Quantum Mechanics - Homework Assignment 9 Alejandro G´ omez omez Espinosa∗ December 3, 2012
1) Fill in the missing parts and keep for your future reference. (Derivations not needed! Use εijk notation where appropiate. [X i , X j ] = 0 [X i , P j ] = i δ δ ij ij ; [X i , P 2 ] = 2i δ δ ij ij P j ;
[P i , P 2 ] = 0;
[P 2 , P 2 ] = 0
[P i , L j ] = i ε jil P l ;
[P 2 , L j ] = 0;
[Li , L j ] = i ijk Lk
[P i , L2 ] = 2i ε jil P l L j ;
[P 2 , L2 ] = 0;
[Li , L2 ] = 0;
[X i , L j ] = i ε jkl X k ; [X i , L2 ] = 2i ε jkl X k L j ;
[P i , P j ] = 0
[L2 , L2 ] = 0
Shankar 12.3.2 Let us try to deduce the restriction on lz from another angle. Consider a superposition of two allowed lz eigenstates: ψ (ρ, φ) = A(ρ)eiφl
z
/
+ B (ρ)eiφl
z
/
(1)
By demanding that upon a 2π rotation we get the same physical state (not necessarily the same state vector), show that lz lz = m , wher where m is an inte integer ger.. By arguin arguingg on the groun grounds ds of symsymmetry that the allowed values of lz must be symmetric about zero, show that these values are either possible le to restric estrict t lz any further ..., 3 /2, /2, /2, 3 /2,... or ..., 2 , , 0, , 2 ,... , .... It is not possib this way.
−
−
−
− −
If we demand a 2π 2π rotation to the state (1 (1), we found that: +2π )l ψ(ρ, φ + 2π 2π ) = A(ρ)ei(φ+2π
= ei2πl
z
/
= ei2πl
z
/
z
/
+2π )l + B (ρ)ei(φ+2π
z
/
A(ρ)eiφl
/
+2π )l + B (ρ)ei(φ+2π
A(ρ)eiφl
/
+ B (ρ)eiφl
z
z
e
then, i2π (lz −lz )/
e
= cos
2π (lz
z
−lz )/
− lz )
lz
− lz = 2lz = m
for m = ..., 3, 1, 1, 3,.... ,....
− −
∗
[email protected]
1
e
=1
= 1;
(l z
⇐⇒
for m = ..., 1, 0, 1,..... ,..... Finally, if lz and lz are allow values, thus:
−
/ −i2πlz /
/ i2π (lz −lz )/
z
that is the same state as (1 (1) times a constant, if: ei2π(l
z
⇒
lz =
m 2
− lz ) = m
(2)
3) Find the ground state energy of a particle of mass µ in 3D confined to the region r0 < r < 2r0 (that is, V = 0 in this region and V = + both inside r0 and outside 2r0 ).
∞
This problem similar to find the eigenstate of a particle in spherical coordinates, but with different boundary conditions. Then, the wave function and the energy value is: ψ (r,θ,φ) r,θ,φ) = (Ajl (kr) kr ) + Bn l (kr)) kr ))Y Y lm (θ, φ),
E =
2 2 k
2µ
but for the ground state: ψ0m = Aj0 (kr) kr )Y 0m (θ, φ) where J 0 (kr) kr ) =
sin(kr sin(kr)) kr
Applying boundary conditions, we found: sin(kr sin(kr 0 ) = 0
nπ = kr 0
⇒
⇒
k=
nπ r0
for n = 1, 2, 3,.... ,.... Therefore, the energy of the ground state is given by: E =
2 2 2 n π 2µr02
4) Find the degeneracy and parity of all energy levels with E < 5 ω for the 3D simple harmonic oscillator using the Y lm -based d solution solution (p. 352), and check check that it agree agreess with the results results expecte expected d base based d on lm -base separation of variables in Cartesian coordinates. Using the results of page 352 for the simple harmonic oscillator, with energy E =
n+
3 2
ω, ω,
n=k+l
Then, the energy levels with E < 5 ω are: E 0 =
3 ω, ω, 2
5 ω, ω, 2
E 1 =
E 2 =
7 ω, ω, 2
E 3 =
9 ω, ω, 2
where the allowed values of n of n are 0, 0, 1, 2, 3. Then, Shankar found that the eigenstates have the values of n, l and m: n=0 l=0 m=0 n=1 l=1 m = 0, 1 n = 2 l = 0, 2 m = 0; 0; 0, 1, 2 n = 3 l = 1, 3 m = 0, 1; 0, 1, 2, 3
± ± ± ± ± ± ±
Here is easy to see that states with n = 1, 2, 3 have degeneracies. n = 1 has 3 degenerate states, n = 2 has 6 degenerate states and, n = 3 has 10 degenerate states. In the case of separation of variables, the energy is given by: E =
3 nx + ny + nz + 2
2
ω
where, looking for the diferent combinations of n of nx , ny , nz , we can verify the previous result: n = 0 = (000) n = 1 = (100, 010 010,, 001) n = 2 = (110, 101 101,, 011 011,, 200 200,, 020 020,, 002) n = 3 = (111, 201 201,, 210 210,, 021 021,, 102 102,, 120 120,, 012 012,, 300 300,, 030 030,, 003)
For the parity, the radial wavefunction is not affected in the case of spherical harmonics, hence the parity operator only acts on Y lm spherical harmonic harmonic term Y lm = P lm (cos θ)eimφ under lm . Then, the spherical parity transformation is: Π(Y Π(Y lm (θ, φ)) = Y lm (π θ, φ + π)
−
Let us split the harmonic term. For the exponential part: im(φ+π ) eim( = eimφ eimπ = ( 1)m eimφ
−
Then, for the associated Legendre polynomials: P lm (cos(π (cos(π Changing
− θ)) = P lm(− cos θ)
− cos θ = −x for simplicity: P lm (
−x)
= = =
l+|m|
−− − −− − − −− 1 1
( x)
2 |m|/2
1
d( x)
2 |m|/2
x
dx d( x)
d
l+|m|
−
dx d( x)
l+|m|
d dx
( 1)2
l+|m|
1
l
( 1)2
l
P lm (x)
= ( 1)l+|m| P lm (x)
−
Join the two results: Π(Y Π(Y lm ) = ( 1)m ( 1)l+|m| Y lm = ( 1)l Y lm
−
−
−
Recalling that n = 2k + l, we can express the last term as ( 1)n Y lm . Now, we see that states with l odd are going to have parity 1 and, if l if l is even the parity is 1.
−
−
In addition, for the case of cartesian coordinates, the only term that affect the parity are the Hermite polynomials, which parity is ( 1)n . Result that confirm the previous result.
−
3
Shankar Shankar Ex. 12.5.2 (3) Contruct the 4 4 matrices J x , J y , and J z for j = 3/2 and verify that [J x , J y ] = i J J z explicitly.
×
Let us write down the matrix elements of J x , J y and J z in the basis J ± :
− − − 3 3 m J x m 2 2
3 J + + J − 3 m m 2 2 2 1 3 3 m J + m + 2 2 2
= =
1 2
=
+
=
2
3 2
2
3 +m 2
3 2
3 2
5 +m 2
m
= =
2
=
2i
3 2
m
1
3 +m 2
δ m ,m+1 ,m+1 +
3 J + J − 3 m m 2 2i 2 1 3 3 m J + m 2i 2 2
3 3 m m+1 + 2 2
3 3 m m 2 2
m+1
− − − 3 3 m J y m 2 2
2
3 +m+1 2
m
− −
3 3 m J − m 2 2
5 2
2
m
δ m ,m− ,m−1
− −
3 3 m J − m 2 2
5 +m 2
2
3 +m 2
δ m ,m+1 ,m+1
5 2
2
m
δ m ,m− ,m−1
3 3 m J z m 2 2
3 3 m m m 2 2 3 3 = m m m 2 2 = m δ δ m ,m =
Now, we now that m = written as:
−l,...,l, l,...,l, then m = − 32 , − 12 , 12 , 32 , and the matrix representation can be √ 3 0 0 0 √ 3 0 2 0 √ 3 J x = 0 2 √ 0 2 0 0 3 0 √ 3 0 0 − 0 √ 3 0 −2 0 i √ 3 J y = 0 2 0 − 2 √ 3 0 0 0 J z =
2
3 0 0 0 4
0 1 0 0
−
0 0 1 0
−
− 0 0 0 3
Finally, let us verify [J [J x , J y ]: [J x , J y ] = J x J y =
J y J x
−√ √ √ −√ − √ √ √ √ − √ −√ √ √ − √ √ − − √ √ − √ √ − − √ √ − − √ − √ √ − − − − 0 3 0 0
i 2 4
0 3 0 0
=
=
3 2 2 0
3
i 2 4
0 1 0
2 3 0 2 3
6 0 0 0
i 2 4
0 2 0 0
0
0 0 2 0
= i J J z
0 3 0 0
3 0
0 2 0 3
0 2 0
3 0
0 0 0 3
0 0 0
2 3 0 1 0 =
0 2 0
0 3 0 0
3
0 0 0 6
3
2
−
3 0 0 0
0 0
0 0 0 3
0
3 2 2 0
0 2 3 0 3
i 2
0 2 0 3
3
0
3 0
3 0 0 1 2 3 0 0 2 3
0 1 0 0
0 0 1 0
2 3 0 1 0
0 0 0 3
0 2 3 0 3
− √
−
Shankar Shankar Ex. 12.5.3 (1) Show that J x = J y = 0 in a state jm jm . Expressing J x and J y in terms of J + and J − , and using Eq (12.5.21a):
jm|J x| jm jm
= =
|
1 jm J + + J − jm jm 2
|
2 = 0
| 1/2 δ jj δ m,m+1 + δ jj δ m,m− (( j + m)( j )( j − m + 1))1/2 m,m+1 (( j − m)( j + m + 1)) m,m−1 (( j
Same procedure for J y , but as J y = J + −2 J is easy to see, from the previous derivation, that J y = 0 too. (2) Show that in these states J x2 = J y2 = 12 2 [ j( j ( j + 1) m2 ] (use symmetry arguments to relate J x2 to J y2 ). By convention, we develop all this relations to rotate about the z-axis, then it is clear that the system is symmetric about the z-axis, then: J x2 = J y2 . Using the total angular momentum: −
−
J 2 = J x2 + J y2 + J z2 ⇒ 2J x2 = J 2 − J z
Thus, 1 2 J J z2 2 1 = jm J 2 J z2 jm jm 2 1 = jm J 2 jm jm jm J z2 jm jm 2 1 2 = j ( j + 1) 2 m2 j( 2 where I used the relations (12.5.17) from Shankar for the last steps.
J x2
=
− | − | | | − | | −
5
(3) Check that ∆J x ∆J y from part (2) satisfies the inequality imposed by the uncertainty principle. [Eq. (9.2.9)] Let us calculate (∆J (∆J x )2 : (∆J (∆J x )2 = J x2 J x 2 = J x2
·
−
Therefore: 2
2
(∆J (∆J x ) (∆J (∆J y ) =
J x2
J y2
=
4
j( j ( j + 1)
− m2 2
4 Now, let us calculate the LHS of the inequality imposed by the uncertainty principle: jm jm |J xJ y | jm
= = = = = = = =
jm | | jm|J xJ y | jm
=
(3)
1 jm (J + + J − )(J )(J + J − ) jm jm 4i 1 2 2 jm J + + J − J + J + J − J − jm jm 4i 1 jm J − J + J + J − jm jm 4i 1 jm J − (( j (( j m)( j )( j + m + 1))1/2 jm jm + 1 4i
| − | | − − | | − | | − | − jm − 1 − jm |J + (((( j + m)( j − m + 1))1/2| jm 2 (( j (( j − m)( j )( j + m + 1) jm | jm jm − ( j + m)( j − m + 1) jm | jm jm ) 4i 2 (( j (( j − m)( j )( j + m + 1) − ( j + m)( j )( j − m + 1)) 4i 2 m − 2i |m|
2
Then, by solving the Schr¨odinger odinger equation, we know that j j
≥ |m|, therefore:
≥ |m| j( j ( j + 1) ≥ |m|(|m| + 1) = m2 + |m| j( j ( j + 1) − m2 ≥ |m| Using this result, we can conclude that ∆J ∆J x ∆J y ≥ |J x J y |. (4) Show that the uncertainty bound is saturated in the state | j, j, ± j . In the result of (3), we found 2 the condition for the inequality is j ( j + 1) − m ≥ |m|, but in the case of m = ± j: j : j ( j + 1) − j 2 = j ⇒ j ≥ ±m |m| ≤ j( Then, the uncertainty is saturated.
6
Shankar Shankar Ex. 13.1.3 Starting from the recursion relation, obtain φ210 (normalized). Let us start from eq. (13.1.10) for the recursion relation: ∞
vEl = ρ
l+1
C k ρk
k=0
where k = n
− l − 1. Then using ψ210, i.e., n = 2, l = 1, we found: v21 = ρ2 C 0 ρ0 = ρ2 C 0
then, the state can be written as: φ210 = where ρ =
r 2a0 .
e−ρ 2 ρ C 0 Y 10 (θ, ψ ) = C 0 ρe−ρ Y 10 (θ, ψ) ρ
(4)
Now, let us normalized the state:
1 =
∗ φ210 φ210 r 2 dr dΩ
8a30 C 02
=
8a30 C 02
=
∞
0
∞
ρ2
e−2ρ ρ2 dρY 10∗ Y 10 dΩ
ρ4
e−2ρ dρ
0
3 = 8a30 C 02 4 3 2 = 6a0 C 0
1 6a30
= C 0
Finally, the state (4 (4) becomes: φ210 =
1 ρe−ρ Y 10 (θ, ψ) 3 6a0
(5) 2
2
Shankar Shankar Ex. 13.1.4 Recall (r )me /k e−kr ecall from from the last chapter [Eq. (12.6.19)] (12.6.19)] that as r , U E E (2mW/ 2 )1/2 ]. Show that this agrees in a Coulomb potential V = e2 /r [ k = (2mW/ agrees with Eq. (13.1.26).
→∞
−
≈
Equation (12.6.19) from Shankar recalls: 2
me
U E E
≈r
k2
e−rk ,
but, W from eq (13.1.23) is given by: W = then:
k=
2mW 2
me4 2 n2
2m me4 m2 e4 me2 1 k= = = = na0 2 2 n2 4 n2 2 n where a0 is the Bohr radius. Then, the relation for the radial part becomes: U E E Rn = r that is the same as equation (13.1.26).
≈
r n e−r/na0 = rn−1 e−r/na0 r
7
(6)
