Download Quantum Mechanics Lecture 4 Dr. Mauro Ferreira

Quantum Mechanics
Lecture 4
Dr. Mauro Ferreira
E-mail: [email protected]
Room 2.49, Lloyd Institute
Postulates of Quantum Mechanics
P3:
The time evolution of the state (wave) functions follows the
Schroedinger equation:
∂
i! Ψ("r, t) = ĤΨ("r, t)
∂t
where H is the Hamiltonian operator and corresponds to
the total energy of the system
P̂ 2
Ĥ =
+ V̂ (!r, t)
2m
!2 2
=−
∇ + V̂ (!r, t)
2m
The solution of the fully deterministic differential equation above
provides ψ(r,t) for all values of r and t.
Solution of partial differential equations (PDE) by the method
of separation of variables
∂2Φ ∂2Φ
+
=0
2
2
∂x
∂y
( Laplace equation )
Φ(x, y) = X(x)Y (y) ( attempted solution )
1 d2 X
1 d2 Y
+
=0
2
2
X dx
Y dy
1 d2 X
1 d2 Y
=−
= const
2
2
X dx
Y dy
The only way these two terms can satisfy the
equation is if each one of them is a constant
1 d2 X
= const
2
X dx
1 d2 Y
= −const
2
Y dy
The two-variable P.D.E. was transformed into two O.D.E.
∂
!2 2
i! Ψ("r, t) = −
∇ Ψ("r, t) + V ("r, t)Ψ("r, t)
∂t
2m
How to solve the Schroedinger Equation:
( V (!r, t) = V (!r) )
∂
!2 2
i! Ψ("r, t) = −
∇ Ψ("r, t) + V ("r)Ψ("r, t)
∂t
2m
Typically solved by the method of separation of variables
Ψ(!r, t) = ψ(!r) φ(t)
(attempted solution)
2 simple ODEs
{
! r) ψ(!r) = E ψ(!r)
H(!
dφ(t)
i!
= E φ(t)
dt
Eigenvalue equation
(Time-independent Schroedinger Eq.)
Note that the separation
constant E is the energy
{
! r) ψ(!r) = E ψ(!r)
H(!
− !i Et
dφ(t)
i!
= E φ(t)
dt
φ(t) = e
Time-dependence of the wave function (conservative potential)
− !i Et
Ψ(!r, t) = ψ(!r) e
Suppose the system is in an eigenstate Ei of the Hamiltonian.
! r) ψE (!r) = Ei ψE (!r)
In other words, H(!
i
i
− !i Ei t
Ψ(!r, t) = ψEi (!r) e
⇒ |Ψ(!r, t)|2 = |ψ(!r)|2
• Probability distribution is independent of time.
• Eigenstate of the Hamiltonian is called a stationary state
Being a linear differential equation, the Schroedinger Eq. must obey
the superposition principle, i.e., a linear combination of solutions is
also a solution.
Consider a state initially described by
Ψ(!r, t = 0) = am ψEm (!r) + an ψEn (!r)
A short time later, the state will have evolved into
− !i Em t
Ψ(!r, t) = am ψEm (!r) e
− !i En t
+ an ψEn (!r) e
Regarding the probability density function
|Ψ(!r, t)|2 = |am ψEm (!r)|2 + |an ψEn (!r)|2
+ a∗m
− !i (En −Em )t
∗
an ψEm ("r) ψEn ("r) e
+ am a∗n
i
∗
!
ψEm ("r) ψEn ("r) e (En −Em )t
It is no longer stationary
Another way of seeing the stationary character of the solutions is by
calculating the uncertainty of the energy E, which is a solution of the
eigenvalue equation Ĥ ψ(x) = E ψ(x)
! ∞
! ∞
!H" =
dx ψ ∗ (x) Ĥ ψ(x) = E
dx ψ ∗ (x) ψ(x) = E
−∞
−∞
Bearing in mind that
Ĥ 2 ψ(x) = Ĥ ( Ĥ ψ(x) ) = Ĥ ( E ψ(x) ) = E 2 ψ(x)
!H 2 " =
!
∞
dx ψ ∗ (x) Ĥ 2 ψ(x) = E 2
−∞
∆H =
!
!H 2 " − !H"2 = 0
If the uncertainty on the energy vanishes, the characteristic
time interval diverges ( ∆E ∆t ≥ !/2 )