phys 434 - assignment 4 - coupled harmonic oscillators

8/13/2019 PHYS 434 - Assignment 4 - Coupled Harmonic Oscillators

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Consider the set of N+1harmonic o scillato rs seen in class, where the 0thoscil lator that we call thedetector) couples to all the other oscillators with the H amiltonian:

H =!0 a0 a0 + !n an

an

n=1

N

" + # a0 + a0( )

1

2!nm

n

an+ a

n

( )n=1

N

" We saw in class how the leading order co ntribution to the proba bility for the detector to get excitedout of va cuum fluctuations from a time t

0to a time t was co mputed using first ord er perturbation

theory:P

00 ! 10= 10 ,1j, 0n"j

U(t,0) 00 , 01,.....,0N2

j=1

N

# = 10 ,1j, 0n"j U(1)

(t,0) 00 , 01,.....,0N2

j=1

N

# +O $3( )

..which is of o rder O(!2 ) and is given by the expression:P

00 !10= 2"

2sin

2 1

2 #

0+#j( )t$% &'

#jm

j #

0+#

j( )2

j=1

N

( +O "3( )

No w, we wish to determine the pro bab ility that the detecto r remains unexcited when coup led to thevacuum state of all the other oscil lators. We will co mpute this to the leading ord er) using secondord er perturbation theory ie; It will be mediated by U(2 )(t,0) ). T his will be d one, given that:

1) The 0 thorder co ntribution, although present in this case, is of co urse not enoug h to get asatisfactory result

2) We need to act twice with the creation and annihilation o perato rs in ord er to excite thevacuum once and g et it bac k to the vacuum aga in, or in other word s, for the first order correctio n tothe time evo lution op erator, we get that:

00

,out U(1)

(t,0) 00

, in =0 where in and out represent arbitrary states for the set o f Nharmonic o scillators.QUEST ION 1 Before computing the explicit final value of the prob ability of the detector remainingin the ground state in the presence o f the vacuum P

00 ! 00), find first what would be the po wer of !

of the leading o rder contribution to P00 ! 00

. Justify yo ur answer, and show it mathematically with theminimum number of calculations possible.We have:

H0 =!

0a

0

a0 + !

na

n

an

n=1

N

"

V = a0+ a

0

( ) 1

2!nm

n

an+ a

n

( )n=1

N

"

Recall the Dyson Series, where we set t0 =0 :

UD(t,0) = I + UD

(1)(t,0) + UD

(2 )(t,0) + UD

(3)(t,0) + .....

Where we have:

UD

(1)(t,0) =!i" dt

1VD

0

t

# (t1) U

D

(2 )(t,0) =!"2 dt

1 dt

2VD

0

t1

# (t1)VD0

t

# (t2 ) U

D

(3)(t,0) = i!

3dt

1 dt

2 dt

3VD

0

t2

" (t1)VD(t2 )VD0

t1

" (t3)0

t

" !

Where VD

is the Potential in the Dirac picture, such that: VD =e

i H0t Ve

!i H0t

In the calculation of P00 ! 00

, we wish to truncate the Dyson Series after the second order. Such that:

UD(t,0) ! I +

UD

(1)(t,0) + UD

(2 )(t,0)

Then in the calculation of P00 ! 00

we have:

P00 !0 0

= 00

,out U(t,0) 00

, in2

out

" = 00 ,out U(t,0) 00 ,01,.....,0N2

out

"


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Note that while summing over all states, we can exploit the use of the kroenecker delta, so as to simplify th

expression:

P00 ! 0 0

= "out, 01,..., 0N

"out, 01,..., 0N

#$

%&

out

' + "out, 01,..., 0N (2 )

out, 01,..., 0N+ )

out, 01,..., 0N

*( )#$ %

&out

' + (4 )out, 01,..., 0N2#

$+%&,out

'

= 1+ (2 )01,..., 0N , 01,..., 0N

+ )01,..., 0N , 01,..., 0N

*#$

%&+ (4 )

out, 01,..., 0N

2

out

'

= 1+ 2(2 Re )01,..., 0N , 01,..., 0N

#$

%& + (

4 )out, 01,..., 0N

2

out

'

Where we have used the fact that !+!"

= 2 Re !( ) , for any complex scalar !"! .

Therefore, we see that in terms of UD

(2 )(t,0) we have the probability:

P00 !0 0

= 1+ 2"2 Re #01,..., 0N , 01,..., 0N

$%

&' + "

4 #out, 01,..., 0N

2

out

(

=1+ 2"2 Re 00

, 01,.....,0

N ) dt1 dt2

VD0

t1

* (t1)VD0

t

* (t2 )$

%+

&

', 00 , 01,.....,0N

$

%++

&

',,+ "4 0

0,out ) dt1 dt2

VD0

t1

* (t1)VD0

t

* (t2 )$

%+

&

', 00 , 01,.....,0N

2

out

(

=1+ 2Re 00

, 01,.....,0

N

UD

(2 )(t,0) 0

0, 0

1,.....,0

N$%

&' + 00 ,out

UD

(2 )(t,0) 0

0, 0

1,.....,0

N

2

out

(

In this question, we are only interested in the leading order of ! , and so we can also more simply write this

as:P

00 ! 0 0=1+ 2"2 Re #

01 ,..., 0N , 01 ,..., 0N

$%

&'+O

"3( )

And so, we see that the leading order term is in this probability is 1 . However, this is the t r v lorder of ! ,which we omit ! We take the first non-trivial power of ! , which is 2 .Therefore, in the leading order contribution to P

00 ! 0

0

, the power of ! is 2 .


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QUES TION 2 Compute the prob abi lity that the detector remains in the ground state in the presenceof vac uum fluctuations as a function o f time, up to the leading o rder in the coup ling strength ! ,using second ord r time-depe ndent perturbation theory and a ssuming that the vacuum state wasprepared at t

0 =0 .

From Question 1, we found that the probability P00 ! 00

is given by:

P00 !0 0

= 1+ 2"2

Re 00

, 01,.....,0

N # dt1 dt2

VD0

t1

$ (t1 )VD0

t

$ (t2 )%

&

'(

)

* 00 , 01,.....,0N%

&

'

'

(

)

*

*

+ "4

00

,out # dt1 dt2VD

0

t1

$ (t1)VD0

t

$ (t2 )%

&

'(

)

* 00, 01,.....,0N

2

out

+

=1+ 2"2

Re 00

, 01,.....,0

N # dt1 dt2

VD0

t1

$ (t1)VD0

t

$ (t2 )%

&'

(

)* 00 , 01,.....,0N

%

&''

(

)**+O "

3( )

,1+ 2Re 00 , 01,.....,0NUD

(2)(t,0) 00 , 01,.....,0N%& ()

Where we have omitted all terms of order O !3( ) for this calculation.

IN order to calculate this probability, we must find UD

(2 )(t,0) , which is given by the formula:

UD

(2 )(t,0) =!"2 dt1 dt2

VD

0

t1

# (t1)VD0

t

# (t2 )

.from our knowledge of the Dyson Series.

Following the calculation given in the lecture notes, we know that VD

(the potential in the Dirac picture) is

given by:

VD = e

i H0t Ve!i H0t

= a0e!i"0t

+ a0

ei"0t( )

1

2"nm

n

ane!i"

nt+ a

n

ei"

nt( )

n=1

N

#

Which allows us to calculate UD

(2 )(t,0) :

UD(2) (t,0 ) =!"2 dt1

0

t

# dt2VD(t1)VD(t2 )0

t1

#

=!"2 dt10

t

# dt21

2$nm

n

a0e! i$0t1

+a0ei$0t1( ) ane

! i$nt1+a

n

ei$nt1( )

n=1

N

%&'(

)(

*+(

,(a0e

! i$0t2+a0

ei$0t2( )

1

2$km

k

ake! i$kt2

+ak

ei$kt2( )

k=1

N

%&'(

)(

*+(

,(0

t1

#

=

!"

2 1

2 $nm

n$

km

kk=1

N

%n=1

N

% dt

10

t

# dt

2 a

0e! i$0t1

+

a

0

ei$0t1

( ) ane! i$

nt1+

an

ei$

nt1

( ) a

0e! i$0t2

+

a

0

ei$0t2

( ) ake! i$

kt2+

ak

ei$

kt2

( )0

t1

#=!"2

1

2 $nm

n$

km

kk=1

N

%n=1

N

% dt10

t

# dt2 a0ane!i $0+$n( )t1

+a0ane! i $0!$n( )t1

+a0a

nei $0!$n( )t1

+a0a

n

ei $0+$n( )t1( ) a0ake!i $0+$k( )t2 + .....(

0

t1

#

.....+a0ake! i $0!$k( )t2

+a0a

kei $0!$k( )t2

+a0a

k

ei $0+$k( )t2 )

=!"2 1

2 $nm

n$

km

kk=1

N

%n=1

N

% dt10

t

# a0ane!i $0+$n( )t1

+a0ane! i $0!$n( )t1

+a0a

nei $0!$n( )t1

+ .....(

.......+a0a

n

ei $0+$n( )t1 ) dt2 a0ake! i $0+$k( )t2 +a0ake! i $0!$k( )t2 +a0akei $0!$k( )t2 +a0akei $0+$k( )t2( )

0

t1

#

=!"2 1

2 $nm

n$

km

kk=1

N

%n=1

N

% dt10

t

# a0ane!i $0+$n( )t1

+a0ane! i $0!$n( )t1

+a0a

nei $0!$n( )t1

+ .....(

.......+a0

an

e

i $0+$n( )t1

) a0ak e

! i $0+$k( )t2

dt20

t1

# +a0ak

e

! i $0!$k( )t2

dt20

t1

# +a0

ak ei $0!$k( )t2

dt20

t1

# +a0

ak

e

i $0+$k( )t2

dt20

t1

#

-

./

0

12

=!"2 1

2 $nm

n$

km

kk=1

N

%n=1

N

% dt10

t

# a0ane!i $0+$n( )t1

+a0ane! i $0!$n( )t1

+a0a

nei $0!$n( )t1

+ .....(

.....+a0a

n

ei $0+$n( )t1 ) a0ak

i

$0 +$k( )e! i $0+$k( )t2

-

./

0

12

0

t1

+a0ak i

$0!$k( )e! i $0!$k( )t2

-

./

0

12

0

t1

+a0a

k

!i

$0!$k( )ei $0!$k( )t2

-

./

0

12

0

t1

+a0a

k

!i

$0 +$k( )ei $0+$k( )t2

-

./

0

12

0

t1-

.

//

0

1

22

=!i"2 1

2 $nm

n$

km

kk=1

N

%n=1

N

% dt10

t

# a0ane! i $0+$n( )t1

+a0ane! i $0!$n( )t1

+a0a

nei $0!$n( )t1

+ .....(

.....+a0a

n

ei $0+$n( )t1 )

a0ak

$0 +$k( ) e

!i $0+$k( )t2-.

010

t1

+

a0ak

$0!$k( ) e

! i $0!$k( )t2-.

010

t1

+

a0a

k

$0!$k( ) !e

i $0!$k( )t2-.

010

t1

+

a0a

k

$0 +$k( ) !e

i $0+$k( )t2-.

010

t1-

./

0

12

.....Calculation Continued On The Next Page....


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And so we have simplified the form of 00, 01,....,0NU

D

(2 )(t,0) 00 , 01,....,0N even further:

00 , 01,....,0NUD

(2 )(t,0) 00 , 01,....,0N =!i"

2 1

2 #nm

n#

km

kk=1

N

$n=1

N

$ 00 , 01,....,0N a0ana0ak 00 , 01,....,0Ne! i #0+#n( )t1 ! e! i #n!#k( )t1

#0 +#k( ) dt1

0

t

% + .....&'(

)(

.....+ 00 , 01,....,0N a0ana0

ak

00 , 01,....,0N

e!i #0!#n( )t1 ! ei #n+#k( )t1

#0 +#k( ) dt1

0

t

% *

+(

,(

Finally, we see that:

00,0

1,....,0

Na

0a

na

0

ak

0

0,0

1,....,0

N = 00,01,....,0N a0an 10 ,01,....,1k,....,0N = !n,k

00,01,....,0N a0ana0

ak 00 ,01,....,0N = 00,01,....,0N a0an

10 ,01,....,1k,....,0N =0

0

,01

,....,0N

00

,01

,....,1k,....,1

n,...,0

N = 0 , if k" n

2 00,0

1,....,0

N0

0,0

1,....,2

n,....,0

N = 0 , if k =n

#$%&%

Which gives us:

00 , 01,....,0NUD

(2 )(t,0) 00, 01,....,0N =!i"

2 1

2 #nm

n#

km

kk=1

N

$n=1

N

$ %n, ke!i #0+#n( )t1 ! e! i #n!#k( )t1

#0 +#k( ) dt1

0

t

& + 0( ) e

!i #0!#n( )t1 ! ei #n+#k( )t1

#0 +#k( ) dt1

0

t

& '

()

*)

+,)

-)

=!i"2 %

n,k

2 #nm

n#

km

kk=1

N

$n=1

N

$ e

! i #0+#n( )t1 ! e!i #n!#k( )t1

#0 +#k( ) dt1

0

t

&

=!i"2 1

2 #nm

n#

nm

n

e!i #0+#n( )t1 ! e!i #n!#n( )t1

#0 +#n( ) dt1

0

t

&n=1

N

$

=!i"2 1

2#nmn (#0 +#n )n=1

N

$ e!i #0+#n( )t1 !1./

0

1

dt1

0

t

&=!i"2

1

2#nm

n(#0 +#n )n=1

N

$ 1

!i(#0 +#n )e!i #0+#n( )t1 ! t

.

/2

0

13

0

t

=!i"2 1

2#nm

n(#0 +#n )n=1

N

$ i

(#0 +#n )e! i #0+#n( )t ! t!

i

(#0 +#n )

.

/2

0

13

=!i2"2 1

2#nm

n(#0 +#n )n=1

N

$ e

!i #0+#n( )t !1./ 0

1(#0 +#n )

+ i t

.

/

22

0

1

33

= "2

e!i #0+#n( )t !1./

01

2#nm

n(#0 +#n )

2 +

i t

2#nm

n(#0 +#n )

.

/

22

0

1

33n=1

N

$

Having found 00

, 01

,....,0N

UD

(2 )(t,0) 0

0

,01

,....,0N

, we now explicitly calculate the probability:

P00 ! 00

= 1+ 2Re 00

, 01,.....,0

N

UD(2 )

(t,0) 00 , 01,.....,0N{ }+O "3( )

= 1+ 2Re "2 e

# i $0+$n( )t #12$

nm

n($0 + $n )

2 +

i t

2$nm

n($0 +$n )

%

&'

(

)*

n=1

N

+,-.

/.

01.

2.+O "

3( )

= 1+ 2"2

Recos ($0 +$n )t( )# isin ($0 +$n )t( )#1

2$nm

n($0 +$n )

2 +

i t

2$nm

n($0 +$n )

%

&'

(

)*

n=1

N

+,-.

/.

01.

2.+O "

3( )

= 1+ 2"2

Recos ($0 +$n )t( )#12$

nm

n($

0+ $

n)

2

3

456

78+ i

t

2$nm

n($

0+$

n)#

sin ($0 +$n )t( )2$

nm

n($

0+$

n)

2

3

456

78%

&'

(

)*

n=1

N

+,-.

/.

01.

2.+O "

3( )

= 1+ 2"2

Recos ($0 +$n )t( )#12$

nm

n($0 +$n )

2

%

&'

(

)*

n=1

N

+ + i t2$

nm

n($0 +$n )

# sin ($0 +$n )t( )2$

nm

n($0 +$n )

2

%

&'

(

)*

n=1

N

+,-.

/.

01.

2.+O "

3( )

= 1+ 2"2 cos ($0 +$n )t( )#1

2$nm

n($0 +$n )

2

n=1

N

+ +O "3( )

= 1+ 2"2

#2sin2 12($0 +$n )t( ){ }2$

nm

n($0 +$n )

2

n=1

N

+ +O "3( )

= 1#2"2 sin

2 12($0 +$n )t( )

$nm

n($0 +$n )

2

n=1

N

+ +O "3( )

Therefore, the probability that the detector (the 0th

oscillator) remains in the ground state in the presence o

vacuum fluctuations, up to order 2 in !(the leading order), is given by:

P00 !0 0

=1" 2#2

sin2 1

2 ($0 +$n )t( )$

nm

n($

0 +$n )2

n=1

N

% +O #3( )


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QUEST ION 3 Compare and co mment on the result ob tained with the probab ility P10 ! 0

0

, computedin class.As computed in Question 2, the probability that the detector (the 0

thoscillator) remains in the ground state

in the presence of vacuum fluctuations is given by:

P00 !0 0

=1" 2#2

sin2 1

2 ($0 +$n )t( )$

nm

n($

0 +$n )2

n=1

N

% +O #3( )

&1"2#

2 sin

2 12 ($0 +$n )t( )

$nmn($0 +$n )2

n=1

N

%

As computed in class, the probability that the detector (the 0th

oscillator) will transition into the first excited

state is given by:

P10 !00

=2"2

sin2 1

2 (#0 +#n )t( )#

nm

n(#

0 +#n )2

n=1

N

$ +O "3( )

% 2"2

sin2 1

2 (#0 +#n )t( )#

nm

n(#

0 +#n )2

n=1

N

$

..where we are omitting the terms of order O !3( ) and higher.

We right away see that:

P00 ! 0

0

"1# P10 ! 0

0

Rearranging this expression, we find that:

P00 !0

0

+ P10 !0

0

"1

Which means that (if we are omitting all the terms of order O !3( ) and higher) that these probabilities sum

to 1 .

In other words, from the above we gather that there are only two possible outcomes for state of the 0th

oscillator; one in which it is in the ground state, and the other in which it is in the first excited state.

Note that if we take P00 ! 0

0

+ P10 ! 0

0

=1 as fact, then there is zero probability that the 0th

oscillator

becomes excited into the second, third, and higher excited states.

We know that !is a small real constant, and so the probability P00 ! 00

is in general, much larger than

P10 ! 00

. (This is because the 1 in P00 ! 00

dominates the behavior)

This makes sense logically. It means that the 0th

oscillator is far more likely to remain in the ground state,

than it is to become excited and transition into the first excited state (even though this probability is non-zero in general).


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10/10

We note that this probabil ity P00 ! 2

0

is certainly non-zero. The fact that the leading order contribution is of

order O !4( ) means that this probabil ity is very small , but definitely non-zero.

The conclusion that P00 ! 2

0

is a non-zero probability may at first seem to contradict the discussion of

Question 3.

Recall from earlier that we found:

P00 !0

0

+ P10 !0

0

"1

Where we have:

P00 ! 00

=1" 2#2

sin2 1

2 ($0 +$n )t( )$

nm

n($0 +$n )

2

n=1

N

% +O #3( ) P10 ! 00 =2"2 sin

2 12 (#0 +#n )t( )

#nm

n(#0 +#n )

2

n=1

N

$ +O "3( )

At first, it seems that the statement P00 !0

0

+ P10 !0

0

"1 implies that P00 ! 20

=0 .

However, we just showed that P00 !20

"0 !

The way to resolve the above paradox is to realize that the statement P00 !0

0

+ P10 ! 0

0

"1 is not actuallytrue. It is only an approximation that has been made, after dropping the terms of order O !

3( ) and higher(in P

00 ! 00and P

10 ! 00).

This means that we actually have:

P00 ! 0

0

+ P10 ! 0

0

+O "3( ) = 1

The higher order terms in P00 ! 00 and P10 ! 00 give a gap in the probabilities such that P00 ! 0 0 + P10 ! 0 0 "1

That means that the terms O !3( ) encompass the probability P00 ! 20 , as well as the probability P00 ! 30 ,

and all other probabilities.

phys 434 - assignment 4 - coupled harmonic oscillators

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