Quantum Mechanics - Homework Assignment 4 Alejandro G´ omez omez Espinosa ∗ October 10, 2012
1) Consider the following operators on a Hilbert space V3 (C )
√
√
0 1 0 1 Lx = 1 0 1 , 2 0 1 0
0 1 Ly = i 2 0
−i 0 i
−
0 i , 0
1 0 Lz = 0 0 0 0
− 0 0 1
a) Obtain eigenvalues and normalized eigenvectors for all the three operators Lx , Ly and Lz . To calculate the eigenvalues we use det(L det(Li λI ) = 0:
−
−
√ 1
λ
0 = det(L det(Lx
√ 1
− λI ) = det
2
0
2
−λ √ 1
2
Then, the eigenvalues are: λ = 0
√ 1 2 0
− − 0
√ 1 = λ3 + λ + λ = λ3 − λ = λ(λ2 − 1) 2 2
λ
2
−1, 0, 1. For the eigenvectors:
√ 1
0
2
√ 1 2
0
√ 1 2
0
x y z
=λ
x y z
Solving this problem for each eigenvalue, we found: λ=
−1 ⇒ √ y2 = −x, √ x2 + √ z2 = −y, √ y2 = −z
λ=0
⇒ √ y2 = 0, √ x2 + √ z2 = 0, √ y2 = 0
⇒ √ y2 = x, √ x2 + √ z2 = y, √ y2 = z Taking the basis 1| = (100), 2| = (010) and 3| = (001)., (001)., the normalize normalized d λ=1
eigenvectors that satisfy this conditions are:
|Lx, −1
=
√ − √ |
|Lx, 0
=
√ | − |
=
√ √ |
|Lx, 1 ∗
| − √ |
1 1 1 1 + 2 3 2 2 2 1 (1 3) 2 1 1 1 1 + 2 + 3 2 2 2
[email protected]
1
|
√ |
Now for Ly :
−i √
− − λ
2
0 = det(L det(Ly − λI ) = det √ i2
0
√ i 2 0
−i = λ(λ2 − 1) √ 2
λ
√ i
0
Then, the eigenvalues are: λ =
− 0
λ
2
−1, 0, 1. For the eigenvectors:
−i √
0
2
−i √
0
x y z
2
√ i 2
0
=λ
x y z
Solving this problem for each eigenvalue, we found:
ix iz iy √ = −y, √ = −z −1 ⇒ −√ iy2 = −x, √ − 2 2 2 −iy ix iz iy λ = 0 ⇒ √ = 0, √ − √ = 0, √ = 0 2 2 2 2 ix iz iy −iy λ = 1 ⇒ √ = x, √ + √ = y, √ = z
λ=
2 2 2 2 The normalized eigenvectors that satisfy this conditions are:
|Ly , −1
=
√ √ |
|Ly , 0
=
√ | |
=
√ − √ |
|Ly , 1
| − √ |
1 i i 1 + 2 3 2 2 2 1 (1 + 3 ) 2 1 i i 1 + 2 + 3 2 2 2
|
√ |
Now for Lz :
− 1
0 = det(L det(Lz
− λI ) = det 0
√ i 2 0
− − − 0 λ
0 0
Then, the eigenvalues are: λ =
λ
0 0
1
= λ(λ
λ
− 1)(−λ − 1)
−1, 0, 1. For the eigenvectors:
−i √
0
2
−i √
0
2
√ i 2
0
x y z
=λ
x y z
Solving this problem for each eigenvalue, we found: λ=
−1 ⇒ x = −x, −z = −z λ = 0 ⇒ x = 0, −z = 0 λ = 1 ⇒ x = x, −z = z
The normalized eigenvectors that satisfy this conditions are:
|Lz , −1 |Lz , 0 |Lz , 1 2
= = =
|3 |2 |1
b) Take the state in which Lz = 1. In this state what are Lx , Lx2 , and ∆Lx . If the system is in the state where Lz = 1, then the system is in the state: Lz , 1 = 1 .
|
|
0
Lx = Lz , 1|Lx Lz , 1
= 1 0 0
=
1 0 0
−i √
2
√ i 2
0
2
Lx2
0
0
−i √
√ i 2
0
0
2
√ i
0
∆Lx =
√ i 2 0
0
Lx2
−i √
| − − √ Lx
2
=
2
0
√ i 2
−i √
2
0
√ i
2
1 2
0=
0
1 0 0
−i √
2
0
0
−i √
2
0
1 0 0
=0
=
1 2
1 2
c) Suppose the system is prepared in the state with Lx = 0, and then L then Lz is measured. What are the possible outcomes and their probabilities? If the system is prepared in the state with Lx = 0, then
|ψ = |Lx, 0 = √ 12 (|1 − |3) Then, Then, if Lz is measured, we know from (a) that the possible outcomes are: Lz = 1, 0, 1. Now, let’s calculate the probabilities:
−
P ( P (Lz =
−1) = ψ|P(Lz = −1)|ψ = 12 (1| − 3|)(|33|)(|1 − |3) = 12 1 P ( P (Lz = 0) = ( 1 2 1 P ( P (Lz = 1) = ( 1 2
| − 3|)(|22|)(|1 − |3) = 0
| − 3|)(|11|)(|1 − |3) = 12
d) Suppose the system is prepared in the state with Lx = 0, and then Lz and Ly are are measu measurred, in that that order order.. What What is the prob probabilit abilityy that that the result resultss of both measurements will be +1 +1? ? From the previous problem (c), we found that after measured Lz the system is in the state: 1 or in 3 ). If we measure Ly = 1 then the probability is:
|
|
P ( P (Ly = 1, Lz = 1) = Lx , 0 P(Lz = 1)P(Ly = 1)P(Lz = 1) Lx , 0
| | √ √ 1 1 = (1|−3|)|11| (i|1 + 2|2 − i|3)(−i1| + 22| + i3|) |11|(|1−|3) 2 4
=
1 8
e) Consider the state
|ψ = 3
1/2 1/2 1/2
in the Lz basis asis.. If L2z is measured in this state and a result +1 is obtained, what is the state after the measurement? How probable was this result? If Lz is measured, what are the outcomes and respective probabilities? The eigenvalues of the operator Lz is:
− 1
0 = det(L det(L2z
− λI ) = det
λ
0 λ
0 0
−
− 0 0
1
=
λ
−λ(1 − λ)2
where the eigenvalues are: λ = 0, 1 and the eigenvectors are: L2z , 0 = 2 and 1 L2z , 1 = √ ( 1 + 3 ). The state with L2z = 1 is: 2
|
|
| |
| |
|ψ = |L2z , 1L2z , 1|ψ = (|11| + 3
3)
1 (1 + 2 + 2
| |
√
|
23 )
|
√
1 2 = 1 + 3 2 2
|
|
and the normalized new state is:
|ψ = √ 2
3
√ 2
1 1 + 3 2 2
|
|
1 = 3
√ |
1 +
√
|
23
Then, the probability that the system is in the state L2z = 1 is: P ( P (L2z = 1) = ψ P(L2z = 1) ψ =
|
√
|
√
1 3 = 1 + 2 + 2 3 (1 1 + 3 3) 1 + 2 + 23 = 4 4 Finally, if we measure Lz , we possible values are Lz = 1, 0, 1 and the probabilities bilities are:
|
P ( P (Lz =
|
−1)
=
P ( P (Lz = 0) = P ( P (Lz = 1) =
| | |
| | |
| −
|
√ 2 |Lz , −1|ψ|2 = 13 3| |1 + 2|3 = 23 √ |Lz , 0|ψ|2 = 13 2| |1 + 2|3 2 = 0 √ 2 |Lz , 1|ψ|2 = 13 1| |1 + 2|3 = 13
f ) A particle is in a state for which the probabilities are P ( P (Lz = 1) = 1/ 1/4, P ( P (Lz = 0) = 1/ 1/2, and P ( P (Lz = 1) = 1/ 1/4. Convin Convincce yourse yourself lf that that the most gener general, al, normalized state with this property is
−
iδ1
iδ2
iδ3
|ψ = e2 |Lz = 1 + e√ 2 |Lz = 0 + e2 |Lz = −1 It was stated earlier on that if ψ is a normalized state then the state eiθ ψ is a physically physically equivale equivalent nt normalized normalized state. Does Does this mean mean that the factors factors eeiδ multiplying the Lz eigenstates eigenstates are irrelevant? irrelevant? (Calculate (Calculate for example example P ( P (Lx = 0).) 0).)
|
4
| i
Let’s take a general state ψ = a 1 + b 2 + c 3 and calculate the probabilities:
|
P ( P (Lz =
−1)
=
P ( P (Lz = 0) = P ( P (Lz = 1) =
| | |
|Lz , −1|ψ|2 = ( 3| (a|1 + b|2 + c|3))2 = |c|2 = 14 |Lz , 0|ψ|2 = (2| (a|1 + b|2 + c|3))2 = |b|2 = 12 |Lz , 1|ψ|2 = (3| (a|1 + b|2 + c|3))2 = |a|2 = 12
where the most general solution for the coeficients are: c=
1 eiδ = , 4 2 3
b=
1 eiδ = , 2 2
1 eiδ = 4 2
2
√
,a =
1
Therefore, the most general state is: iδ1
iδ2
iδ3
|ψ = e2 |Lz = 1 + e√ 2 |Lz = 0 + e2 |Lz = −1 Now, let’s calculate the probability of L of Lx = 0: P ( P (Lx = 0) = Lx , 0 ψ
|
1 iδ = e 8
1
| |
iδ3
−e
2
e−iδ
1
1 = 8
| − | | − 1
− e−iδ
3
=
3
e
iδ1
1 1 8
iδ2
1 +e
ei(δ −δ 3
1
)
√
iδ3
22 +e
|
3
2
− ei−(δ −δ ) + 1 3
1
1 = (1 cos(δ cos(δ 3 δ 1 )) 4 then, we know that this factors are not irrelevant.
−
|
−
2) Read the discussion on pp. 133-134 of Shankar about the statistical operator (or ”density matrix”). Show that the time-dependent statistical operator ρ(t) obeys the evolution equation dρ( dρ(t) i = [H, ρ(t)] dt You may start from ρ(t) = i pi ψi (t) ψi (t) , where the statistical weights pi are independent of time, and where each ψi (t) obeys the time-dependent Schr¨ odinger equation.
−
|
|
|
Let’s calculate the LHS: dρ( dρ(t) d = dt dt
pi ψi ψi
i
|
| =
pi
i
d ( ψi ) d ( ψi ) ψi + ψi dt dt
| | | |
Then we now that the time-dependent Schr¨ orinder orinder equation equation is: d H ψ = i ψ dt
|
| ⇒ − i H |ψ = dtd |ψ
and the adjoint Schr¨ orinder orinder equation:
| −i dtd ψ| ⇒ i ψ|H = dtd ψ|
H ψ =
5
Replacing this in the previous relation: dρ dt
=
− | i
pi
= =
i
H
−
H ψi ψi + ψi
| | ψi|H
i
pi ψi ψi
|
i
i
i
|−
pi ψi ψi H
i
| |
− [H, ρ(t)]
Shank Shankar Ex. 4.2.2 Show that for a real wave function ψ(x), the expectation value momentum P = 0. (Hint: Show that the probabilities for the momenta p are equal.) Generalize this result to the case ψ = cψr , where ψr is real and c an arbitrary (real or complex) constant. (Recall that ψ and α ψ are physically equivalent.)
±
| | Let’s compute the probabilities for ± p: p: P ( P ( p) p) = | p|ψ|2 = ψ| p p p|ψ ∞ = dx (ψ|xx| p p) p|xx|ψ
−∞
= =
P ( P ( p) p) =
−
=
∞ 1 dx ψ ∗ (x)eipx e−ipx ψ (x) 2π −∞ ∞ 1 dx ψ 2 (x) since ψ ∗ (x) = ψ(x) 2π −∞
p ψ
2
= ψ
p
p ψ
|− | | | | −| −− −| | | ∞
dx ( ψ x x
p ) p x x ψ
−∞
= =
∞ 1 dx ψ∗ (x)e−ipx eipx ψ(x) 2π −∞ ∞ 1 dx ψ2 (x) sinc ince ψ∗ (x) = ψ (x) 2π −∞
As the probabily for p and p are equal, equal, then then P = 0. Also Also,, for for the prev previou iouss relations, is easy to see that the result is maintained if we replace ψ by cψ. cψ.
−
Shank Shankar Ex. 4.2.3 Show that if ψ if ψ (x) has mean momentum P , eip tum P + p0 .
First compute P
P
= =
ψ P ψ
| | | | | | − ∞
dx ψ x x P x x ψ
−∞
=
∞
i
−∞
6
dx ψ∗
∂ ψ ∂x
o
x/ ψ (x)
has momen-
Then, lets calculate this mean value when ψ = eip
o
P
= = = =
−i −i
∞
x/ ψ (x)
− − − − −∞ ∞ −∞ ∞
−i −∞ P + p0
dx dx
ipo x
exp
ipo x
exp
ψ∗
∂ψ dx ψ∗
∞
ip0
i
∂x
∂ ipo x exp ψ ∂x ∂ψ ipo x exp ψ ∂x
ψ∗
ip0
exp
ipo x
ψ
dx ψ∗ (x)ψ (x)
−∞
4) A particle in 1D has a wavefunction ψ (x) = x ψ = 1/(x2 + a2 ) where a is a positive real real constant. onstant. Obtain Obtain a corr correc ectly tly normalize normalized d version version of this wavefunction wavefunction and com2 2 pute X , X , P , and P . Also compute the correctly normalized ψ ( p) p) = p ψ . (Note: In this kind of problem, you are allowed to quote results fro definite or indefinite integrals that you obtain from books, or from programs like Maple or Mathematica, but please briefly cite your source.)
|
|
To correctly normalized this wavefunction, first calculate ψ ψ :
|
ψ|ψ
= =
∞
−∞ ∞ −∞
dx ψ x x ψ =
| |
dx
∞
dx ψ∗ (x)ψ(x)
−∞
1 π = (x2 + a2 )2 2a3
where here, and in the following, all the solutions of the integrals are taken from Mathematica1 . Thus, the system is in the normalized state:
x|ψ =
2a3 1 2 π x + a2
Now, let’s calculate the X , X 2 , P , and P 2 :
X = ψ|X |ψ ∞ = dx ψ|xx|X |xx|ψ =
∞ ∞ ∞
= 1
dx ψ∗ (x)xψ( xψ (x)
2a3 ∞ x dx 2 =0 π ∞ (x + a2 )2
Wolfram Alpha LLC. 2012. Wolfram—Alpha. http://www.wolframalpha.com (access July 10, 2012).
7
X 2
= = =
ψ X 2 ψ
|
∞
∞ ∞ ∞
= =
P
=
∞
∞ ∞
= =
=
= =
dx ψ x x P x x ψ
dx ψ∗ (x)
ψ P 2 ψ
|
∞
∞ ∞ ∞
=
d i ψ(x) dx ∞ 2ia3 ∞ 1 d 1 dx 2 2 2 π x +a dt x + a2 ∞ 2ia3 ∞ 1 2x dx π x2 + a2 (x2 + a2 )2 ∞ 2ia3 ∞ 2x dx 2 =0 π (x + a2 )3 ∞
=
=
2a3 ∞ x2 dx 2 π ∞ (x + a2 )2 π 2a3 = a2 2a π
ψ P ψ
=
=
dx ψ∗ (x)x2 ψ(x)
| | | | | | − − − −
=
P 2
| dx ψ|xx|X 2 |xx|ψ
| dx ψ|xx|P 2 |xx|ψ d2 ψ(x) dx2 ∞ 1 d 2x dx x2 + a2 dt (x2 + a2 )2 ∞ ∞ 1 2 2x2 dx x2 + a2 (x2 + a2 )2 (x2 + a2 )3 ∞ 3π π 4a3 8a5
−
dx ψ ∗ (x)
2a3 2 π 2a3 2 π 2a3 2 π
2
−
Finally, the wavefunction in the momentum representation is: ψ( p) p) = p ψ =
|
∞
−∞
dx p x x ψ =
| |
∞
−∞ 8
dx ψ p (x)∗ ψ(x) =
1 2π
√
−
ipx exp 2a3 ∞ dx π −∞ x2 + a2
