Creation and annihilation operators


by Lindsayyyy
Tags: annihilation, creation, operators
Lindsayyyy
Lindsayyyy is offline
#1
Nov9-12, 09:46 AM
P: 208
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
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grzz
grzz is offline
#2
Nov9-12, 10:19 AM
P: 939
(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
Lindsayyyy is offline
#3
Nov9-12, 11:15 AM
P: 208
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
grzz is offline
#4
Nov9-12, 11:30 AM
P: 939

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
Lindsayyyy is offline
#5
Nov9-12, 11:36 AM
P: 208
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
Fightfish is offline
#6
Nov9-12, 11:45 AM
P: 595
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
Lindsayyyy is offline
#7
Nov9-12, 11:48 AM
P: 208
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
Fightfish is offline
#8
Nov9-12, 11:50 AM
P: 595
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
grzz is offline
#9
Nov9-12, 11:53 AM
P: 939
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
Lindsayyyy is offline
#10
Nov9-12, 11:55 AM
P: 208
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.


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