Special Topics In OneDimensional Quantum Mechanics
Selected Exercises In Spatial and Momentum Translations
Spiros Konstantogiannis
Special Topics Topics In One-Dimensional Quantum Mechanics Selected Exercises In Spatial and Momentum Translations Spiros Konstantogiannis Physicist, M.Sc.
Copyright © 2017 by Spiros Konstantogiannis. All rights reserved. No part of this eBook may be reproduced, in any form or by any means, without the express written permission of the writer.
2
Contents
Contents Contents ................................ ................................................ ................................. .................................. ................................. .............................. .............. 3 Preface Preface .................................. .................................................. ................................. .................................. ................................. .............................. .............. 4 A note on the notation... notation.................... .................................. ................................. ................................. .................................. ...................... ..... 5 1. Spatial Spatial translation translation operators operators ................................ ................................................. ................................. .............................. .............. 7 2. Momentum translation operators.......................................................................15 3. Exercises Exercises ................................ ................................................. .................................. ................................. ................................. ........................17 .......17 The coherent states of the quantum harmonic oscillator (QHO) as the result of the action of spatial and momentum translation operators on the QHO ground state ................................ ................................................ ................................. .................................. .................................. ................................3 ...............34 4 The charged QHO in in a homogeneous homogeneous electric field .................. ......... .................. ................... ................67 ......67 The most general unitary operator that can be constructed by a linear combination of the position position and momentum momentum operators .................. ........ ................... .................. ...........88 ..88 Momentum translations in the infinite potential well.........................................89 Momentum translations in the 1-d attractive Coulomb potential......................108 Momentum translations in the 1-d attractive delta potential.............................115 4. Selected Selected references references .................................. ................................................... ................................. ................................. ......................120 .....120
3
Preface
The present eBook is a – hopefully hopefully successful – attempt attempt to present, in the framework of quantum mechanics, the spatial and momentum translation operators in one dimension, through a series of specifically designed exerc ises and applications. The eBook is divided into two parts. In the first two sections, the translation operators are defined and their basic properties are proved. It follows a series of problems with detailed solutions, covering a number of aspects o f translations applied to various onedimensional quantum systems, such as the harmonic oscillator, the infinite well, the attractive delta potential, and the attractive Coulomb potential, with particular emphasis on the harmonic oscillator, which serves as a model system for both spatial and momentum translations. The reader is assumed to have a basic knowledge of the postulates and the mathematical formalism of quantum mechanics, including the Dirac notation and the ladder operator method of the harmonic oscillator.
4
A note on the notation
We reserve
0
and
p0 to
denote, respectively, the length and momentum scale of the
quantum harmonic oscillator (QHO). Thus, we use spatial and momentum translations.
5
x1 and p1 to
denote, respectively,
Exercises
æ ip1 x1 ö as we know – does does not change the ÷ is a constant phase that – as è h ø
The factor exp ç physical
content
æ ip1 x1 ö ˆ ˆ ÷ T x T p h è ø
exp ç
1
1
y
of
the
quantum
states.
Thus, the
states
Tˆ p1T ˆx1
y
and
are physically equivalent. In other words, the action of the
ˆ on an arbitrary quantum state is physically the same as the action of the product Tˆ x1 T p1 ˆ on the same state. product Tˆ p1T x1
Another way of proving (1) is by using the property v of the momentum translation ˆ ö æ ipx operators, for f ( pˆ ) = Tˆ x1 ( pˆ ) = exp ç - 1 ÷ . Indeed, we have è h ø
æ i ( pˆ + p1 ) x1 ö æ ipˆ x1 ip1 x1 ö † Tˆ p1 Tˆx1 ( pˆ ) Tˆp1 = Tˆx1 ( pˆ + p1 ) = exp ç e x p = ÷ ç- h - h ÷ = h è ø è ø æ ipˆ x ö æ ip x ö æ ip x ö = exp ç - 1 ÷ exp ç - 1 1 ÷ = T ˆ x exp ç - 1 1 ÷ è h ø è h ø è h ø 1
That is
æ ip x ö æ ip x ö † † Tˆ p1 Tˆx1Tˆp1 = Tˆx1 exp ç - 1 1 ÷ Þ Tˆp1Tˆp1 Tˆx1Tˆp1 = Tˆp1 Tˆx1 e xp ç - 1 1 ÷ Þ h ø è è h ø {
1
æ ip x ö exp ç 1 1 ÷ è h ø 14243
Þ Tˆ x Tˆp 1
1
æ ip x ö = Tˆp Tˆx Þ Tˆp Tˆx = exp ç 1 1 ÷Tˆx Tˆp è h ø 1
1
1
1
1
1
Constant, it can be moved to the left
The coherent states of the quantum harmonic oscillator (QHO) as the result of the action of spatial and momentum translation operators on the QHO ground state 9) The coherent states of the QHO are the eigenstates of the annihilation i ö pˆ ÷ . They are states of minimum positionç xˆ + 2h è mw ø æhö momentum uncertainty ç ÷ , thus the respective wave functions, in both the è2ø
operator
aˆ =
mw æ
position and the momentum representation, are Gaussian wave forms. In coherent states, the time evolution of the position and momentum expectation values follows the classical laws, i.e. the formulas giving the time evolution of the position and momentum of a classical harmonic oscillator. We ll show that the coherent states are generated by the action of the product Tˆ p1 T ˆx1 on the ground state 0 of the QHO, for all the values of the spatial and ’
momentum translations,
and p1 , respectively. From the exercise 8, it is obvious that the same applies for the product Tˆ T ˆ too. 1
x1
Solution
34
p1
Exercises
We begin by writing the annihilation operator as a function of the length scale h
x0 =
and momentum scale p0 = mhw of the QHO. We have
mw
æ ö ç ÷ 1 æ xˆ mw æ i pˆ pˆ ö ö 1 ç xˆ ÷ ˆ ˆ aˆ = x p i i + = + = + ç ÷ ç ÷ 2h è mw ø p0 ø 2ç mhw ÷ 2 è x0 h ç ÷ è mw ø That is 1 æ xˆ
aˆ =
ç 2è
0
+i
pˆ ö
÷ (1)
p0 ø
Observe the symmetry and simplicity in (1). The factor
1 2
can be thought of as a
kind of normalization constant. Now, we would like to know how the annihilation operator is transformed transfor med by the ˆ , i.e. we want to calculate the product action of the product Tˆ p1T x1
(
Tˆ 1Tˆ x1
†
)
aˆ ( xˆ, pˆ )Tˆp1Tˆx1 (the annihilation operator is a function – actually a linear
function – of of the t he position and momentum operators). We have †
(Tˆ Tˆ ) 1
x1
aˆ ( xˆ, pˆ ) Tˆp1Tˆx1 = Tˆx1 †Tˆp1 †aˆ ( xˆ, pˆ )Tˆp1Tˆx 1
† Using the property Tˆ p1 f ( xˆ , pˆ ) Tˆp1 = f ( xˆ , pˆ + p1 ) of the momentum translation
operators for f ( xˆ , pˆ ) = aˆ ( xˆ, pˆ ) , we obtain
Tˆ p1 † aˆ ( xˆ, pˆ ) Tˆp1 = aˆ ( xˆ , pˆ + p1 ) Thus
(
Tˆ p1Tˆx1
†
)
† aˆ ( xˆ, pˆ ) Tˆp1Tˆx1 = Tˆx1 aˆ ( xˆ, pˆ + p1 )Tˆx1
† Now, using the property Tˆ x1 f ( xˆ, pˆ ) Tˆx1 = f ( xˆ + x1 , pˆ ) of the spatial translation
operators for f ( xˆ , pˆ ) = aˆ ( xˆ , pˆ + p1 ) , we obtain
(
Tˆ p1Tˆx1
†
)
aˆ ( xˆ, pˆ ) Tˆp1Tˆx1 = aˆ ( xˆ + x1 , pˆ + p1 ) (2)
From (1) we obtain aˆ ( xˆ + x1 , pˆ + p1 ) =
1 æ xˆ + x1
ç 2è
+i
0
pˆ + p1 ö p0
1 æ xˆ pˆ ö 1 æ x1 p1 ö ÷= ç +i ÷+ ç +i ÷ p0 ø p0 ø 2 è x0 2 è x0 ø
That is aˆ ( xˆ + x1 , pˆ + p1 ) = aˆ ( xˆ, pˆ ) +
1 æ
ç 2è
1 0
+i
p1 ö
÷ (3)
p0 ø
35
Exercises
Substituting (3) into (2) yields †
(Tˆ Tˆ ) p1
x1
1 æ x1 p1 ö aˆ ( xˆ , pˆ ) Tˆp1Tˆx1 = aˆ ( xˆ , pˆ ) + + i ç ÷Þ p0 ø 2 è x0 †
Þ Tˆ p Tˆx (Tˆp Tˆx 1
1
1
1
)
æ 1 æ x1 p1 ö ö aˆ ( xˆ, pˆ ) Tˆp1Tˆx1 = Tˆp1Tˆx1 ç aˆ ( xˆ , pˆ ) + + i ç ÷÷ ç p0 ø ÷ø 2 è x0 è
But
(
Tˆ p1Tˆx1 Tˆp1Tˆx1
)
†
= Tˆp Tˆx Tˆx † Tˆp † = Tˆp Tˆp † = 1 1
1
1
{
1
1
1
1
ˆ is also a unitary operator. As the product of two unitary operators, Tˆ p1 T x1
Thus 1 æ x1 p1 ö aˆ ( xˆ , pˆ ) Tˆ p1Tˆx1 = Tˆp1Tˆx1 aˆ ( xˆ, pˆ ) + ç + i ÷ Tˆp1Tˆx1 p0 ø 2 è x0 Or 1 æ x1 p1 ö ˆ ˆ aˆTˆ p1Tˆx1 = Tˆp1Tˆx1 aˆ + + i ç ÷ T p Tx (4) p0 ø 1 1 2 è x0 where aˆ = aˆ ( xˆ, pˆ ) . If
l ¢
is an arbitrary eigenstate of aˆ with eigenvalue
l ¢ Î £
(the annihilation
operator is not Hermitian, therefore its eigenvalues is generally complex numbers), then acting on l ¢ by both members of (4) yields aˆTˆ p1Tˆx1
l¢
1
= l ¢Tˆ p Tˆx 1
= Tˆp Tˆx aˆ l ¢ + l¢
+
1
1 æ x1
ç
2 è x0
1
+i
1 æ x1
ç
2 è x0
+i
p1 ö ˆ ˆ ÷ T p Tx l ¢ = p0 ø 1 1
æ 1 æ x1 p1 ö ˆ ˆ p1 ö ö ˆ ˆ ¢ ¢ T T = + + i l l ç ÷ p x ç ÷ ÷÷ T p1Tx1 l ¢ ç p0 ø 1 1 x p 2 0 øø è 0 è
That is
æ 1 æ x1 p öö aˆTˆ p1Tˆx1 l ¢ = ç l ¢ + + i 1 ÷ ÷ Tˆp1Tˆx1 l ¢ (5) ç ç p0 ø ÷ø 2 è x0 è ˆ From (5) we deduce that Tˆ p1 T x1
eigenvalue
l ¢ +
1 æ x1
ç 2è
0
+i
l ¢
is an eigenstate of the annihilation operator, with
p1 ö
÷.
p0 ø
ˆ is unitary, and thus it preserves the norms of the states on which it The operator Tˆ p1 T x1
acts. Therefore ˆ l Tˆ p1 T x1
=
l
=1
36
Exercises
ˆ Thus, the state Tˆ p1 T x1
eigenvalue
1 æ x1
l ¢ +
ç 2è
l ¢ =
0
p1 ö
÷ , i.e.
p0 ø
p1 ö ç + i ÷ (6) p0 ø 2 è x0
0 we obtain
p1 ö ç + i ÷ (7) p0 ø 2 è x0
The parameter
1 –
the parameter
p1
i.e. the spatial translation – can i.e. can be any real number and so can be
p1 ö
i.e. the momentum translation. Thus, the eigenvalue
–
x
p
1 + i 1 can be any complex number (including zero). ç ÷= p0 ø 2 è x0 2 0 2 p0 1 æ x1 p1 ö + Therefore, the eigenstate i ç ÷ is an arbitrary eigenstate of the p0 ø 2 è x0 1
+i
+i
1 æ x1
Tˆ p1 Tˆx1 0 =
1 æ
is a normalized eigenstate of the annihilation operator, with
1 æ x1
Tˆ p1 Tˆx1 l ¢ = l ¢ +
Choosing
l ¢
annihilation operator aˆ . ˆ The operator Tˆ 1T , acting on the ground state of the QHO, generates all the x1
eigenstates of aˆ . ˆ Since the state Tˆ p1 T 0 is an arbitrary eigenstate of aˆ , we can write x1 ˆ (8) Tˆ p1 T x1 0 = l
where
l =
1 æ x1
ç
2è
0
+i
p1 ö
÷.
p0 ø
That is Re l = = Im l = =
1
2
(9) 0
p1
2 p0
(10)
The equation (8) tells us that the coherent states are generated by applying spatial and momentum translations to the ground state of the QHO. Particularly, as it can be seen by (9) and (10), the application of spatial translations to the ground state generates coherent states with real eigenvalues, while the application of momentum translations generates coherent states with imaginary eigenvalues. The combined application of spatial and momentum translations generates coherent states with complex eigenvalues.
37
Exercises
As shown in the exercises 1 and 6, the spatial translation operators leave unchanged the position-uncertainty, the momentum uncertainty, and, obviously, the positionmomentum uncertainty product, and so do the momentum translation operators. The QHO ground state is a Gaussian function, meaning that the position-momentum uncertainty product is minimum, i.e.
h
2
ˆ
, in in the state 0 . Sin Since ce the the acti action on of T x1 does ˆ
not change either the position or the momentum uncertainty, the state T x1 0 is also a ˆ
state of minimum position-momentum uncertainty. Again, since the action of T p1 leaves unchanged the position uncertainty and the momentum uncertainty, the state ˆ
ˆ
T p1 T x1 0 is still a state of minimum position-momentum uncertainty. Therefore, the
eigenstates of the annihilation operator are states of minimum position-momentum uncertainty, and moreover, in all those eigenstates, both the position uncertainty and the momentum uncertainty are respectively equal to the position uncertainty and the momentum uncertainty in the QHO ground state, that is
( D ) l = ( D x ) 0
(11)
( D p ) l = ( Dp ) 0
(12)
and
( D xDp ) l = ( DxDp ) 0 =
h
2
(13)
10) Calculate the arbitrary eigenstate l – i.e. the arbitrary coherent state – in the position representation, and express it in terms of the position and momentum expectation values. Solution
In the exercise 9, we showed that Re l = =
and Im l = =
1
2
0
p1
2 p0
Projecting the eigenstate
)=
eigenfunction y l ( ˆ
ˆ
ˆ
l = T p1T x1 0 , where l =
1 æ x1
ç
2è
l onto the
axis
“
”
ˆ
ˆ
ˆ
ˆ
{
{
y l ( x )
y 0 ( x )
ˆ
Þ y l ( ) = T p ( x) Tx ( x )y 0 ( x) (1) 1
The wave function y 0 (
)=
÷ , i.e.
p0 ø
, we obtain the respective respect ive
representat ion. Then, Then, we have x l in the position representation.
ˆ
1
0
p1 ö
.
l = T p1Tx1 0 Þ x l = x T p1Tx1 0 = T p1 ( x ) Tx1 ( x ) x 0 Þ ˆ
+i
stat e of the QHO, that is x 0 is the ground state
1
æ 1 ö4 æ x2 ö y 0 ( x ) = ç (2) exp ç 2 ÷ 2 ÷ p x 2 0 ø è 0 ø è
38
Exercises
h
where x0 =
mw
is the length scale of t he QHO.
We have shown – refer refer to exercise 1 – that that the action of Tˆ x1 ( x ) on an arbitrary wave function y (
Tˆ x1
) of an arbitrary system shifts the ( x )y ( x ) = y ( x - x1 ) . Applying that to (2) yields
argument x
by - x1 , i.e.
1
4 æ ( x - x1 ) 2 ö æ ö 1 Tˆ x ( x )y 0 ( x ) = ç exp ç ÷ (3) 2 ÷ 2 ç ÷ p x 2 x 0 è 0 ø è ø 1
On the other hand, in the position representation, the momentum translation operator æ ip x ö is a position-dependent phase, i.e. Tˆ p1 ( x ) = exp ç 1 ÷ . Thus, (1) is written as
è
h
ø
1
æ ( x - x1 ) 2 ö æ 1 ö4 æ ip1 x ö y l ( x ) = ç exp ç ÷ exp çç - 2 x 2 ÷÷ (4) 2 ÷ p x h è ø è 0 ø 0 è ø The spatial translation x1 is equal to the position expectation value in the state l and the momentum translation p1 is equal to the momentum expectation value in the state l , i.e. 1
= xˆ
p1 = pˆ
l
(5) l
(6)
Thus, (4) becomes 1
æ i p l æ 1 ö4 ç y l ( x ) = ç e x p 2 ÷ ç p x h è 0 ø è where l =
1 æ xˆ
ˆ
æ xö ç x - xˆ l ÷ exp ç 2 ÷ 2 x0 ç ø è
(
)
2
ö ÷ ÷ (7) ÷ ø
ö l ç ÷. +i p0 ÷ 2 çè x0 ø l
pˆ
The relation (7) is the expression of the arbitrary coherent state l in the position representation in terms of the position pos ition and momentum expectation values values in that state. Let us now prove (5) and (6). First proof Since l is an eigenstate of the t he annihilation operator aˆ , with eigenvalue l , l , we have aˆ l = l l Thus, the expectation value of o f aˆ in the state l is aˆ
l
= l aˆ l = l l l = l l l = l {
1
39
Exercises
That is aˆ
= l (8)
l
1 æ ˆ
Using the expression aˆ =
+i
pˆ ö
÷ , the expectation value of aˆ in an arbitrary p 0 0 ø state – not not necessarily in an eigenstate of o f aˆ – will will be pˆ ö pˆ ö 1 æ ˆ +i ç ÷ ç +i ÷ = p0 ø p1 ø 2 è x0 2 è x0
1 æ xˆ
aˆ =
Thus, in the eigenstate aˆ
1 æ xˆ
ç
=
l
ç
2è
l
2 çè x0
pˆ
+i
,
l
l
p1
ö ÷ (9) ÷ ø
Comparing (8) and (9) we obtain 1 æ xˆ
ç
l =
l
pˆ
+i
2 çè x0
l
p1
ö ÷ (10) ÷ ø
Now, comparing (10) with with 1 æ xˆ
pˆ
ç +i 2 çè x0 l
Since xˆ xˆ pˆ
l
l
l
l
p1
, pˆ
l
l =
1 æ
ç
2è
1 0
+i
p1 ö
÷ , we obtain
p0 ø
ö 1 æ p ö 1 ÷= +i 1 ÷ ç ÷ p0 ø 2 è x0 ø
, x1 , p1 are all real, from the last equation we obtain
= x1 = p1
which are (5) and (6). Second proof The expectation value of an observable Oˆ ( xˆ, pˆ ) in the state
Oˆ
=
l
l
Since
l
Oˆ
l
is
(11)
ˆ 0 , the bra = Tˆ p T x 1
l
1
l
is
l
= 0 (Tˆ p T ˆx 1
1
†
) (be careful to take the adjoint of
the product, not the product of the two adjoints!). Thus, we have l
ˆ † (12) = 0 Tˆ x †T p 1
1
The expectation value of Oˆ is then written as
Oˆ
l
= 0 Tˆ x †Tˆp †Oˆ Tˆp Tˆx 0 (13) 1
1
1
1
40
Exercises
Using the property vi of the spatial and momentum tra nslation operators, we have ˆ ( xˆ , pˆ + p ) Tˆ = Oˆ ( xˆ + x , p ˆ + p1 ) Tˆ x1 †Tˆp1 †Oˆ ( xˆ , pˆ ) Tˆp1 Tˆx1 = Tˆx1 † O 1 x1 1
That is † † Tˆ x1 Tˆp1 Oˆ ( xˆ , pˆ ) Tˆp1 Tˆx1 = Oˆ ( xˆ + x1 , pˆ + p1 ) (14)
Thus, (13) is written as
Oˆ ( xˆ, pˆ )
l
= Oˆ ( xˆ + x1 , pˆ + p1 )
0
(15)
Applying (15) to the position and momentum operator, ˆ and pˆ , respectively, we obtain xˆ
l
= xˆ + x1
l
= pˆ + p1
= xˆ
0
0
+ x1 = x1
and pˆ
0
= pˆ
0
+ p1 = p1
Thus 1
= xˆ
p1 = pˆ
l
l
We remind that in the ground state of the QHO, the position and momentum expectation values are both zero. Third proof We’ll calculate the position and momentum expectation values in the position d representation, where ˆ = x and pˆ = - ih . dx In the position representation, the position expectation value in the state l is written as ¥
xˆ
l
=
ò dxy
* l
( x ) xy l ( x )
-¥
where the wave function y l ( x ) is given by (4), i.e. 1
æ ( x - x1 ) 2 ö æ 1 ö4 æ ip1 x ö y l ( x ) = ç exp ç ÷ exp çç - 2 x 2 ÷÷ 2 ÷ p x h è ø 0 è 0 ø è ø Thus 1
xˆ
l
æ ( x - x1 ) 2 ö æ 1 ö2 ¥ dxx exp ç =ç ÷ (16) 2 ÷ 2 ò ç ÷ p x x 0 è 0 ø -¥ è ø
We’ll calculate the t he integral doing the variable change
¢ = x - x1 (17) 41
Exercises
Then, we have
= x¢ + x1 dx = dx ¢ x¢ : - ¥, ¥ Thus, the integral in (16) takes the form for m
æ ( x - x1 )2 ö ¥ æ x ¢2 ö ò-¥ dxx exp çç - x0 2 ÷÷ = -ò¥ dx¢ ( x¢ + x1 ) exp çè - x0 2 ÷ø = è ø ¥ ¥ æ ¢2 ö æ x¢ 2 ö = ò dx¢x¢ exp ç - 2 ÷ + x1 ò dx ¢ exp ç - 2 ÷ è x0 ø è x0 ø -¥ -¥ ¥
æ ¢2 ö The first integral is zero because the integrand x¢ exp ç - 2 ÷ is an odd function – as a s è 0 ø æ x¢2 ö the product of the odd function ¢ with the even function exp ç - 2 ÷ – and the è 0 ø integration is done over a symmetric interval. The second integral is a Gaussian integral and referring, for instance, to https://en.wikipedia.org/wiki/Gaussian_integral we see that it is equal to
p
2 0
. Thus
æ ( x - x1 ) 2 ö 2 ò-¥ dxx exp çç - x0 2 ÷÷ = x1 p x0 (18) è ø ¥
Substituting (18) into (16), we obtain xˆ
l
= x1
In the same way, the momentum expectation value in the position representation is written as, in the state l , ¥
pˆ
¥
d ö æ = ò dxy l ( x ) ç -i h ÷y l ( x ) = -i h ò dxy l * ( x )y l¢ ( x ) dx ø è -¥ -¥ *
l
The derivative of the wave function y l ( y l¢ (
)=
ip1 h
) is obtained from (4),
æ x - x1 ö æ ip1 x - x1 ö = y x y l ( x) ( ) ç l 2 ÷ 2 ÷ h x x 0 0 è ø è ø
y l ( x ) + ç -
Thus, the momentum expectation value beco mes ¥
pˆ
l
æ ip x - x ö = -ih ò dxy l * ( x ) ç 1 - 2 1 ÷y l ( x ) = x0 ø è h -¥
= ( -ih )
ip1 h
¥
ò
2
dx y l ( x ) +
-¥ 4 1 4244 3
ih x0 2
¥
ò
dx ( x - x1 ) y l ( x )
-¥
1
42
2
Exercises
That is p ˆ
l
= p1 +
ih x0
2
¥
ò
2
dx ( x - x1 ) y l ( x ) (19)
-¥
( - x1 ) y l ( x )
Since the integrand
ih
number, so the quantity
2
2
is a real function, the integral in (19) is a real
¥ 2
ò
dx ( x - x1 ) y l ( x ) is an imaginary number, which
x0 -¥ must be zero because the momentum expectation value must be real. This means that ¥
the integral
ò
2
dx ( x - x1 ) y l ( x ) must be zero. Let us verify that.
-¥
From (4) we obtain 1
æ ( x - x1 )2 ö æ 1 ö2 y l ( x ) = ç exp ç ÷ (20) 2 ÷ 2 ç ÷ p x x 0 è 0 ø è ø 2
2
Observe that the function y l ( x ) is not even. Using (20), the integral in (19) is written writte n as ¥
æ 1 ö d x x x y x = ( ) ( ) ç 1 l 2 ÷ ò-¥ p x 0 è ø 2
1 2 ¥
æ ( x - x1 ) 2 ö ò-¥ dx ( x - x1 ) exp çç - x02 ÷÷ è ø
æ ( x - x1 ) 2 ö ÷ Doing again the variable change (17), the integral ò dx ( x - x1 ) exp ç 2 ç ÷ x 0 -¥ è ø ¥ æ ¢2 ö æ ¢2 ö becomes ò dx¢x ¢ exp ç - 2 ÷ , which is zero because the integrand x¢ exp ç - 2 ÷ is è x0 ø è x0 ø -¥ ¥
¥
now odd (as product of an odd with an even funct ion). Thus
ò
2
dx ( x - x1 ) y l ( x ) = 0 ,
-¥
as we expected. Therefore, from (19), the momentum expectation value is p ˆ
l
= p1
11) In the exercise 9, we showed that an arbitrary coherent state l of the QHO, with eigenvalue l , is generated by the action of a spatial translation operator T x1 and a momentum translation operator T p1 on the ground state of ˆ
the QHO, i.e.
ˆ
l = T p1T x1 0 , with ˆ
ˆ
l =
1 æ x1
p1 ö ç + i ÷ , where x0 , p0 are, p0 ø 2 è x0
respectively, the length and momentum scale of the QHO. Assume a QHO in an arbitrary arbitrary coherent state l . i) Show that the probability of finding the QHO in an arbitrary energy 2
eigenstate n is given by a Poisson distribution with parameter l , i.e. with
43
