Quantum Mechanics - Homework Assignment 3 Alejandro G´ omez omez Espinosa ∗ October 8, 2012
1) Consider two point charges of identical mass m confined to a 2D plane with an external potential 12 γ (xi2 + yi2 ) affecting both particles ( i = 1, 2), and a radial (e.g., Coulomb) interaction k/ interaction k/ r1 r2 between the particles.
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a) Choosing x1 , y1 , x2 , y2 coordinates, write the Lagrangian and obtain the Lagrange equations of motion. In this coordinates, the kinetic energy for the two particles is: 1 T = m x˙ 21 + y˙ 12 + x˙ 22 + y˙ 22 2
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For the potential energy, we have to define r1 r2 = (x1 x2 )2 + (y (y1 + y2 )2 . Then: 1 k V = γ (x21 + y12 + x22 + y22 ) + 2 (x1 x2 )2 + (y (y1 + y2 )2
| − |
|
Thus, the Lagrangian: 1 L = T V = m x˙ 21 + y˙ 12 + x˙ 22 + y˙ 22 2
−
−
−
|
−
|
1 γ (x21 +y12 +x22 +y22 ) 2
−|
(x1
−
k x2 )2 + (y (y1 + y2 )2
To solve Euler-Lagrangian equations, let’s solve only for x1 and then we can easily have the other terms: ∂L ∂x 1
d dt
− d dt ∂L ∂ ˙ ∂ x˙ 1
∂L ∂ ˙ ∂ x˙ 1 ∂L ∂x 1
k (x1
=
−γx1 −
=
d (mx˙ 1 ) = mx ¨1 dt
((x ((x1
− x2)
(y1 + y2 )2 )3/2 − x2)2 + (y
= 0 = mx ¨1 + γx 1 +
k(x1 ((x ((x1
− x2)
(y1 + y2 )2 )3/2 − x2)2 + (y
Finally our equations of motion are: mx ¨1 + γx 1 +
k(x1
− x2)
=0
(y1 + y2 )2 )3/2 − x2)2 + (y k(y1 − y2 ) my¨1 + γy 1 + =0 ((x ((x1 − x2 )2 + (y (y1 + y2 )2 )3/2
∗
((x ((x1
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1
|
mx ¨2 + γx 2
−
my¨2 + γy 2
−
k(x1 ((x ((x1 ((x ((x1
−
− x2)
=0
− y2)
=0
x2 )2 + (y (y1 + y2 )2 )3/2 k(y1
(y1 + y2 )2 )3/2 − x2)2 + (y
b) Choose center-of-mass coordinates X,Y,x,y instead and repeat. To transform the previous coordinates in center-of-mass coordinates, we choose: x = x1
− x2, |r − r | = 1
X =
2
y = y1
− y2
x22 + y22
mx1 + mx2 (x1 + x2 ) = , m+m 2
Y =
(y1 + y2 ) 2
Then, the kinetic and potential energy, replacing in the results of (a): ˙ 2 + Y ˙ 2 ) + 1 m(x˙ 2 + y˙ 2 ) T = m(X 4 V =
k
1 + γ (X 2 + Y 2 + x2 + y 2 ) x2 + y 2 2
We have the Lagrangian from L = T
d dt
∂L ∂x ∂L ∂ ˙ ∂ x˙
V . Then, solving for x: − V .
=
− (x2 +kxy2)3/2 − γx
=
1 mx ¨ 2
where the procedure is similar for y . Thus, solve for X :
d dt
∂L = γX ∂X ∂L ¨ = 2mX ˙ ∂ X
where the procedure is the same for Y . Y . Finally our equations of motions are: 1 kx mx ¨+ 2 + γx = 0 2 (x + y2 )3/2 1 ky my¨ + 2 + γy = 0 2 (x + y2 )3/2 ¨ 2mX
− γX = 0 ¨ − γY = 0 2mY
2
c) Transform further to radial coordinates for the relative coordinate ( x, x, y r, φ) and the center-of-mass coordinate ( X, X, Y R, Θ) and repeat. Replacing x = r cos φ, y = r sin φ, X = R cosΘ and Y = R sin Θ, we have have the potential potential and kinetic kinetic energy: energy: ˙ 2 ) + m (r˙ 2 + r2 φ˙ 2 ) T = m(R˙ 2 + R2 Θ 4 k 1 V = + γ (R2 + r 2 ) r 2 Solving Solving the Euler-Lagr Euler-Lagrange ange equations for r :
→
→
d dt
∂L ∂r ∂L ∂ ˙ ∂ r˙
=
− rk2 − γr + m2 rφ˙ 2
=
m r¨ 2
Next for R:
d dt
∂L ∂R ∂L ∂ R˙
˙2 2mRΘ −γR + 2mR
=
¨ = 2mR
Then for φ: ∂L ∂φ d dt
∂L ∂ φ˙
= 0 d dt
=
mr2 φ˙ 2
where is similar for Θ. Then the equations of motion are: m k m ˙2 r¨ + 2 + γr rφ = 0 2 2 r ¨ + γR 2mRΘ ˙2 =0 2mR
−
mr 2 φ˙ = a, ˙ = A, mR2 Θ
−
with a constant with A constant
d) Write Write Hamilton Hamilton’s ’s equatio equations ns of motion motion for case (c). Comment Comment on cyclic cyclic coor coordidinates and describe the behavior of the system in general terms. From (c), we have that the coordinates φ and Θ are cyclic coordinates because ∂L momentum in this coordinates is conserved conserved.. Thus, Thus, ∂q = 0. It means that the momentum our previous Lagrange is: i
A2 L = m R˙ 2 + 2 2 m R
m 2 a2 + r˙ + 2 2 4 m r
−
k r
− 12 γ (R2 + r2)
The Hamiltonian of our system would be build from: H =
pi q ˙i
i
−L 3
where pi =
∂L ∂ ˙ ∂ q ˙i
Calculating the momentum: m r˙ 2
⇒
pR = 2mR˙
⇒
pr =
2 pr m pR R˙ = 2m
r˙ =
Then, the Hamiltonian has the form: H = pr
2 pR m 4m2
− − 2 pr pR + pR m 2m
pr2 p2 A2 + R m 4m mR2 To solve the equations of motion: H =
−
4 pr2 m2
−
a2 k 1 + + γ (R2 + r 2 ) 2 4mr r 2
2
a k 1 − 4mr + + γ (R2 + r 2 ) 2 r 2
−
−
−
−
R˙ = p˙R =
−
m 4
∂H 2 pr = ∂p r m ∂H a2 k = + γr 2mr3 r2 ∂r ∂H pR = ∂p R 2m ∂H 2a2 = + γR ∂R mr 3
r˙ = p˙ r =
A2 mR2
−
2) a) Shankar Shankar Ex. 2.7.1 2.7.1 Show that (i)
{w, λ} = −{λ, w} {w, λ}
=
−
∂w ∂λ ∂q i ∂p i
i
=
i
=
−
∂λ ∂w ∂q i ∂p i
∂w ∂λ ∂p i ∂q i
−
∂λ ∂w ∂p i ∂q i
−{λ, w}
(ii)
{w, λ + σ} = {w, λ} + {w, σ} {w, λ + σ}
=
i
=
i
= =
∂w ∂ (λ + σ ) ∂q i ∂p i
−
∂w ∂ (λ + σ ) ∂p i ∂q i
∂w ∂λ ∂w ∂σ + ∂q i ∂p i ∂q i ∂p i
−
∂w ∂λ ∂q i ∂p i
− i {w, λ} + {w, σ} 4
∂w ∂λ ∂p i ∂q i
∂w ∂λ ∂p i ∂q i
+
i
−
∂w ∂σ ∂p i ∂q i
∂w ∂σ ∂q i ∂p i
−
∂w ∂σ ∂p i ∂q i
(iii)
{w,λσ} = {w, λ}σ + λ{w, σ} {w,λσ}
= =
i
∂w ∂ (λσ) λσ) ∂q i ∂p i
i
∂w ∂q i
∂σ ∂λ λ + σ ∂p i ∂p i
∂w ∂σ ∂q i ∂p i
= λ =
−
∂w ∂ (λσ) λσ ) ∂p i ∂q i
− ∂w ∂p i
∂w ∂σ ∂p i ∂q i
− i λ{w, σ} + {w, λ}σ
∂σ ∂λ λ + σ ∂p i ∂p i
+
i
∂w ∂λ ∂q i ∂p i
−
∂w ∂λ ∂p i ∂q i
σ
b) For a 2D system with H = px2 + py2 + a(x2 + y2 ), show that xpy yp x is a constant of the motion by explicitly computing its Poisson bracket with H . Just for simplicity, let’s call Q = xpy ypx . Then, the Poisson bracket is:
−
−
{Q, H }
∂Q ∂H ∂Q ∂H ∂Q ∂H ∂Q ∂H + ∂x ∂p x ∂p x ∂x ∂y ∂p y ∂p y ∂y = py (2 p (2 px ) + y (2ax (2ax)) px (2 p (2 py ) x(2ay (2ay)) = 0 =
−
−
−
As Q, H = 0, therefore xpy
{
}
−
− ypx is a constant of the motion.
3) The classical Hamiltonian for a particle in an electromagnetic field is given by Shankar Eq. (2.6.2): 1 q A(r, t))2 + qφ( = (p qφ(r, t) 2m c Applying Hamilton’s equations of motions, obtain expressions for r˙ and p˙ .
H
p˙ = r˙ =
−
− − − −
∂H q ∂ A = p ∂ r 2c ∂ r ∂H 1 q = p A ∂ p m c
−
q ∂φ A + q c ∂ r
=
q ∂ A p 2c ∂ r
q A c
− −
∂φ q ∂ r
4) Consider two operators whose matrix representations in a 3D vector space are A=
−
1 0 0
0 0 1 0 0 2
−
,
B=
1 0 0 0 0 3 0 3 0
1 1 a) The system is initially in state ψ = √ 1 + √ 2 . If A were were to be measur measureed, 2 2 what would be the possibl possiblee al lowed lowed values(s) values(s) of a and their probabili probabilities?. ties?. If B were to be measured, what would be the possible allowed value(s) of b and their probabilities? Let’s start with the operator A. If A were to be measured, we have a probability to found one of the eigenv eigenvalues of this operator. operator. Therefore, Therefore, the eigenvalues eigenvalues of this matrix are:
det(A det(A
|
− aI ) =
− −
|
1 a 0 0
5
|
−
0 0 1 a 0 =0 0 2 a
− −
(1
− a)2(2 − a) = 0
a = 2, 1
⇒
−
With this eigenvalues, we calculate the eigenvectors using A a = a a . Then Then,, 1 1 the normalized eigenvectors are: a(2) = 3 and a(−1) = √ 2 1 + √ 2 2 . Now, the probability of each eigenvalue is:
| |
a(−1)|ψ =
a(2)|ψ =
1 √
1 √ 2 1 √
1 √
2
|
0
2
| |
2
| |
=1
0
1 √ 2 1 √
0 0 1
=0
2
0
|a(−1)|ψ|2 = ψ|a(−1)a(−1) |ψ = 1 |a(−1)|ψ|2 = 0
P ( P (a(−1) ) = P ( P (a(2) ) =
That makes sense because the system is already in such state with a = For the case of the operator B, let’s repeat the same procedure:
− 1
det(B det(B
− bI ) =
b
−
0 b 3
0 0
−1.
0 3 =0 b
−
− b)b2 − 9(1 − b) = b3 − b2 − 9b − 9 = 0 ⇒ b = 1, 3, −3 1 where the eigenvectors are: |b(1) = |1, |b(3) = √ (|2 + |3) and |b(−3) = 2 1 √ (−|2 + |3). 2 (1
P ( P (b(1) ) = P ( P (b(3) ) = P ( P (b(−3) ) =
2
|b(1)|ψ|
2
|b(3)|ψ|
= ψ b(1) b(1) ψ =
| |
=
|b(−3)|ψ|2 =
1 2
1 2
1 2
=
1 2
1 4
=
√ √ 1 2
1 2
=
1 2
1 4
b) Compute the expectation value ψ B ψ for the same state ψ in two ways: (i) from the probabilities probabilities computed computed in part (a); and (ii) directly directly from the matrixvector product product ψ B ψ .
| |
|
| |
(i)
A =
i
P ( P (ai )ai = P ( P (a(−1) )a(−1) + P ( P (a(2) )a(2) = 0 + 1( 1) =
−
B =
i
1 1 1 1 P ( P (bi )bi = (1) + (3) + ( 3) = 2 4 4 2
−
6
−1
(ii)
ψ|A|ψ =
ψ|B|ψ =
1 √
1 √
2
1 √
2
1 0 0
0
2
1 √
2
1 √ 2 1 √
− − 0 0 1 0 0 2
1 0 0 0 0 3 0 3 0
0
2
1 = √ 2
1 √
2
0
1 √ 2 1 √ 2
1 = √ 2
1 √ 2 1 √
− − 0
2
=
0
1 √
2
0
0
1 √
2
0 3 √
=
1 2
2
c) As a result of the measurement of B the system is found to have b = 3. What is the new state state of the system? system?.. In this state, state, what are the allowe allowed value( value(s) s) of A and their probabilities? If we found the measurement of B is equals to 3, the new state of the system 1 will be: ψ = √ ( 2 + 3 ). In this state the allowed values of A are the same: 2 a = 1, 2. Then, the probabilities of this values are:
−
|
| |
1 1 P ( P (a(−1) ) = ψ P(a(−1) ) ψ = ( 2 + 3 )( 1 1 + 2 2 )( 2 + 3 ) = 2 2
|
|
| | | | | | | |
1 1 P ( P (a(2) ) = ψ P(a(2) ) ψ = ( 2 + 3 )( 3 3 )( 2 + 3 ) = 2 2
|
|
| | | | | | d) In the context of part (c), A is found to have value a value a = −1. What is now the new state of the system? We have to calculate the projection of this value in the new state:
| | √ | |
P(a(−1) )
|ψ = (|11| + 2
2)
1 1 (2 + 3 ) = 2 2 2
√ |
e) In general, if A is measured and found to have value a = 1 when the system is in an unknown initial state, is the final state of the system known after the measurement?. No, because if we only know the value of a, the only thing that we can guess is the projection operator. operator. As we don’t know the initial state we cannot determine determine the final state.
−
f ) Find a state of the system that is a definite state of both observables A and B. From the matrix representation of the operators, the only definite state of both 1 observables is: 1 = 0 0
|
7
−1
