Quantum Harmonic Oscillator
Problem 15.7 mω
= = √ + = √ − + − =
Introducing a dimensionless parameter ξ
x ,
(a) Show that the operators aˆ and aˆ † can be written as 1 ∂ , ξ aˆ ∂ξ 2 1 ∂ † . (15.64) ξ aˆ ∂ξ 2 (b) Show that the time-independent Schr¨odinger equation becomes 2 E ∂ 2φ 0. (15.65) ξ 2 φ 2 ∂ξ ω (c) Show that the wave function φ1 ( x ) of the n 1 energy state can be written as 2 (15.66) φ1 ( x ) 2ξ A 1 e−ξ /2 . (d) Find the normalization constant A 1 . (e) Using (a) as the representation of the operators aˆ and aˆ † , verify the commutation relation aˆ , aˆ † 1.
=
=
=
Solution (a) Using the relations
= =
mω
aˆ aˆ †
and the fact that
2
mω
2
ˆ x
+ i √ 2m1 ω pˆ ,
x ˆ
− i √ 2m1 ω pˆ ,
=
mω x ,
ξ
and that
aˆ
(15.68)
=− =− =− = √ + √ = √ + = √ − ∂
p ˆ
we obtain
(15.67)
∂ ∂ξ
i ∂ x
1
ξ
i ∂ξ ∂ x
1
2 2 Similarly, we can show that aˆ
†
mω
i
1
∂
m 2 ω ∂ξ
1
2
ξ
2
∂
∂ξ
.
mω ∂
ξ
∂ξ
,
∂
∂ξ
(15.69)
.
(15.70)
(15.71)
137
138
Quantum Harmonic Oscillator
Solution (b) ¨ We start from the time-independent Schrodinger equation ˆφ H
= E φ ,
(15.72)
where
ˆ H
aˆ † aˆ
= ω
+ 21
.
(15.73)
Using the results from part (a)
= √ + = √ − 1
aˆ aˆ †
∂
ξ
2 1
∂ξ ∂
ξ
2
∂ξ
, ,
(15.74)
we have 1
aˆ † aˆ φ
= 2
1
=2
1
=2
1
=2
− + = − + − − + −− − −− ∂
ξ
ξ
∂ξ
2
ξ φ 2
ξ φ 2
ξ φ
ξ ξ
∂
∂
∂ξ
∂ξ
∂φ
∂ 2φ
φ
ξ
ξ
2
∂
∂ξ
ξφ
∂ξ 2
∂φ
∂ 2φ
∂ξ
∂ξ 2
.
∂ξ 2
∂φ
+ ∂ξ
∂ 2φ
(ξ φ )
φ
∂ξ
φ
∂ξ
∂φ
1
(15.75)
Hence, the time-independent Schrodinger ¨ can be written as ˆφ H
1
− E φ = 2 ω
ξ 2 φ
∂ 2φ
− φ − ∂ξ
2
+ − φ
= 0,
E φ
(15.76)
which can be rewritten as ∂ 2φ ∂ξ 2
2
2 E
− ξ φ + ω φ = 0,
(15.77)
and finally ∂ 2φ ∂ξ 2
+
2 E
2
− ξ ω
= φ
0.
(15.78)
Quantum Harmonic Oscillator
Solution (c) The wave function φ1 ( x ) is of the form (see Section 15.1 of the textbook, Eq. (15.41)): mω 2 (15.79) φ1 ( x ) x φ0 ( x ).
√
=
Since
=
mω x
ξ
we have where
=
and
φ1 ( x )
φ0 ( x )
√
2
2φ0 (0)ξ e−ξ /2
= φ (0)e− 0
ξ 2 /2
= 2 A ξ e− 1
= √ 12 φ (0).
A 1
mω 2 x 2
,
,
(15.80) (15.81) (15.82)
0
Solution (d) From the normalization condition
+∞ | | = −∞ +∞ − | | = −∞ = +∞ − | | = −∞ +∞ − = √ −∞ √ | | = | | = √ φ1 ( x )
2
1,
dx
(15.83)
and from part (c), we have 4 A 1
However,
2
ξ 2
ξ 2 e
1.
dx
(15.84)
d ξ , mω so we have the normalization condition of the form dx
4 Since
mω
A 1
2
ξ 2 e
ξ 2 e
we have
4
mω
ξ 2
A 1
1
d ξ
21
2
ξ 2
2
π
d ξ
1.
(15.85)
(15.86)
π,
(15.87)
1,
(15.88)
.
(15.89)
from which we find
A 1
1
2
mω
π
1 4
139
140
Quantum Harmonic Oscillator
Solution (e) Consider the action of the commutator on a wave function φ : aˆ , aˆ † φ .
(15.90)
From the definition of the commutator and part (a), we have aˆ aˆ †
†
− aˆ aˆ
= + − − − + = + − − − + = − + − − − + + = − + + = 1
φ
ξ
2 1
ξ
2
ξ
1
2
2
2
∂
ξ
∂ξ ∂
∂ξ
∂
∂φ
∂
∂ξ
∂ξ
∂φ
∂
∂ξ
∂ξ
∂φ
∂ξ
2ξ
∂
∂ξ
φ
∂ξ
2
2ξ
∂ξ
ξ
∂φ
ξ
ξ
∂
∂ξ
ξφ
∂ξ
ξ
∂φ
ξφ
∂ξ
ξ φ
ξ φ
1
∂
(ξ φ )
(ξ φ )
∂φ
2φ
∂ξ
∂ 2φ ∂ξ 2
∂ 2φ ∂ξ 2
φ.
(15.91)
Hence aˆ , aˆ †
=
1.
(15.92)
Problem 15.8
Calculate the expectation value x ˆ and the variance (fluctuations) x x ˆ2 ˆ 2 of the position operator of a one-dimensional σ harmonic oscillator being in the ground state φ0 ( x ), using
= −
(a) Integral definition of the average. (b) Dirac notation, which allows to express x ˆ in terms of aˆ , aˆ † , and to apply the result of the Tutorial Problem 15.5. (c) Show that the average values of the kinetic and potential energies of a one-dimensional harmonic oscillator in an energy eigenstate φn are equal.
|
Quantum Harmonic Oscillator
Solution (a) The wave function of the position operator of a one-dimensional harmonic oscillator being in the ground state is of the form φ0 ( x )
β x 2
= Ae−
,
(15.93)
where mω
=
β
and
2
1 4
1 4
= = +∞ − = | | = mω
A
2β
π
π
.
(15.94)
Thus, the expectation value of the position operator is
+∞ ∗ = x ˆ
−∞
φ0 ( x ) x φ0 ( x ) d x
A
2
2β x 2
x e
dx
0, (15.95)
−∞
since the function under the integral is an odd function. Calculate now x ˆ 2 . From the definition of the expectation value, we have
2
x ˆ
Since
+∞ ∗ = −∞
+∞ = | | −∞ =
φ0 ( x ) x 2 φ0 ( x ) d x
+∞
A
2
x 2 e−2β x dx
−∞
2
x 2 e−2β x dx .
2
1
π
4β
2β
(15.96)
,
(15.97)
we have for the variance 2
2
= x ˆ − x ˆ = | A|
σ
1
2
4β
π
−0 2β
1 2
= 2β π
1
1 . = 2β 4β π
4β
(15.98)
Solution (b) Using the representation of x ˆ in terms of aˆ , aˆ † , we have 1 = 2√ aˆ + aˆ β
†
ˆ x
.
(15.99)
Hence, we can write the expectation value of x ˆ as 1
†
√
x ˆ = 2√ β φ |aˆ + aˆ |φ = 2 0
0
1
β
†
| | + φ |aˆ |φ
φ0 aˆ φ0
0
0
.
(15.100)
141
142
Quantum Harmonic Oscillator
Since
| = 0,
aˆ φ0
aˆ † φ0
| = | φ ,
φ |φ = 0,
and
1
0
1
(15.101)
we have for the expectation value
x ˆ = 0.
(15.102)
Calculate now x ˆ2 :
x ˆ = 41β φ |(aˆ + aˆ )(aˆ + aˆ )|φ 2
†
0
1
†
0
†
= 4β φ |aˆ aˆ |φ + φ |aˆ aˆ |φ + φ |aˆ aˆ |φ + φ |aˆ aˆ |φ . 0
0
†
0
0
0
0
† †
Since
| = 0, aˆ aˆ |φ = 0, aˆ aˆ |φ = aˆ |φ = |φ , √ aˆ aˆ |φ = 2|φ ,
0
0
(15.103)
aˆ aˆ φ0 †
0
†
0
† †
| = 0, we obtain
1
0
0
2
(15.104)
and φ0 φ2
x ˆ = 41β .
(15.105)
= x ˆ − x ˆ = 41β ,
(15.106)
2
Hence, the variance is 2
σ
2
which is the same value as predicted in part (a).
Solution (c) The kinetic and potential energies of the harmonic oscillator are defined as 1 2 1 ˆE k ˆ ˆ , ˆ 2. (15.107) p V mω2 x 2m 2 Consider first the kinetic energy. Since
=
p ˆ
=
=− − i
mω
2
aˆ
aˆ † ,
(15.108)
Quantum Harmonic Oscillator
we have 2
ˆ p
=−
mω
2
Hence,
aˆ †
aˆ † .
− − aˆ
aˆ
(15.109)
E ˆ = φ | E ˆ |φ = − 14 ω φ |aˆ aˆ |φ − φ |aˆ aˆ |φ − φ |aˆ aˆ |φ + φ |aˆ aˆ |φ . k k
n
k k
n
n
†
n
n
n
†
n
n
† †
n
Since
=
n
(15.110)
| n(n − 1)|φ − , aˆ aˆ |φ = n|φ , aˆ aˆ |φ = (n + 1)|φ , aˆ aˆ |φ = (n + 1)(n + 2)|φ + , aˆ aˆ φn †
n
†
† †
n 2
n
n
n
n
(15.111)
n 2
| = δ , we obtain 1 1 1 1 ˆ E = − 4 ω[0 − (n + 1) − n + 0] = 4 ω(2n + 1) = 2 ω n + 2 and φn φm
nm
k k
.
(15.112) Consider now the expectation value of the potential energy. Since 1
= = 2
ˆ x we have ˆ V
+ 2
mω
1
ˆ φn φn V
2
aˆ
aˆ † ,
2
(15.113)
aˆ †
2
= | | = | + | = | | + | | + | | + | | 1
4
ω
2
mω
φn aˆ aˆ φn
φn aˆ † aˆ φn
Hence,
4mω
φn
φn
φn aˆ aˆ † φn
φn aˆ † aˆ † φn
1
aˆ
1
.
1
V ˆ = 4 ω[0 + (n + 1) + n + 0] = 4 ω(2n + 1) = 2 ω
(15.114)
+ 1
. 2 (15.115)
n
Thus,
E ˆ = V ˆ . k k
(15.116)
143
144
Quantum Harmonic Oscillator
Problem 15.9 Show that the non-zero minimum energy of the quantum harmonic oscillator, E ω/ ω/ 2, is the consequence of the uncertainty relation between the position and momentum operators of the oscillator.
≥
(Hint: Use the uncertainty relation for the position and momentum operators in the state n 0 and plug it into the average energy of the oscillator. Then, find the minimum of the energy with respect to δ x .) .)
=
Solution Take ake the the squa squarre of the the unce uncert rtai aint nty y relat elatio ion n for δ x and δ p in the the stat state e n 0:
=
2
2
2
≥
, (15.117) 4 and and the the expr expres essi sion on for for the the aver averag age e ener energy gy of the the harm harmon onic ic osci oscilla llato torr δ x δ p
(15.118) E = 21m pˆ + 21 mω x ˆ . Since x ˆ = 0 and pˆ = 0, we have δ x = x ˆ and δ p = pˆ , so 2
2
2
2
2
2
2
that we can write
E = 21m δ p + 21 mω δ x . 2
2
2
(15.119)
From this expression, we have δ p2
2
2
2
= 2m E − m ω δ x ,
(15.120)
and substitut substituting ing it into Eq. (15.117), (15.117), we get 2
δ x
From this, we find
2
2
2
− m ω δ x
2m E
2
2
≥
4
E ≥ 8mδ x + 21 mω δ x .
2
2
2
.
(15.121)
(15.122)
Since δ x 2
= 21 m ω ,
(15.123)
Quantum Harmonic Oscillator
we have
E ≥ 4ω + 4ω = 2ω .
(15.124)
Thus, starting from the uncertainty relation for the position and moment momentum um opera operato tors rs of the quantu quantum m harmon harmonic ic oscilla oscillato torr, the minimum energy of the oscillator is ω/ ω/2.
145
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Chapter 16
Quantum Theory of Hydrogen Atom
Problem 16.2 The The norm normali alize zed d wave wave func functi tion on of the the gr grou ound nd stat state e of a hydro ydroge genn-lik like e atom with nuclear charge Z e has the form
| = Ae−
β r
,
(16.1)
where A and β are real constants, and r is is the distance between the electron and the nucleus. The Hamiltonian of the atom is given by ˆ H
2
2
Z e2 1
= = − 2m ∇ − 4π ε
0
r
.
(16.2)
Show that (a) A 2 β 3 /π . Z /ao , where ao is the Bohr radius. (b) β (c) The The ener energy gy of the elec electr tron on is E where e E 0 Z 2 E 0 , wher e2 /(8π ε0 ao ). (d) The expectation values of the potential potential and kinetic energies are 2 E and E , respectively.
= =
=−
−
Problems and Solutions in Quantum Physics Zbigniew Ficek c 2016 Pan Stanford Publishing Pte. Ltd. Copyright ISBN 978-981-4669-36-8 (Hardcover), 978-981-4669-37-5 (eBook) www.panstanford.com
=
148
Quantum Theory of of Hydrogen Atom
Solution (a) The constant A is found from the normalization condition 1
Hence, A 2
= | ≡ | |
2 d V
= 4π A
2
= 4π A
2
∞
r 2 e−2βr dr
0
2
(2β )3
π A 2
=
β3
.
(16.3)
3
= β /π .
Solution (b)
|
We find find β from from the the cond condit itio ion n that that the the wav wave func functi tion on is a solu soluti tion on to the stationary stationary Schr¨ Schrodinger ¨ equation for a hydrogen-like atom 2
2
− 2m ∇ | + V (r )| = E | . (16.4) We see that we have to evaluate ∇ | . Since the wave function is 2
given in the spherical coordinates, we have ∂2
2 ∂
∂ r 2
r ∂ r
− ∇ | = ∇ − = + − = = − − | − − + − | = − − − − | = 2
A2
A
2
2
e
β
A2
β r
2β
2
r
e
β r
β
β r
e
2β
2
.
r
(16.5)
Hence, the Schrodinger ¨ equation (16.4) takes the form 2
2β
β2
2m
V (r )
r
E
0,
(16.6)
which after substituting, the explicit form of V V ( x ) reduces to 2
β2
2m
2β
α
r
r
E
0,
(16.7)
where α Z e2 /(4π ε0 ). Since is different from zero, the left-hand side of Eq. (16.7) will be zero only if
= |
2
− 2m
β2
−
2β r
−
α
r
− E = = 0.
(16.8)
We know that the energy E of the electron in a given energy state of the hydrogen atom is a constant independent of r . Therefore, the
Quantum Theory of Hydrogen Atom
terms dependent on r in Eq. (16.8) must add to zero. This happens when 2
β
m
− α = 0,
(16.9)
which gives β
m Z e2
m
= α = 2
2
Z
=a.
4π ε0
(16.10)
o
Solution (c) Since the terms dependent on r in Eq. (16.8) are equal to zero, we have 2
2
me4
2
2
= = − 2m β = − Z 2 (4π ε ) = − Z E ,
E
2
2
0
(16.11)
0
2
= e /(8π ε a ).
where E 0
0 o
Solution (d) The expectation value of the potential energy is
= ∗
V (r )
V (r ) d V
= −4αβ
3
2
= −4π A α
∞ 0
2
1
r e−2β r dr
2
= −αβ = − m β = 2 E . (2β )
(16.12)
2
The expectation value of the kinetic energy is
E
k k
E k k d V
= −4π A
2
2m 2
= −4π A = −4π A
2
2m 2 2
2m
2 3
m
2
2m
∞ − ∞ − − − = 2
= −4π A
= −2β
∞
2
= ∗
r 2 e
β r
r 2 e
β2
2
(2β )3
− − ∞ − 2β
β2
r
2β r
dr
0
∇ e−
0
0
β2
r 2 e−β r
2
β r
dr
β r
e
2β
r e−2β r dr
0
2β
1
1
1
4β
2β
2
1
(2β )2
β
2 2
2
2
= β = − E . 2m m
(16.13)
149
150
Quantum Theory of Hydrogen Atom
We see that
E = − 12 V (r ).
(16.14)
k
This result is a special case of the virial theorem, which states that for a system in a stationary state in a potential V (r ) proportional to r n ,
E = 2n V (r ).
(16.15)
k
Thus, for the electron in a hydrogen-like atom, the potential is inversely proportional to r (n 1), which gives the result in Eq. (16.14). The virial theorem holds also in classical physics, and as we know from classical mechanics, the result (16.14) applies, e.g., to a satellite orbiting the Earth.
=−
Problem 16.3 ˆ Consider the angular momentum operator L
= rˆ × pˆ . Show that
ˆ is Hermitian. (a) The operator L (Hint: Show that the components L x , L y , L z are Hermitian). ˆ (b) The components of L ( L x , L y , L z ) do not commute. ˆ2 (c) The square of the angular momentum L commutes with each of the components L x , L y , L z . (d) In the spherical coordinates, the components and the square of the angular momentum can be expressed as
= −i
L x
= −i
L y
−
sin φ
cos φ
cos φ cos θ ∂ − sin θ ∂θ ∂φ ∂
∂
− ∂θ
sin φ cos θ ∂ sin θ
∂φ
,
,
= −i ∂φ∂ ,
L z
L2
2
= −
1
∂2
sin2 θ ∂φ 2
1
∂
+ sin θ ∂θ
sin θ
∂ ∂θ
.
(16.16)
Quantum Theory of Hydrogen Atom
Solution (a) ˆ is Hermitian if the components ˆ , ˆ , ˆ are The operator L L x L y L z Hermitian, i.e., if
ˆ† L x
= Lˆ ,
ˆ† L y
= Lˆ ,
x
ˆ† L z
= Lˆ .
y
(16.17)
z
ˆ x , L ˆ y , L ˆ z in terms of the position First, we find the components L and momentum operators, which are Hermitian. Since the angular momentum operator can be written as
Lˆ = Lˆ i + Lˆ j + Lˆ k = rˆ × pˆ = y ˆ pˆ − z ˆ pˆ i + (ˆ z pˆ − x ˆ p ˆ ) j + x ˆ p ˆ − y ˆ pˆ k , x
y
z
= − = − = z
y
x
z
we find that ˆ x L
ˆ pˆ z y
ˆ y L
y
x
z
ˆ x : Consider L ˆ† L x
ˆ pˆ z y
ˆ pˆ y z
†
( y ˆ pˆ z )
†
ˆ pˆ y z
Since
[ y ˆ , pˆ z ] we find that ˆ† L x
†
ˆ z † y ˆ † p
− = = = z ˆ , pˆ y
ˆ p ˆ y x
y
. (16.19) x
† † y
z
y
(16.20)
0,
z
† y
− y ˆ pˆ
− pˆ z ˆ = pˆ y ˆ − pˆ z ˆ .
(16.21)
= pˆ y ˆ − pˆ z ˆ = y ˆ pˆ − z ˆ pˆ = Lˆ . ˆ = L ˆ and L ˆ = Similarly, we can show that L z
(16.18)
x
ˆ z L
= (ˆ z pˆ − x ˆ p ˆ ) ,
ˆ pˆ y , z
=
y
(16.22)
x
ˆ is ˆ z . Hence, L L
† z
y
Hermitian.
Solution (b) Consider a commutator ˆ x , L ˆ y . L
(16.23)
Using the expressions (16.19), we obtain ˆ x , L ˆ y L
=
− z ˆ pˆ (ˆ z pˆ − x ˆ p ˆ ) − (ˆ z pˆ − x ˆ p ˆ ) y ˆ pˆ − z ˆ pˆ = y ˆ pˆ z ˆ pˆ − y ˆ pˆ x ˆ p ˆ − z ˆ pˆ z ˆ pˆ + z ˆ pˆ x ˆ p ˆ − z ˆ pˆ y ˆ pˆ + z ˆ pˆ z ˆ pˆ + x ˆ p ˆ y ˆ pˆ − x ˆ p ˆ z ˆ pˆ . (16.24) y ˆ pˆ z z
y
x
x
z
z
x
x
z
z
y
y
x
x
z
z
y
z
z
z
y
z
y
151
152
Quantum Theory of Hydrogen Atom
Since [ y ˆ , pˆ z ] we have
=
= ˆ , pˆ y z
= x ˆ p ˆ y ˆ pˆ
y ˆ pˆ z x ˆ p ˆ z
Thus,
z
ˆ x , L ˆ y L
z
[ x ˆ , pˆ z ]
= [ˆ z , pˆ ] = 0,
(16.25)
x
= z ˆ pˆ z ˆ pˆ .
and
z ˆ pˆ y z ˆ pˆ x
y
(16.26)
ˆ p ˆ z z ˆ pˆ y . x
(16.27)
x
= + − − == − + − − − = − = = = = ˆ pˆ z z ˆ pˆ x y
ˆ pˆ y x ˆ p ˆ z z
ˆ pˆ x y ˆ pˆ z z
Since [ z ˆ , pˆ z ] i , we can replace pˆ z z ˆ by z ˆ pˆ z i and obtain ˆ x , L ˆ y L ˆ ( z ˆ pˆ z i ) pˆ x z ˆ pˆ y x ˆ p ˆ z z ˆ pˆ x y ˆ pˆ z x ˆ ( z ˆ pˆ z i ) p ˆ y y ˆ z . ˆ p ˆ y y ˆ pˆ x (16.28) i x i L
−
Consequently,
ˆ x , L ˆ y L
Similarly, we can show that ˆ y , L ˆ z ˆ x L i L
ˆ z . i L
ˆ z , L ˆ x L
and
(16.29)
ˆ y . i L
(16.30)
Solution (c) Since ˆ2 L we find that ˆ 2, L ˆ x L
= Lˆ + Lˆ + Lˆ , 2 x
ˆ 2, L ˆ L x x
ˆ 2, L ˆ L y x
ˆ 2, L ˆ x L
ˆ2L ˆ L y x
2 y
2 z
(16.31)
ˆ 2, L ˆ L z x
ˆ 2, L ˆ L y x
ˆ 2, L ˆ L z x . (16.32)
ˆ x L ˆ 2. L z
(16.33)
= + + = + = − + − = − + − = − − + + − = − − + − = − − − + +
Thus,
ˆ x L ˆ2 L y
ˆ2L ˆ L z x
Using the commutation relations of (b), we then find ˆ 2, L ˆ x ˆ2L ˆ x L ˆ x L ˆ2 L ˆ2L ˆ x L ˆ x L ˆ2 L L y
y
z
ˆ y L ˆ x L ˆ y L
ˆ z i L
ˆ y L ˆ x ) L ˆ y (L
ˆ y L ˆ z i L
ˆ z L ˆ y i L
ˆ x L ˆ y L
z
ˆ x L ˆ2 L y
ˆ x L ˆ2 L y
ˆ y L ˆ z i L
ˆ z L ˆ x L ˆ z L
ˆ z L ˆ x ) L ˆ z (L 2 ˆ x L ˆ y L
+i Lˆ Lˆ − Lˆ Lˆ z y
ˆ y i L
ˆ x L ˆ2 L z
ˆ x L ˆ2 L z
ˆ x L ˆ z L
ˆ y L ˆ z i L
2 x z
= 0.
Similarly, we can show that ˆ 2, L ˆ y L
(16.34) ˆ 2, L ˆ z L
= =
0.
(16.35)
Quantum Theory of Hydrogen Atom
Solution (d) In spherical coordinates
= r sin θ cos φ , y = r sin θ sin φ , z = r cos θ ,
x
= +
+
(16.36)
and r x 2 y 2 z 2 . ˆ x , which in cartesian coordinates can be written as Consider L ∂ ∂ ˆ x ˆ pˆ z z ˆ pˆ y . (16.37) L y i y z ∂ z ∂ y Using the chain rule, we can express the derivatives in terms of the spherical components as ∂ ∂ ∂ r ∂ ∂θ ∂ ∂φ , ∂ z ∂ r ∂ z ∂θ ∂ z ∂φ ∂ z ∂ ∂ ∂ r ∂ ∂θ ∂ ∂φ . (16.38) ∂ y ∂ r ∂ y ∂θ ∂ y ∂φ ∂ y Since z y r x 2 y 2 z 2 , θ arccos , φ arctan , (16.39) r x we find that z x 2 y 2 1 ∂ r ∂θ ∂φ cos θ , sin θ , 0, r r r ∂ z ∂ z ∂ z y yz 1 ∂ r ∂θ sin θ sin φ , cos θ sin φ , 2 2 r r ∂ y ∂ y r z 1 cos φ ∂φ x , 2 2 ∂ y x y r sin θ 1 ∂ r ∂θ x x z sin θ cos φ , cos θ cos φ , ∂ x ∂ x r r r 2 z 2 y 1 sin φ ∂φ . (16.40) x 2 y 2 r sin θ ∂ x Consequently, 1 ∂ ∂ ∂ ∂ ˆL x /( i ) sin θ y z r sin θ sin φ cos θ ∂ z ∂ y ∂ r ∂θ r 1 1 cos φ ∂ ∂ ∂ cos θ sin φ r cos θ sin θ sin φ ∂ r ∂θ r r sin θ ∂φ cos θ cos φ ∂ ∂ 2 2 sin θ cos θ sin φ sin θ ∂θ ∂φ cos θ cos φ ∂ ∂ sin φ . (16.41) sin θ ∂θ ∂φ
=
= +
−
=−
−
=
+
+
=
+
+
+
=
= = = =
=−
=
+
= + = = =
=− = √ − =
=
= √ − =
= − + = − − = −
=− =−
−
=
+
+
−
−
+
−
153
154
Quantum Theory of Hydrogen Atom
Similarly,
− = z ∂∂ x − x ∂∂ z ∂ φ ∂ = r cos θ sin θ cos φ ∂∂r + 1r cos θ cos φ ∂θ − 1r sin sin θ ∂φ
ˆ y /( i ) L
−r sin θ cos φ
=
sin2 θ
cos θ
2
+ cos
1
∂
∂
sin θ − ∂ r ∂θ r
θ cos φ
cos θ sin φ ∂ − sin θ ∂φ ∂θ ∂
∂ θ sin φ ∂ = cos φ ∂θ − cossin , θ ∂φ
(16.42)
and
− = x ∂∂ y − y ∂∂ x
ˆ z /( i ) L
= r sin θ cos φ
sin θ sin φ
−r sin θ sin φ =
sin2 φ
2
+ cos
∂ r
sin θ cos φ
φ
1
∂
1 cos φ ∂
∂
+ r cos θ sin φ ∂θ + r sin θ ∂φ 1
∂ ∂ r
1 sin φ ∂
∂
+ r cos θ cos φ ∂θ − r sin θ ∂φ
∂
∂ = . ∂φ ∂φ
(16.43)
Having the angular momentum components L x , L y , and L z in the spherical coordinates, we can find L2 in the spherical coordinates: ˆ2 L
= Lˆ + Lˆ + Lˆ . 2 x
2 y
2 z
(16.44)
ˆ x 2 : Calculate L 2 ˆ x /( L
2
− )
= − × −
sin φ
2
= sin +
φ
cos θ cos φ ∂
∂
− ∂θ
sin φ
∂ ∂θ
−
sin θ ∂φ cos θ cos φ ∂ sin θ
∂φ
∂2
∂ cos θ ∂ sin φ cos φ + ∂θ ∂θ sin θ ∂φ 2
cos θ cos φ ∂ sin θ
∂φ
sin φ
∂
+ ∂θ
cos2 θ cos φ ∂ sin2 θ
∂φ
cos φ
∂
. ∂φ (16.45)
Quantum Theory of Hydrogen Atom
ˆ 2: Next we calculate L y 2 ˆ y /( L
2
− )
=
cos φ
×
∂θ
cos φ
2
= cos −
∂
φ
− ∂
∂θ
cos θ sin φ ∂
sin θ ∂φ cos θ sin φ ∂
−
sin θ
∂φ
∂2
∂ cos θ ∂ sin φ cos φ − ∂θ ∂θ sin θ ∂φ 2
cos θ sin φ ∂ sin θ
∂φ
cos φ
cos2 θ sin φ ∂
∂
+ ∂θ
sin2 θ
∂φ
sin φ
∂
, ∂φ (16.46)
ˆ 2: and L z ˆ x 2 /( L
∂2
2
− ) = ∂φ
.
2
(16.47)
Hence, ˆ 2 /( L
− ) = Lˆ /(− ) + Lˆ /(− ) + Lˆ /(− ) ∂ = sin φ + cos φ ∂θ θ ∂ ∂ ∂ ∂ cos φ sin φ sin φ cos φ + cos − sin θ ∂φ ∂θ ∂φ ∂θ 2
2 x
2
2 y
2
+ sin
2
θ
2 z
cos φ
2
∂ ∂φ
cos φ
2
2
cos2 θ sin2 θ 1
= sin
2
+ 1
∂2
∂2
∂
+ sin φ ∂φ sin φ ∂φ
∂φ
∂φ 2
+ ∂φ
∂
sin θ ∂θ sin θ
∂2 ∂φ 2
∂2
θ ∂φ 2
1
+
∂
2
+
1 ∂ + sin θ ∂θ θ ∂φ 2
∂
cos2 θ ∂ 2
cos θ ∂
= ∂θ + sin θ ∂θ + sin
=
2
2
2
cos2 θ
∂2
2
∂
∂θ
Problem 16.4 Particle in a potential of central symmetry
sin θ
.
∂
∂θ
(16.48)
155
156
Quantum Theory of of Hydrogen Atom
A particle of mass m move movess in a potent potential ial of centr central al symme symmetry try,, i.e., V ( x , y , z ) ). The energy of the particle is given by the V (r ). Hamiltonian
=
2
ˆ H
= = − 2m ∇ + V ˆ (r ). 2
(16.49)
ˆ ˆ commut Show Show that that H commutes es with with the angular angular moment momentum um L of of the the particle.
Solution The angular momentum of the particle can be written as
Lˆ = Lˆ i + Lˆ j j + + Lˆ k k , x
y
(16.50)
z
and then the commutator splits into three commutators ˆ ˆ,L H
= + + + ˆ,L ˆ x i H
ˆ,L ˆ y j H j
ˆ,L ˆ z k H k .
(16.51)
ˆ when H ˆ commutes with L ˆ commutes with the components Thus, H ˆ x , L ˆ y , and L ˆ z . L ˆ,L ˆ x , which can be written as the Consider the commutator H sum of two commutators
= − ˆ,L ˆ x H
2
2m
2
∇ +
= −
ˆ (r ), ˆ x V ), L
2
2
2m
∇
+
ˆ x ,L
ˆ (r ), ˆ x . V ), L
(16.52)
First, we consider the commutator:
−
2
2
ˆ ,L ∇ 2m
= − ∇
ˆ x L
2
x
2
2m
ˆ x . ,L
(16.53)
Since
= −i
∂
y ∂ z
∂
− z ∂ y
,
(16.54)
Quantum Theory of Hydrogen Atom
we find
−
2
ˆ ∇ ,L 2m 2
2
3
= − ∇ = ∇ − = ∇ − − − ∇ = + + − − − + + = − + − +
x
2
2m 3
i
2m
∂ z ∂ y
∂ y ∂ z
2
∂2
∂2
∂2
2m
∂ x 2
∂ y 2
∂ z 2
∂
∂
y ∂ z
3
i
2m
∂ , y ∂ z
i
∂ y ∂ z
2
3
i
ˆ x ,L
z ∂ y
∂2 ∂
2m
∂2 ∂
∂ z ∂ y
∂ y ∂ z
∂2
∂2
∂ x 2
∂ y 2
∂ z 2
∂2 ∂
z 2 ∂ x ∂ y
∂ ∂2
2
∂ z ∂ y
∂2
∂2 ∂
y 2 ∂ x ∂ z
∂ z ∂ y
y 2 ∂ y ∂ z
∂ ∂2
∂3
z 3 ∂ y
∂3
∂3
y 3 ∂ z
∂ ∂2
− z ∂ z ∂ y − y ∂ z ∂ x − y ∂ z ∂ y − y ∂ z + z ∂ y ∂ x 2
2
∂3
∂ ∂2
+ z ∂ y + z ∂ y ∂ z 3
2
3
= + i
2m y
y
∂2 ∂ ∂ x 2 ∂ z
∂2 ∂
2
∂ ∂2
+ −
− ∂ z ∂ x
∂ ∂2
∂ y 2 ∂ z
z
2
− ∂ z ∂ y
z
2
3
∂ ∂2 ∂ y ∂ x 2
∂2 ∂
2
∂2 ∂
− ∂ x
2
∂ ∂2
∂ z 2 ∂ y
− ∂ y ∂ z
2
∂ y
.
(16.55)
∂ /∂ x , ∂/∂ ∂ /∂ y , and ∂/∂ ∂ /∂ z commute with each other, we obtain Since ∂/∂
−
2
ˆ ∇ ,L 2m 2
x
Similarly, we can show that 2
= 2
0.
(16.56)
− ∇ = − ∇ = = − − =− − − − =− − − + − + =− −
2
ˆ y ,L
2
ˆ z ,L
0. (16.57) 2m 2m Consider now the commutator commutator involving involving the potential energy ∂ ∂ ˆ (r ), ˆ x ˆ (r ), ), L ), y V i V z ∂ z ∂ y
∂ ˆ (r ) y i V ∂ z ∂ i V y ∂ z ∂ V i z ∂ y
∂ z ∂ y
∂ V y ∂ z
∂ V z ∂ y
∂ V y ∂ z
.
∂ y ∂ z
∂ V z ∂ y
∂ ˆ (r ) z V ∂ y
yV y V
∂
∂ z
zV
∂
∂ y
(16.58)
157
158
Quantum Theory of of Hydrogen Atom
ˆ (r ) depend Since V dependss only only on r , we have
∂ V z ∂ y
− y ∂∂ z V
=
∂ V ∂ r z ∂ r ∂ y
− y ∂∂V r ∂∂ z r .
(16.59)
If r is is the positio position n of an arbitrary arbitrary point in the x , y , z coordinates, we have
= = + x 2
r
and then
∂ r
y , r
=
∂ y
y 2
2
+ z ,
∂ r ∂ z
(16.60)
z . r
=
(16.61)
Hence, ∂ V ∂ r z ∂ r ∂ y
−
∂ V ∂ r y ∂ r ∂ z
∂ V
y z ∂ r r
=
Thus,
ˆ (r ), ˆ x ), L V
=
Similarly, we can show that ˆ (r ), ˆ y ), L V
In summary, since
− we have
2
2
∇ 2m
−
z y r
=
0.
0.
ˆ (r ), ˆ y ), L V
(16.63)
= = = − ∇ = − ∇ = = = = = 2
ˆ x ,L
2
2m
ˆ (r ), ˆ x ), L V
ˆ (r ), ˆ y ), L V
0.
2
ˆ y ,L
2m
ˆ (r ), ˆ z ), L V
ˆ ˆ,L H
(16.62)
0.
2
ˆ z ,L
0,
(16.64)
0, (16.65)
(16.66)
Problem 16.6 Transition dipole moments
The electron in a hydrogen hydrogen atom can be in two states of the form
√
2 N e−r /ao ,
1 (r )
=
2 (r )
= 4N a r e−
r /(2ao )
o
cos θ ,
(16.67)
Quantum Theory of Hydrogen Atom
=
2
2
+ +
2
1 2
=
=
where r ( x y z ) , cos θ z / r , N 1/ 2π ao3 , and ao is the Bohr radius. Using the spherical coordinates, in which
= r sin θ cos φ , y = r sin θ sin φ , z = r cos θ ,
x
(16.68)
and
= ∞
2π
π
dV
0
0
r 2 sin θ dr d θ d φ ,
(16.69)
0
(a) show that the functions 1 (r ), 2 (r ) are orthogonal. (b) Calculate the matrix element (1 (r ), rˆ 2 (r )) of the position operator rˆ between the states 1 (r ) and 2 (r ). (The matrix element is related to the atomic electric dipole moment between the states 1 (r ) and 2 (r ), defined as ˆ 2 (r )) e(1 (r ), rˆ 2 (r )).) (1 (r ), µ (c) Show that the average values of the kinetic and potential 1 V . energies in the state 1 (r ) satisfy the relation E k 2
=
=−
Solution (a) Two functions are orthogonal when the scalar product
= ∗ √ − = √ ∞ = √ ∞ =
(1 (r ), 2 (r ))
1 (r )2 (r )dV
= 0.
(16.70)
Calculate the integral
∗
1 (r )2 (r )dV
2 N 2
r cos θ e
4ao
4ao
2π
2π
π
2 N 2
0
0
π
0
dV
r 3 sin θ cos θ e−3r /2ao dr d θ d φ
0
2 N 2
4ao
3r /2ao
r 3 sin θ cos θ e−3r /2ao dr d θ .
0
(16.71)
Consider the integral over θ : π
0
sin θ cos θ d θ .
(16.72)
159
160
Quantum Theory of Hydrogen Atom
= x . Then, cos θ d θ = dx and the integral takes the form sin θ cos θ d θ = (16.73) xdx = 0.
Let sin θ
0
π
0
0
Thus, the functions 1 (r ), 2 (r ) are orthogonal.
Solution (b) From the definition of the matrix element, we have 1 (r ), rˆ 2 (r ) ∗ (r )rˆ 2 (r )dV
= = ∗ + ∗
1
+ ∗
1 (r ) x 2 (r )dV
i
1 (r ) y 2 (r )dV
j
1 (r ) z 2 (r )dV .
k
(16.74)
In spherical coordinates, and substituting the explicit forms of the functions 1 (r ), 2 (r ), the integrals take the form 1 (r ), rˆ 2 (r )
√ ∞ − = √ ∞ − + √ ∞ − + = = = √ ∞ √ ∞ = √ ∞ − = i
j
k
Since
4ao
r 4 sin2 θ cos θ cos φ e
0
0
0
4ao
dr d θ d φ
2π
r 4 sinθ cos 2 θ e
0
3r /2ao
0
π
2 N 2
dr d θ d φ
2π
r 4 sin2 θ cos θ sin φ e
0
3r /2ao
0
π
2 N 2
4ao
2π
π
2 N 2
0
3r /2ao
dr d θ d φ .
(16.75)
0
2π
2π
sin φ d φ
0
cos φ d φ
0,
(16.76)
0
the x and y components of the matrix element are zero. Hence 2π π 2 N 2 ˆ 1 (r ), r 2 (r ) k r 4 sin θ cos 2 θ e−3r /2ao dr d θ d φ 4ao 0 0 0 2π k 2π k
π
2 N 2
4ao
0
0
π
2 N 2
4ao
r 4 sin θ cos 2 θ e−3r /2ao dr d θ
4
r e
0
3r /2ao
dr
sin θ cos2 θ d θ .
0
We will calculate separately the integrals over r and θ .
(16.77)
Quantum Theory of Hydrogen Atom
=
Let cos θ
1
1
π
cos2
2
3
−1
0
Thus,
− sin θ d θ = dx , and the integral over θ gives 1 2 = 3 . (16.78) θ sin θ d θ = x dx = x 3
x . Then,
=
1 (r ), rˆ 2 (r )
1
√ π 2 N
∞
2
k
−
r 4 e−3r /2ao dr .
(16.79)
3ao 0 The remaining integral over r is easily evaluated, e.g., by parts, and gives ∞ 4! , (16.80) r 4 e−β r dr 5 β 0
= 3/2a . Hence, substituting for N = 1/ 2π a , we get √ √ π 2 N 4! π 2 N 24a × 32 = (r ), rˆ (r ) = k k 3a 3a 243 β √ = 1282432a k . (16.81)
where β
=
o
1
2
2
2
5
o
3 o 5 o
o
o
Solution (c) The function 1 (r ) can be written as 1 (r )
√ where A = 2 N and α = 1/a .
= Ae−
αr
,
(16.82)
o
Consider the kinetic energy 2 1 2 2 ˆ k ˆ , E p 2m 2m where ∂2 ∂2 ∂2 2 . 2 2 2 ∂ x ∂ y ∂ z Hence, the average kinetic energy in the state 1 (r ) is
=
∇ ≡
E ˆ
1 (r ) ˆE k 1 (r )dV 2
= − 2m A =−
4π 2 2m
+
(16.84)
2
2π
0
0
A 2
2
dr d θ d φ r 2 sin θ e−αr
2
2
αr
2
e
∇ e−
αr
∇ e−
0
dr r 2 e
0
dV e−αr
= − 2m A
∞ ∞ − ∇ − π
2
(16.83)
+
= ∗
k
= − ∇
αr
.
αr
(16.85)
161
162
Quantum Theory of Hydrogen Atom
∇ e− : = − αr x e− 2
First, calculate ∂ −αr e ∂ x ∂2
−αr e ∂ x 2
αr
αr
∂
= −α ∂ x
,
− =− x r
αr
e
α
y 2
2
2
+ z − α x r r 3
2
− e
αr
.
(16.86)
Similarly, ∂2
−αr e ∂ y 2 ∂2 ∂ z 2
x 2
=− =− α
e−αr
Thus,
∂2
∇ e−αr = 2
∂ x 2
2
3
x 2
α
2
+ z − α y r r 2
2
2
+ y − α z r r 3
e−αr +
2
∂2 ∂ y 2
= − 2r α e− + α αr
e−αr +
=−
2π 2
1
2
αr
e
αr
,
.
+α
=
2
3 o
2
3
−αr e ∂ z 2 (16.88)
2
∞ 2
A
dr r 2 e−2αr
0
o 2
2
(16.87)
∂2
2
π = − m A 2α 4α α 8α . 2mα = 1/(π a ) and α = 1/a , we get A π π 1 ˆ E = 2mα = 2m π a a = 2ma . 2
2
Since A 2
e
e−αr .
2
Hence, the average kinetic energy is ∞ 2π 2 2 ˆ k 2α E A dr r e−2αr m 0
− − +
− −
(16.89)
2
k
3 o
o
2 o
(16.90)
Consider now the potential energy defined as η e2 1 ˆ V , (16.91) r 4π ε0 r where η e2 /(4π ε0 ). The average potential energy in the state 1 (r ) is given by 1 ˆ V V 1 (r )dV dV e−αr e−αr 1∗ (r ) ˆ η A 2 r
=−
=−
=
=
= −4π η A
=−
2
∞
dr r e−2αr
0
= −π η π1a a = − aη . 3 o
2 o
0
= −4π η A
2
1 4α 2 (16.92)
Quantum Theory of Hydrogen Atom
Since η
e2
= 4π ε
and
ao
0
4π ε0 2
=
me2
,
(16.93)
we finally obtain e2
η
V ˆ = − a = − 4π ε 0
1 0
ao2
ao
e2
= − 4π ε
1 4π ε0 2 ao2
0
me2
2
= − ma
2 o
.
(16.94) ˆ calculating the average value from the definition We have found V of the quantum expectation (average) value. However, there is a ˆ , simply by using the relation much quicker way to find V
V ˆ = E − E ˆ = E − E ˆ , k
(16.95)
k
where
= − 4π ε
E
e2
1
0
2ao
e2
1
= − 4π ε
0
2ao2
e2 4π ε0 2
1
= − 4π ε
ao
0
2ao2
me2
2
= − 2ma
2 o
(16.96)
is the total energy of the electron in the state 1 (r ) . Thus, 2
2
2
V ˆ = E − E ˆ = − 2ma − 2ma = − ma k
2 o
2 o
2 o
.
(16.97)
Hence,
V ˆ = −2 E ˆ
E ˆ = − 12 V ˆ .
i.e.,
k
k
(16.98)
Problem 16.7
= 1 and n = 2,
The wave functions of the electron in the states n l 1, m 0 of the hydrogen atom are
=
=
=
100
1
=
210
π ao3
e−r /ao ,
1
32π ao3
r ao
e−r /(2ao ) cos θ ,
(16.99)
where ao is the Bohr radius.
r 2 r 2 of the position (a) Calculate the standard deviation σ 2 of the electron in these two states to determine in which of these states, the electron is more stable in the position.
= −
163
164
Quantum Theory of Hydrogen Atom
(b) The electron is found in the state
=
8
e−2r /ao .
π ao3
(16.100)
Determine what is the probability that the state is the ground state (n 1) of the hydrogen atom.
=
Solution (a) Calculate first the standard deviation in the state 100 . From the definition of expectation value, we find
= ∗ = ∗ r
4π
2
r
3 o
2
100r 100 dV
4 3ao4
dr r e−2r /a
= πa
100r 100 dV
∞ ∞
3
=a
o
3 o
0
4π
= πa
3 o
dr r e−2r /a 4
o
0
= 23 a , o
8
4 3ao5
=a
3 o
4
2 o
= 3a .
(16.101)
Thus, the standard deviation in the state 100 is
= r − r = 3a − 49 a = 43 a . (16.102) In the case of the state , the average values r and r are given 2 σ 100
2
2
2 o
2 o
2 o
2
210
by the following double integrals:
= ∗ = ∗ r 2
r
∞ ∞ π
= 2π
210r 210 dV
dr
0
0
∗ r 3 210, d θ sin θ 210
π
2
210r 210 dV
= 2π
dr
0
0
∗ r 4 210 . d θ sin θ 210 (16.103)
Substituting the explicit form of 210 , we get for r
∞ ∗ ∞ ∞ − π
r = 2π
d θ sin θ 210r 3 210
dr
0
0
π
2π
= 32π a
dr
5 o
0
d θ cos 2 θ sin θ r 5 e−r /a0
0
π
1
= 16a
5 o
5
dr r e
0
d θ cos 2 θ sin θ .
r /a0
0
(16.104)
Quantum Theory of Hydrogen Atom
Since 1
π
=
2
= 32 ,
x 2 dx
d θ cos θ sin θ
−1
0
(16.105)
and
∞
dr r 5 e−r /a0
0
6 o
= 120a ,
(16.106)
we have
r = 161a
2
5 o
3
120ao6
= 5a .
(16.107)
o
Similarly, for r 2 , we get
r = 2π 2
∞ ∗ ∞ ∞ − ∞ − = π
d θ sin θ 210r 4 210
dr
0
0
π
2π
= 32π a
dr
5 o
0
d θ cos 2 θ sin θ r 6 e−r /a0
0
π
1
6
= 16a
dr r e
5 o
= 161a
0
0
2
5 o
3
d θ cos 2 θ sin θ
r /a0
dr r 6 e
r /a0
0
1
7 720 a o 5
2 o
= 30a .
24ao
(16.108)
Hence, the standard deviation in the state 210 is 2 σ 210
2 o
2 o
2 o
= 30a − 25a = 5a .
(16.109)
2 2 Since σ 100 , the electron is more stable in the state 100 than < σ 210 in the state 210 .
Solution (b) The probability is determined by the scalar product of the state and the state 100 . In other words, the probability tells us to what extent the state overlaps with the state 100 . In the modern terminology, it is called fidelity. From the definition of the scalar product of two states, we have in the spherical coordinates
165
166
Quantum Theory of Hydrogen Atom
∞
√ 8 2π
∞
= 4π dr r ∗ = π a dr r e− √ √ 8 2 2a 16 2 = a 27 = 27 ≈ 0.84. (16.110) Thus, with probability P = 0.84, the state can be considered the (, 100 )
2
100
0
3 o
3 o
ground state of the hydrogen atom.
3 o
0
2
3r /ao
Chapter 17
Quantum Theory of Two Coupled Particles
Problem 17.1 Suppose that a particle of mass m can rotate around a fixed point A , such that r constant and θ π/ 2 constant.
=
=
=
(a) Show that the motion of the particle is quantized. (b) Show that the only acceptable solutions to the wave function of the particle are those corresponding to positive energies ( E > 0) of the particle.
Solution (a) Consider the rotation in spherical coordinates. Since r and θ are constant, the rotation depends only on the azimuthal angle φ . In this case, the Schr¨odinger equation simplifies to 2
− 2mr
2
∂ 2 ∂φ 2
= E ,
Problems and Solutions in Quantum Physics Zbigniew Ficek c 2016 Pan Stanford Publishing Pte. Ltd. Copyright ISBN 978-981-4669-36-8 (Hardcover), 978-981-4669-37-5 (eBook) www.panstanford.com
(17.1)
168
Quantum Theory of Two Coupled Particles
which can be written as ∂ 2
=−
2mr 2 E
(17.2) . 2 ∂φ 2 We see that the wave function of the rotating mass satisfies a simple differential equation of harmonic motion ∂ 2
2
∂φ 2
= −β ,
(φ )
= Ae
(17.3)
whose solution is i βφ
,
(17.4)
2
2mr E where A is a constant and β 2 . 2 Since in rotation (φ ) (φ 2π ), we find that this is satisfied when
=
=
+
e2π iβ
= 1, i.e., when β is an integer (β = 0, ±1, ±2, . . .).
(17.5)
Hence, β 2 is not an arbitrary number but an integer. This shows that the energy in the rotation is quantized.
Solution (b) ¨ For E < 0, the Schrodinger equation takes the form ∂ 2
=
2mr 2 E
| | = β 2
β 2 > 0.
2 ∂φ 2 The solution to the above differential equation is of the form
(φ )
= Ae + B e− βφ
βφ
.
(17.6)
(17.7)
This is a damped function that does not describe rotation. Thus, it is not an acceptable solution to the wave function of the rotating mass. ¨ For E > 0, the Schrodinger equation is of the form ∂ 2
=−
2mr 2 E
2
= −β
β 2 > 0.
∂φ 2 2 The solution to the above differential equation is of the form (φ )
i βφ
= Ae
,
(17.8)
(17.9)
which describes rotation. Thus, it is an acceptable solution to the wave function of the rotating mass.
Chapter 18
Time-Independent Perturbation Theory
Problem 18.1 In an orthonormal basis, a linear operator Aˆ is represented by the matrix 2λ 1 λ , (18.1) Aˆ 1 λ λ where λ is a small real parameter (λ 1). The operator Aˆ can be written as the sum of two operators, Aˆ Aˆ 0 λ ˆ V , where
=
+
=
+
= + ˆ = 21 . V 11
01 , (18.2) Aˆ 0 10 Using the first-order perturbation theory, find the eigenvalues and eigenvectors of Aˆ in terms of the eigenvalues and eigenvectors of Aˆ 0 . Notice that Aˆ 0 is of the same form as the x -component of the electron spin, σ ˆ x .
Solution The unperturbed states are the eigenstates of the operator Aˆ 0 . Since Aˆ 0 ˆ x , the unperturbed eigenstates and the corresponding σ
=
Problems and Solutions in Quantum Physics Zbigniew Ficek c 2016 Pan Stanford Publishing Pte. Ltd. Copyright ISBN 978-981-4669-36-8 (Hardcover), 978-981-4669-37-5 (eBook) www.panstanford.com
170
Time-Independent Perturbation Theory
eigenvalues are (for details see Tutorial Problem 13.4) (0) 1
|φ
(0) 2
|φ
= √ = √ 1
1 1
2
1
1 1
−
2
(0)
,
= 1.
E 1
(0)
,
E 2
= −1.
(18.3)
(0)
is equal to the
The first-order correction to the eigenvalue E 1 ˆ in the state φ (0) . Hence, expectation value of V 1
|
(1)
E 1
(0) 1
1
(0) 1
= φ |V ˆ |φ = 2 (1 1) 1
= 2 (1 1)
= 3 2
1
(3
2
21 11
1 1
+ 2) = 25 .
(18.4) (0)
Similarly, the first-order correction to the eigenvalue E 2 is (1)
E 2
(0) 2
1
(0) 2
= φ |V ˆ |φ = 2 (1 − 1) 1
= 2 (1 − 1)
= 1 0
1 2
21 11
1 1
−
.
(18.5)
Thus, the eigenvalues of Aˆ to the first order in λ are E 1
(0) 1
(1) λ E 1
2
(0) 2
(1) λ E 2
= E + E = E +
= 1 + 25 , = −1 + 21 λ
λ.
(18.6)
Calculate now the first-order corrections to the eigenvectors. (0) The first-order correction to the eigenvector φ1 is
|
(0) 2 (0) 1
(0) 1 (0) 2
1 2
(1
− 1) 21 11 1 − (−1)
1 1
ˆ |φ φ | V | = |φ = |φ E − E 3 (1 − 1) 2 1 1 = | φ = (3 − 2) |φ = |φ . 2 4 4 (1) φ1
1 2
(0) 2
(0) 2
(0) 2
(0) 2
(0) 2
(18.7)
Time-Independent Perturbation Theory
(0)
| is
Similarly, the first-order correction to the eigenvector φ2 (0) 1 (0) 2
(0) 2 (0) 1
ˆ |φ φ | V |φ = |φ = E − E (1) 2
1 2
(1 1)
=
1 0
−2
(0) 1
1 2
(1 1)
21 11
1 1
− |φ (−1) − 1
|φ = − 14 |φ . (0) 1
(0) 1
(0) 1
(18.8)
Thus, the eigenvectors of Aˆ to the first order in λ are 1
(0) φ1
2
(0) 2
(1) λ φ1
(0) φ1
(1) 2
(0) 2
1
λ
(0) 2
λ
(0) 1
|φ = | + | = | + 4 |φ , |φ = |φ + |φ = |φ − 41 |φ . λ
(18.9)
171
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Chapter 19
Time-Dependent Perturbation Theory
Problem 19.1 Consider a two-level atom represented by the spin operators σ ˆ ± , σ ˆ z , interacting with a one-dimensional harmonic oscillator, represented by the creation and annihilation operators aˆ † and aˆ . The Hamiltonian of the system is given by 1 1 1 ˆ ˆ + aˆ σ ˆ − aˆ † . (19.1) ω0 ˆ σ z ω0 aˆ † aˆ H i g σ 2 2 2 The Hamiltonian can be written as
=
+
+
ˆ H
−
−
= H ˆ + V ˆ ,
where ˆ0 H
= 2 ω ˆσ + ω
ˆ V
0 z
1
= − 2 i g
(19.2)
0
1
0
aˆ † aˆ
1
+2
,
+− − ˆ aˆ σ
ˆ aˆ † . σ
(19.3)
ˆ 0 are product states The eigenstates of H
|φ = |n|1, |φ − = |n − 1|2, n
|
n 1
(19.4)
where n is the photon number state of the harmonic oscillator and 1 , 2 are the energy states of the atom.
||
Problems and Solutions in Quantum Physics Zbigniew Ficek c 2016 Pan Stanford Publishing Pte. Ltd. Copyright ISBN 978-981-4669-36-8 (Hardcover), 978-981-4669-37-5 (eBook) www.panstanford.com
174
Time-Dependent Perturbation Theory
(a) Write the state vector of the system in terms of the eigenstates ˆ 0. of H (b) Assume that initially at t 0, the system was in the state φn . Find the probability, using the time-dependent perturbation theory, that after a time t , the system can be found in the state φn−1 .
=
|
|
Solution (a) The state vector of the system is given by
=
| (t ) where E m is E n
cm (t )e−
i
E m t
|φ , m = n, n − 1, (19.5) is the energy of the state |φ . The energy of the state |φ
m
m
m
n
1
= φ | H ˆ |φ = 2 ω 1|σ ˆ |1 + ω n
0
0
n
0
z
| n
aˆ † aˆ
1
+2
| n
= − 12 ω + n ω + 21 ω = n ω . (19.6) The energy of the state |φ − is ˆ |φ − E − = φ − | H = 21 ω 2|σ ˆ |2 + ω n − 1| aˆ aˆ + 21 |n − 1 0
0
0
0
0
n 1
n 1
n 1
0
0
n 1
z
†
= 21 ω + (n − 1) ω + 21 ω = n ω . (19.7) = E − , i.e., the states |φ and |φ − are degenerate. 0
0
0
0
Thus, E n n 1 n n 1 Hence, the state vector of the system is of the form
| (t )
=
cm (t )e−i nω0 t φm ,
|
m
m
= n, n − 1.
(19.8)
The unknown coefficients cm (t ) can be determined using the timedependent perturbation theory. We shall limit our calculations to the first-order corrections. ˆ is independent of time, and Since the interaction Hamiltonian V cn(0) (0), the first-order corrections to the assuming that cn(0) (t ) amplitudes c m (t ) are
=
(1) (t ) cm
V mk cn(0) (0)
= − E − E m
k
i ωmk t
e
−
1 .
(19.9)
Time-Dependent Perturbation Theory
Hence, the first-order correction to cn (t ) is (0)
cn(1) (t )
E n
− E −
n 1
i ωn, n−1 t
e
−
1 .
(19.10)
= E − = n ω , we get using the Taylor expansion V − c − (0) i (1 + i ω − t + . . . − 1) =− c − (0)V (t ) =− ω −
Since E n cn(1)
=−
V n, n−1 cn−1 (0)
n 1
0
(0) n, n 1 n 1
(0) n 1
n, n 1
n, n 1
n, n 1 t .
−
(19.11)
The explicit value of the matrix element V n, n−1 is
= φ |V ˆ |φ − = − 12 i g1|n|σ ˆ + aˆ |2|n − 1 + 21 i g1|n|σ ˆ −aˆ |2|n − 1 (19.12) = 0 + 21 i g√ n = 21 i g√ n,
V n, n−1
n
n 1
†
where we have used the results
1|σ ˆ +|2 = 0, 1|σ ˆ −|2 = 1, n|aˆ |n − 1 = 0, n|aˆ |n − 1 = √ n. †
(19.13)
Thus, cn(1) (t )
1
=2
(0) cn 1 (0) g
−
√ nt .
(19.14)
Consider now the first-order correction to cn−1 (t ), which is given by (1) cn 1 (t )
−
=− =−
V n−1, n cn(0) (0) E n−1
− E
n
V n−1, n cn(0) (0) ωn 1, n
= − i c
i ωn−1, n t
e
(1
−
− 1
+ i ω − t + . . . − 1) n 1, n
(0) n (0) V n 1, n t .
(19.15)
−
Calculating the value of the matrix element V n−1, n , we get V n−1, n
= φ − |V ˆ |φ = − 12 i g2|n − 1|σ ˆ +aˆ |1|n + 21 i g2|n − 1|σ ˆ −aˆ |1|n √ √ 1 1 = − 2 i g n + 0 = − 2 i g n. (19.16) n 1
n
†
Thus, (1) cn 1 (t )
−
= − 12 c
(0) n (0) g
√ nt .
(19.17)
175
176
Time-Dependent Perturbation Theory
Hence, in the first order, the coefficients cn (t ) and cn−1 (t ) are
√ 1 = + c (t ) c (0) c − (0) g nt , 2 √ 1 c − (t ) = c − (0) − c (0) g nt . 2 n
n 1
n
n 1
n 1
(19.18)
n
Thus, the state vector of the system, in the first order of the interaction, is of the form
| (t )
=
cn (0)
1
+ 2c −
n 1
√ (0) g nt |φ + c
n
√ − (0) − 2 c (0) g nt |φ − .
1
n 1
n
n 1
(19.19)
Solution (b) The probability of a transition that after time t the system, initially at t 0 in the state φn , will be found in the state φn−1 is given by the (1) absolute square of the amplitude cn−1 (t ):
=
P n→n−1 (t )
=|
and
V n
(1) cn 1 (t ) 2
−
Since
| = − −− − − = −
i ωn−1, n t
e
1
E n−1
E n
2
1, n
i ωn−1, n t
e
E n
1
1
2
4sin
2
ωn
1
,
1, n t
− E = ω −
n 1, n ,
n
2
.
(19.20)
(19.21)
(19.22)
the transition probability simplifies to P n→n−1 (t )
2 2
= | V − | t n 1, n
2
sin 12 ωn 1, n t 2 1 t ω n n 1, 2 2
−
−
.
(19.23)
Since the states φn and φn−1 are degenerate in energy, i.e., ωn−1, n the function sin2 x / x 2 1, and then
=
2 2
| V − | t = 1 g nt . P → − (t ) = 4 n 1, n
n
n 1
2
2
2
= 0,
(19.24)
The probability is proportional to the strength of the interaction, g2 , number of photons in the field, n , and the square of interaction time, t 2 . Since V n−1, n 2 V n, n−1 2 , we see that the probability of transitions between the two states is the same in either direction.
|
| =|
|
Chapter 20
Relativistic Schr¨ odinger Equation
Problem 20.1 Show that the Klein–Gordon equation for a free particle is invariant under the Lorentz transformation. The Lorentz transformation is given by x
= γ ( x − β ct ), y = y , z = z , ct = γ (ct − β x ),
−
1/2
where γ 1 β2 is the Lorentz factor, β velocity an observed moves.
= −
(20.1)
= u/c, and u is the
Solution In order to show that the Klein–Gordon equation for a free particle is invariant under the Lorentz transformation, we have to demonstrate that the equation has the same form in both (t , x , y , z ) and (t , x , y , z ) coordinates. Problems and Solutions in Quantum Physics Zbigniew Ficek c 2016 Pan Stanford Publishing Pte. Ltd. Copyright ISBN 978-981-4669-36-8 (Hardcover), 978-981-4669-37-5 (eBook) www.panstanford.com
178
¨ Relativistic Schr odinger Equation
Let us start from the Klein–Gordon equation in the ( t , x , y , z ) coordinates
+
m2 c2
2
= 0.
(20.2)
and using the Lorentz transformation (20.1), we shall demonstrate that in the (t , x , y , z ) coordinates it has the form m2 c2
+
2
= 0.
(20.3)
We see that to demonstrate the invariance of the Klein–Gordon equation under the Lorentz transformation, it is enough to show that . Since
=
1 ∂ 2
= c
2
∂ t 2
∂ 2
2
∂ 2
∂ 2
∂ 2
2
2
∂ z 2
− ∇ = ∂ (ct ) − ∂ x − ∂ y − 2
,
(20.4)
we have to find how in the above equation, the second-order derivatives given in the (t , x , y , z ) coordinates transform to those in the (t , x , y , z ) coordinates. Consider the first-order derivative over time. Since ct is a function of ct and x , we apply the chain rule and obtain ∂ ∂ (ct )
∂ ∂ (ct )
∂ ∂ x
= ∂ (ct ) ∂ (ct ) + ∂ x ∂ (ct ) .
(20.5)
From Eq. (20.1), we have ∂ (ct ) ∂ (ct )
∂ x
= γ ,
∂ (ct )
= −βγ .
(20.6)
Hence, ∂
∂ ∂ = − . γ βγ ∂ (ct ) ∂ (ct ) ∂ x
(20.7)
Then the second-order derivative over time is 2
∂ ∂ (ct )2
= ∂ (∂ct ) ∂∂ (ct ) 2
= γ
2
∂
=
γ
∂
∂ − βγ ∂ (ct ) ∂ x 2
2
γ
∂
∂ − βγ ∂ (ct ) ∂ x
∂ ∂ − − +β β β ∂ (ct ) ∂ (ct )∂ x ∂ x ∂ (ct ) 2
2∂
2
∂ x 2
. (20.8)
¨ Relativistic Schr odinger Equation
Consider now the first-order derivative over x : ∂ ∂ ∂ x ∂ ∂ (ct ) ∂ x
= ∂ x ∂ x + ∂ (ct )
∂ x
.
(20.9)
Since ∂ (ct ) ∂ x
∂ x
= −βγ ,
∂ x
= γ ,
(20.10)
we get ∂
∂ ∂ = . γ − βγ ∂ x ∂ x ∂ (ct )
(20.11)
Then, the second-order derivative over x is ∂ 2 ∂ x 2
∂ ∂
= ∂ x ∂ x
2
= γ Since y are
2
=
∂
∂ x 2
γ
∂
∂
− βγ ∂ (ct ) ∂ x 2
γ
∂ ∂ x
∂
− βγ ∂ (ct )
2
∂ − β ∂ x ∂∂ (ct ) − β ∂ (ct )∂ x + β
2
∂
2
∂ (ct )2
.
(20.12)
= y and z = z , the second-order derivatives over y and z ∂ 2
=
∂ 2
∂ 2
=
∂ 2
, . ∂ y 2 ∂ y 2 ∂ z 2 ∂ z 2 Collecting the results (20.8), (20.12), and (20.13), we get
∂ 2
∂ 2
∂ 2
∂ 2
2
2
∂ z 2
= ∂ (ct ) − ∂ x − ∂ y − 2
2
= γ
2
−γ
∂ 2
∂ 2
∂ (ct )2
∂ 2 ∂ x 2
∂ 2
∂ 2
− β ∂ (ct )∂ x − β ∂ x ∂ (ct ) + β ∂ 2
∂ 2
− β ∂ x ∂ (ct ) − β ∂ (ct )∂ x + β 2
2
∂ 2
2
2
2 2∂
2
∂ x 2
(20.13)
∂ 2
∂ (ct )2
−
∂ 2
∂ 2
2
2
− ∂ z = γ (1 − β ) ∂ (ct ) − γ (1 − β ) ∂ x − ∂ y 2
∂ 2
− ∂ z
2
2
.
However, γ 2 (1
∂ 2 ∂ y 2
(20.14) 2
− β ) = 1. Therefore ∂ ∂ ∂ ∂ − − − = = . ∂ (ct ) ∂ x ∂ y ∂ z 2
2
2
2
2
2
2
2
(20.15)
This shows that the Klein–Gordon equation has the same form in both coordinates. In other words, the Klein–Gordon equation is invariant under the Lorentz transformation.
179
180
¨ Relativistic Schr odinger Equation
Problem 20.2 Act on the Dirac equation E
with the operator
β mc2
(20.16)
2
(20.17)
− · − cα p
= 0
+ cα · p + β mc
E
to find under which conditions the Dirac equation satisfies the relativistic energy relation E 2
2
2
2 4
= c p + m c .
(20.18)
ˆ α z k ˆ is a three-dimensional Hermitian operator Here, α α x iˆ α y j and β is a one-dimensional Hermitian operator. The operator β does not commute with any of the components of α .
= + +
Solution This tutorial problem follows closely the derivation of the Dirac equation presented in the textbook. We particularly feel that the derivation should be discussed in details especially that the Dirac equation is not usually introduced at the basic level of quantum mechanics, but at the advanced level. We adopt here a simple vectorial formalism and show that the basic concepts of the relativistic Schrodinger ¨ equation can be easily understood in terms of the vector analysis and matrix multiplication. Let us act on the equation β mc2
(20.19)
2
(20.20)
− · − E
cα p
= 0
from the left with the operator
+ cα · p + β mc .
E
We then have β mc2
β mc2
− · − + · + E
cα p
E
c α p
= 0.
(20.21)
One might worry why the quantities E cα p β mc2 and E cα p β mc2 are called operators if they involve scalars and vectors only. The reason is that α is a three-dimensional (vector) matrix and
·+
− · −
+
¨ Relativistic Schr odinger Equation
as such α can be treated as the matrix representation of an operator αˆ . Performing the multiplication of the terms in Eq. (20.21), we get
2
E
2
2
3
2
2 4
− c (α · p ) − mc [(α · p ) β + β (α · p )] − β m c
= 0.
(20.22) The scalar product α p, appearing in the above expression, can be written in terms of components
·
· = α p + α p + α p .
α p
x x
y y
(20.23)
z z
Squaring this expression gives 2
·
(α p )
=
+ α p + α p α p + α p + α p = α p + α p + α p + α α + α α p p + α α + α α p p + (α α + α α ) p p . α x p x
y y
2 2 x x
z z
2 2 y y
2 2 z z
y z
z y
y z
x x
y y
z z
x y
z x
y x
x y
x z
z x
(20.24)
Using the results of Eqs. (20.23) and (20.24) in Eq. (20.22) yields
2
α x 2 p x 2
2
y z
3
+
2 2 α y p y
α z 2 p z 2
+
+ α α + α α p p + α α + α α p p + (α α + α α ) p p −mc (α β + βα ) p + α β + βα p + (α β + βα ) p (20.25) −β m c = 0. E
−c
2
z y
x
2 4
y z
x
z x
x
x y
x z
y
y
y x
z x
y
z
x y
z
z
We require this equation to be equal to
2
E
2
2
2 4
− c p − m c
= 0.
(20.26)
Comparing terms in Eqs. (20.25) and (20.26), we find the following. Since p2
2 x
2 y
2 z
= p + p + p ,
(20.27)
we see that the second term in Eq. (20.25), that multiplied by c2 , will be equal to p2 if α x 2
2 y
2 z
= α = α = 1,
(20.28)
and
+ α α = [α , α ]+ = 0, α α + α α = [α , α ]+ = 0, (α α + α α ) = [α , α ]+ = 0. α x α y
y x
x
y
y z
z y
y
z
z x
x z
z
x
(20.29)
181
182
¨ Relativistic Schr odinger Equation
The third term in Eq. (20.25), that multiplied by mc3 , is absent in Eq. (20.26). Therefore,
+ βα = 0, α β + βα = 0, α β + βα = 0. α x β
x
y
y
z
z
(20.30)
Finally, comparing the fourth term in Eq. (20.25) with Eq. (20.26), we see that β2
= 1.
(20.31)
Thus, the Dirac equation satisfies the relativistic energy relation under the condition that the four relations (20.28)–(20.31) are simultaneously satisfied. Under these conditions, the Dirac equation ¨ can be treated as the relativistic form of the Schrodinger equation.
Chapter 21
Systems of Identical Particles
Problem 21.1 Consider a system of three identical and independent particles.
(a) What would be the level of degeneracy if particle 1 of energy n1 2 would be distinguished from the other two particles? (b) What would be the level of degeneracy if the distinguished particle has energy n1 1?
=
=
Solution (a) If a single excitation is present in the system of three identical particles, there are three combinations possible of which of the particles is excited, (n1 2, n2 1, n3 1), (n1 1, n2 2, n 3 1), and (n1 1, n 2 1, n3 2). Thus, for three identical particles, the degeneracy of the single excitation level is three. If the excited particle is distinguished from the other two particles, then there is only one combination possible ( n1 2, n2 1, n3 1). Hence, in this case, the level of degeneracy is one.
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Problems and Solutions in Quantum Physics Zbigniew Ficek c 2016 Pan Stanford Publishing Pte. Ltd. Copyright ISBN 978-981-4669-36-8 (Hardcover), 978-981-4669-37-5 (eBook) www.panstanford.com
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184
Systems of Identical Particles
Solution (b)
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If the distinguished particle is in its ground state (n1 1), the level of degeneracy would be two, as there are two possible combinations of n2 and n 3 with the single excitation: (n1 1, n2 2, n 3 1) and (n1 1, n2 1, n3 2).
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Problem 21.2 Two identical particles of mass m are in the one-dimensional infinite potential well of dimension a. The energy of each particle is given by E i
=
π 2 2
ni2
2 i
= n E . 2ma 2
(21.1)
0
(a) What are the values of the four lowest energies of the system? (b) What is the degeneracy of each level.
Solution (a) The total energy of the two particles is
= E + E =
E
1
2
(n21
+
2 2 2 π n2 ) 2ma2
2 1
2 2
= (n + n ) E . 0
(21.2)
Hence 2 1
2 2
= (n + n )
E / E 0
(21.3)
determines the energies of the system. The first lowest energy level is for n1 n2 1 at which E / E 0 2. The second lowest energy level is for either (n1 2, n2 1) or (n1 1, n2 2) and the energy of this level is E / E 0 5. The third lowest energy level is for n1 n2 2 at which E / E 0 8. The fourth lowest energy level is for either (n1 3, n2 1 ) o r (n1 1, n2 3) and the energy of this level is E / E 0 10.
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Systems of Identical Particles
Solution (b)
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n2 For n1 1, there is only one wave function 11 , so the degeneracy of the first lowest level is one. There are two sets of n ’s numbers (n1 2, n 2 1) or (n1 1, n2 2), which determine the second lowest energy level. The wave functions corresponding to those combinations are 21 and 12 . Therefore, the degeneracy of this level is two. There is only one set of numbers n1 n2 2, which determines the third lowest energy level. Therefore, the degeneracy of the level is one. For the fourth lowest energy level, there are two sets of numbers (n1 3, n2 1) or (n1 1, n2 3). Thus, there are two wave functions 31 and 13 corresponding to those combinations. Therefore, the degeneracy of the fourth lowest energy level is two.
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Problem 21.3 Redistribution of particles over a finite number of states
(a) Assume we have n identical particles that can occupy g identical states. The number of possible distributions, if particles were bosons, is given by the number of possible permutations
+ g − 1)! t = . (21.4) n!( g − 1)! = 2 and g = 3 give t = 6. However, this is (n
For example, n true only for identical bosons. What would be the number of possible redistributions if the particles were fermions or were distinguishable? (b) Find the number of allowed redistributions if the particles were: (i) (ii) (iii) (iv)
Identical bosons. Identical fermions. Non-identical fermions. Non-identical bosons.
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Ilustrate this with the example of n 2 independent particles that can be redistributed over five different states.
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186
Systems of Identical Particles
Solution (a) Since two fermions cannot occupy the same state, only three redistributions are possible: (1, 1, 0), (1, 0, 1), (0, 1, 1). If the two particles are distinguishable, then each has three available states and then the total number of redistributions is 3 3 9.
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Solution (b)
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(i) According to Eq. (21.4), for n 2 identical bosons, there are 5 states. The allowed t 15 allowed distributions over g distributions are
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11000
01010
20000
10100
01001
02000
10010
00110
00200
10001
00101
00020
01100
00011
00002
(21.5)
where, e.g., 11000 represents the system state in which each of the first and second states contains one particle, while the remaining states contain none. There are 15 possible distributions of which 10 have the two particles in different states and 5 have the two particles in the same state. (ii) Two identical fermions cannot occupy the same state. Therefore, the five-system states in the right column of Eq. (21.5) are not allowed. Thus, there are 10 allowed system states for identical fermions. (iii) For two non-identical fermions, each of the states with two particles in different states, left and middle columns, is doubly degenerated. Therefore, there are 20 allowed system states for non-identical fermions. (iv) For two non-identical bosons, each of the states with two particles in different states is doubly degenerated. The right column is also allowed for non-identical bosons, so there are 25 allowed system states for non-identical bosons.
