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Lorentz action on creation/annihilation operators 
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#1
Feb813, 04:01 PM

P: 12

Hi,
I'm currently reading the book "Quantum Field Theory for Mathematicians" by Ticciati and in section 2.3 he mentions that the Lorentz action on the free scalar field creation operators [itex] \alpha(k)^\dagger [/itex] is given by [tex] U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger = \alpha(\Lambda k)^\dagger .[/tex] Can someone tell me how to show this, or at least how to get started? 


#2
Feb813, 04:33 PM

HW Helper
P: 2,685

[tex]U(\Lambda)k>=\Lambda k>[/tex] By writing a momentum eigenstate in terms of creation operators and the vacuum you should be able to make use of this definition and unitarity of operators to obtain the transformation rule for creation operators. 


#3
Feb813, 04:49 PM

P: 12

Thanks for the hint. However, I'm still stuck/puzzled. I have
[tex] U(\Lambda)\alpha(k)^\dagger 0\rangle = U(\Lambda)k\rangle = \Lambda k\rangle = \alpha(\Lambda k)^\dagger 0\rangle ,[/tex] and so, I would think [itex] U(\Lambda)\alpha(k)^\dagger = \alpha(\Lambda k)^\dagger [/itex]. I'm not sure how the other transformation comes into play. 


#4
Feb813, 05:13 PM

HW Helper
P: 2,685

Lorentz action on creation/annihilation operators
[tex] U(\Lambda)k>=\Lambda k>[/tex] Your first instinct is correct. So, you can rewrite this line as: [tex]U(\Lambda)\alpha(k)^\dagger 0\rangle=\alpha(\Lambda k)^\dagger0\rangle[/tex] Now, remember that [itex]U^{\dagger} U = 1[/itex], and that the vacuum is invariant. (i.e. All observers agree that there are no particles.) 


#5
Feb813, 05:21 PM

P: 12

I'm sorry for being so annoying but I still don't see it. When you say the vacuum is invariant do you mean [itex] U(\Lambda)0\rangle = 0\rangle [/itex] and similar for [itex] U(\Lambda)^\dagger [/itex]?



#7
Feb813, 05:34 PM

P: 12

Ok, so then I can say [itex] U(\Lambda)k\rangle = \Lambda k\rangle [/itex] implies [itex] U(\Lambda)\alpha(k)^\dagger 0\rangle = \alpha(\Lambda k)^\dagger 0\rangle [/itex] and since the vacuum is invariant, [itex] U(\Lambda)^\dagger 0\rangle =
0\rangle [/itex], we have [itex] U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger 0\rangle = \alpha(\Lambda k)^\dagger 0\rangle .[/itex] However, wouldn't we also be able to say, for ex., [itex]U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger U(\Lambda)^\dagger 0\rangle = \alpha(\Lambda k)^\dagger 0\rangle [/itex] or, for ex., [itex] U(\Lambda)\alpha(k)^\dagger 0\rangle = \alpha(\Lambda k)^\dagger U(\Lambda) 0\rangle ?[/itex] 


#8
Feb813, 05:43 PM

HW Helper
P: 2,685

[itex]U(\Lambda)0>=0>[/itex] implies [itex]<0U^{\dagger}(\Lambda)=<0[/itex] not [itex] U(\Lambda)^\dagger 0\rangle = 0\rangle [/itex] Use unitarity to insert lorentz operators inbetween the creation operator and the vacuum ket. Then make use of the invariance of the vacuum. 


#9
Feb813, 05:56 PM

P: 12

Ok, but now I have more questions :( . But first, let me just make sure I understand how to answer my original question. Skipping ahead, we have [itex] U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger U(\Lambda)0\rangle = \alpha(\Lambda k)^\dagger 0\rangle [/itex] which implies [itex] U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger = \alpha(\Lambda k)^\dagger [/itex]. My next question is, what happens if you act on the vacuum ket with [itex] U(\Lambda)^\dagger [/itex]. Also, why not stop once you know that [itex] U(\Lambda)\alpha(k)^\dagger 0\rangle = \alpha(\Lambda k)^\dagger 0\rangle [/itex], implying [itex]U(\Lambda)\alpha(k)^\dagger = \alpha(\Lambda k)^\dagger[/itex]; that is, why do you want to conjugate (if this is the correct usage of the word) [itex]\alpha(k)^\dagger [/itex]. Finally, I should be able to find the analogous transformation for the annihilation operator, perhaps starting with bras, correct (Ticciati doesn't mention it so I'm not sure if something different might happen)?



#10
Feb813, 06:15 PM

HW Helper
P: 2,685

Remember that we want to preserve the normalization of our kets. i.e. : [tex]\langle k k\rangle = \langle \Lambda k \Lambda k\rangle = 1[/tex] This will imply operator transformations like [itex]U(\Lambda )a^{\dagger}(k)U(\Lambda )^{\dagger}=a^{\dagger}(\Lambda k)[/itex] if the kets transform like [itex]U( \Lambda )k\rangle=\Lambda k\rangle[/itex] 


#11
Feb813, 06:27 PM

P: 12

Thanks so much for your help.



#12
Feb813, 06:29 PM

Emeritus
Sci Advisor
PF Gold
P: 9,359

##U(\Lambda)## is unitary, so ##U(\Lambda)^\dagger=U(\Lambda)^{1}=U(\Lambda^{1})## is another Lorentz transformation operator.



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