Register to reply

Creation and annihilation operators

by Lindsayyyy
Tags: annihilation, creation, operators
Share this thread:
Lindsayyyy
#1
Nov9-12, 09:46 AM
P: 210
Hi all

1. The problem statement, all variables and given/known data
Show:

[tex] (a^\dagger a)^2=a^\dagger a^\dagger a a +a^\dagger a[/tex]

wheres:
[tex] a= \lambda x +i \gamma p [/tex]
[tex] a^\dagger= \lambda x -i \gamma p [/tex]




2. Relevant equations
-

3. The attempt at a solution

Well, I haven't got much.

I just tried to use the stuff given, put it into my equation and solve it, but I don't get to the right side.

I calculated a+a first

[tex] a^\dagger a ={\lambda}^2x^2 + \frac {1}{2} I + \gamma^2 p^2[/tex]

But when I now try to calculate the square of that term I get lost. If I square it I get to:

[tex] (a^\dagger a)^2= \lambda^4x^4+\gamma^4p^4 +\lambda^2 \gamma^2 (x^2p^2+p^2x^2)-\lambda^2 x^2 -\gamma^2 p^2 +\frac 1 4 I[/tex]


Can anyone help me with this? I don't know what to do now/ If I'm on the right way.

Thanks for your help

edit: I is the identity matrix
Phys.Org News Partner Science news on Phys.org
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
grzz
#2
Nov9-12, 10:19 AM
P: 950
(a[itex]^{+}[/itex]a)[itex]^{2}[/itex]
= a[itex]^{=}[/itex]aa[itex]^{+}[/itex]a
=a[itex]^{+}[/itex](aa[itex]^{+}[/itex])a.

Now try to get aa[itex]^{+}[/itex] in terms of a[itex]^{+}[/itex]a.
Lindsayyyy
#3
Nov9-12, 11:15 AM
P: 210
Quote Quote by grzz View Post
(a[itex]^{+}[/itex]a)[itex]^{2}[/itex]
= a[itex]^{=}[/itex]aa[itex]^{+}[/itex]a
=a[itex]^{+}[/itex](aa[itex]^{+}[/itex])a.

Now try to get aa[itex]^{+}[/itex] in terms of a[itex]^{+}[/itex]a.
what does your a^= mean? I don't know this sign.

Thanks for your help

edit:

[tex] aa^\dagger=\lambda^2 x^2 + \frac 1 2 I +\gamma^2 p^2[/tex]

grzz
#4
Nov9-12, 11:30 AM
P: 950
Creation and annihilation operators

I used the plus sign instead of the dagger sign.

You are supposed to see a with + as a superscript wherever you see a with ^ as a superscript.

Sorry for the trouble in the notation used.
Lindsayyyy
#5
Nov9-12, 11:36 AM
P: 210
I see a = not a + but I think it's just a typing mistake.

so what you mean

[tex]...= a^\dagger a a^\dagger a=a^\dagger (aa^\dagger) a[/tex]

is that what you meant?

I calculated my aa+ (see my post) and a+a is the same, though there's a minus infront of the I
Fightfish
#6
Nov9-12, 11:45 AM
P: 623
Quote Quote by Lindsayyyy View Post
I calculated my aa+ (see my post) and a+a is the same, though there's a minus infront of the I
"The same"? That difference in sign makes all the difference. Can you now relate [itex]aa^{\dagger}[/itex] to [itex]a^{\dagger}a[/itex]?
Lindsayyyy
#7
Nov9-12, 11:48 AM
P: 210
With "the same" I meant the rest of the terms. Unlucky word choice from my side I guess.
What do you mean with "can you now relate aa+ to a+a"
do you mean that I should calculate [tex] (aa^\dagger) (a^\dagger a)[/tex]

thanks for the help
Fightfish
#8
Nov9-12, 11:50 AM
P: 623
No, I meant the relationship between [itex]aa^{\dagger}[/itex] and [itex]a^{\dagger}a[/itex] ie express one of them in terms of the other.
grzz
#9
Nov9-12, 11:53 AM
P: 950
Yes that is what I meant.

Allow me to use a+ to mean a with + as a superscript.

So aa+ = a+a + 1.

Now one more step.
Lindsayyyy
#10
Nov9-12, 11:55 AM
P: 210
Do you mean something like

[tex] a^\dagger a = 2\lambda^2 x^2 +2 \gamma^2 x^2 -a a^\dagger[/tex]

edit: nevermind, fail lol.

I guess you meant what grzz just said. I have to eat dinner now. I'll try later or tomorrow and come back when I have further questions (I'm pretty sure I will have some :( )

Thanks so far for the help guys.


Register to reply

Related Discussions
Exponential of creation/annihilation operators Quantum Physics 2
Commuting creation and annihilation operators Quantum Physics 2
Creation/annihilation operators Quantum Physics 16
Creation and Annihilation Operators in QM General Physics 3