mLE
The general solution is
11
sin)0,()0,(n
nn
n L
xnBxx
At t = 0 the general solution is
Particular solution to the time dependent Particular solution to the time dependent SchrSchrödinger equationdinger equation
2
22
1 2mLE
2
22
3 2
9
mLE
1st Eigenvalue 3rd Eigenvalue
1st Eigenfunction
3rd Eigenfunction
2
22
2 2
4
mLE
2nd Eigenvalue
tE
L
xtE
L
xtE
L
x
Ltx 321 cos
3sin3
1cos
2sin3
1cossin
3
12),(
-1.5
-1
-0.5
0
0.5
1
1.5
displacement from x = 0 to x = L
tota
l a
mp
litu
de
Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation
tE
L
xtE
L
xtE
L
x
Ltx 321 cos
3sin3
1cos
2sin3
1cossin
3
12),(
2
22
2 2
4
mLE
2
22
3 2
9
mLE
and Superposition at t = 0
where2
22
1 2mLE
In this particular example Ψ(x,t) is composed of eigenstates with different parity (even and odd). Therefore Ψ(x,t) does not have a definite parity and P(x,t) oscillates from side to side.
www.falstad.com/mathphysics.html
2. Mixed states (namely, superpositions of energy eigenstates) do not have a definite energy but have a probability of being in any one of the energy states when measured.
3. The probability densities of mixed states vary with time as do therefore the < x >.
1. Energy eigenstates (namely, states with definite energy) are stationary states: they have constant probability densities and definite energies.
The expectation value is interpreted as the average value of x that we would expect to obtain from a large number of measurements.
NB. In order to have time dependence in any observable such as position, it is necessary for the wavefunction to contain a superposition of states with different energies.
For example if we have the single eigenfunction within
an infinite potential well then
and . Notice how there is no time dependence.
tiEeL
xBtx 2
2sin),( 22
tiEeL
xBtx 2
2sin),( 2
*2
L
xxBtxxtx
2sin),(),( 22
22*2
Just for Quantum Mechanics course
This is because the probability density for a mixed state varies with time, whereas for a pure state it is constant in time. Pure states are known as stationary states.
This means < x > is invariant with timewww.falstad.com/mathphysics.html
But if we add another eigenfunction for example:
The complex conjugate is written as:
A superposition of eigenstates having different energies is required in order to have a time dependence in the probability density and therefore in < x >.
Just for Quantum Mechanics course
),(),(),( 21 txtxtx
),(),(),( *2
*1
* txtxtx
),(),(),(),(),(),( 21*2
*1
* txtxtxtxtxtx
tiEtiE eL
xBe
L
xBtxtxtx 21
2sinsin),(),(),( 2121
tiEtiE eL
xBe
L
xBtxtxtx 21
2sinsin),(),(),( 21
*2
*1
*
tiEtiEtiEtiE e
L
xBe
L
xBe
L
xBe
L
xBtxtx 2121
2sinsin
2sinsin),(),( 2121
*
tiEtiEtiEtiE eeee
L
xB
L
xB
L
xB
L
xBtxtx 2121
2sinsin
2sinsin),(),( 21
222
221
*
Therefore:
Let eigenfunctions be:
So:
tEEitEEi ee
L
xB
L
xB
L
xB
L
xBtxtx )()(
2122
222
1* 2121
2sinsin
2sinsin),(),(
Non zero so long as E1≠ E2