Physics 200A Homework 2.2
Mark Derdzinski
October 15, 2013
Problem 2 Solution
Part (a)
We begin with the Lagrangian density:
dΨ
~2 dΨ∗
− Ψ∗ (U − E)Ψ .
L=−
2m dx
dx
(1)
The Euler-Lagrange Equation corresponding to Ψ∗ is given by
d ∂L
∂L
=0
∗ −
dΨ
dx ∂ dx
∂Ψ∗
~2 dΨ
d
−
− (−Ψ(U − E)) = 0
=⇒
dx
2m dx
~2 d2 Ψ
=⇒ −
+ U Ψ = EΨ
2m dx2
(2)
Thus we recover the SE for Ψ. Part (b)
Consider the probability P = Ψ∗ Ψ. It is invariant under the following symmetry:
ıθ
Ψ→e~Ψ
Ψ∗ → e
−ıθ
~
Ψ∗
Such a complex phase rotation preserves the Lagrangian, so we can apply Noether’s Theorem. In
the infitesimal limit where θ → δθ,
ı
Ψ → Ψ + Ψ(δθ) = Ψ + (δθ)QΨ
~
ı
∗
∗
Ψ → Ψ − Ψ∗ (δθ) = Ψ∗ + (δθ)QΨ∗
~
Where we identify the generators of our coordinates,
1
ı
Ψ
~
ı
= − Ψ∗
~
QΨ =
QΨ∗
By Noether’s Theorem, the quantity
=−
∂L
∂ dΨ
dx
!
QΨ −
∂L
∗
∂ dΨ
dx
!
QΨ∗
ı~ dΨ∗
ı~ dΨ ∗
Ψ−
Ψ
2m dx
2m dx dΨ dΨ∗
~
Ψ∗
=
+
Ψ
2mı
dx
dx
=
which we identify as the probability current, is conserved (i.e.
We now return to the time derivative of probability:
(3)
∂
∂x
= 0).
∂P
∂(Ψ∗ Ψ)
=
∂t
∂t
∂Ψ∗
∂Ψ
=
Ψ + Ψ∗
∂t
∂t
∂
Using the explicit expression for E = ı~ ∂t
in the SE, we can rearrange Eqn. 2 to solve for
dΨ
1
~2 d2 Ψ
=
−
+ UΨ
dt
ı~
2m dx2
(4)
∂Ψ
∂t :
(5)
If U ∈ <, plugging this and the conjugate equation into Eqn. 4 yields the continuity equation:
1 ~2 d2 Ψ∗
~2 d2 Ψ ∗
∂P
∗
∗
=
Ψ − UΨ Ψ −
Ψ + UΨ Ψ
∂t
ı~ 2m dx2
2m dx2
2 ∗
2
~ d Ψ
∗d Ψ
=
Ψ
−
Ψ
2mı dx2
dx2
∗
~ d dΨ
d
∗ dΨ
=
Ψ−Ψ
=−
2mı dx dx
dx
dx
Thus the time derivative of probability is equal to the divergence of the probability current,
which, by Noether’s theorem, is zero. 2
(6)
5
6
7
8
9
5
Consider a system with Lagrangian in Cartesian coordinates which is given
by:
1X
L=
ma ẋ2a + ẏa2 + ża2 − U
2 a
Now, if this is more conveniently expressed in terms of generalized coordinates (q1 , q2 , . . . , qs ), use the transformations:
x = f (q1 , q2 , . . . , qs )
X ∂f
q̇c
ẋ =
∂qc
c
(a) What is the most general form of L in terms of the q’s?
T is given in general by
!2
X ∂fa
1X
ḃc ,
ma
T =
2 a
∂qb
(1)
b
where the sums a and b are over the number of particles and generalized
coordinates respectively. U (x) simply becomes U (f ).
(b) Specify the form of the kinetic energy as fully as possible.
If f does not depend on time explicitly, the expression for T in generalized
coordinates will have the form
1X
T =
Mjk q̇j q̇k .
(2)
2
j,k
A comparison between (1) and (2) shows that
Mjk =
X
ma
a
∂fa ∂fa
·
.
∂qj ∂qk
Note also that Mjk is symmetric in its indicies.
(c) Derive the Lagrangian EOMs
X ∂fa ∂U
∂L
=−
·
≡ Qm
∂qm
∂qm ∂fa
a
∂L
= Mjm q̇j
∂ q̇m
dMjm
d ∂L
= Mjm q̈j +
q̇j
dt ∂ q̇m
dt
10
(3)
where dMjm /dt is can be obtained using the chain rule. The EOM is then
Mjm q̈j +
dMjm
q̇j = Qm
dt
11
(4)