
#1
Nov912, 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 



#2
Nov912, 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. 



#3
Nov912, 11:15 AM

P: 208

Thanks for your help edit: [tex] aa^\dagger=\lambda^2 x^2 + \frac 1 2 I +\gamma^2 p^2[/tex] 



#4
Nov912, 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. 



#5
Nov912, 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 



#6
Nov912, 11:45 AM

P: 595





#7
Nov912, 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 



#8
Nov912, 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.




#9
Nov912, 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. 



#10
Nov912, 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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