Lorentz action on creation/annihilation operators

In summary, the Lorentz action on the free scalar field creation operators \alpha(k)^\dagger is given by U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger = \alpha(\Lambda k)^\dagger. This can be shown by using the definition of how momentum eigenstates transform and the unitarity of operators, as well as the invariance of the vacuum state. The analogous transformation for annihilation operators can be derived by taking the hermitian conjugate of the equation for creation operators.
  • #1
jackson1
12
0
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?
 
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  • #2
jackson1 said:
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?

Momentum eigenstates [itex]|k>[/itex] transform (by definition given how the momentum, k, transforms) according to:

[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
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
jackson1 said:
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.

You want to start from this line:
[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)^\dagger|0\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
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]?
 
  • #6
jackson1 said:
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]?

Yes, that's it.
 
  • #7
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
jackson1 said:
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]

Not quite. Sorry, I misread your above post.

[itex]U(\Lambda)|0>=|0>[/itex] implies [itex]<0|U^{\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
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
jackson1 said:
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].

Correct.

My next question is, what happens if you act on the vacuum ket with [itex] U(\Lambda)^\dagger [/itex].

I honestly don't remember. Perhaps someone else can chime in on this one? I never thought about it as the non conjugated Lorentz operator is what we care about when defining how transformations affect kets.

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].

You are confusing how kets transform and how operators transform. That definition of an operator transformation is inconsistent with kets that transform like [itex]U(\Lambda)|k\rangle=|\Lambda k\rangle[/itex]

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]

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)?

It follows directly from the work you just did. Just take the hermitian conjugate of the equation we just derived.
 
  • #11
Thanks so much for your help.
 
  • #12
##U(\Lambda)## is unitary, so ##U(\Lambda)^\dagger=U(\Lambda)^{-1}=U(\Lambda^{-1})## is another Lorentz transformation operator.
 
  • #13
Fredrik said:
##U(\Lambda)## is unitary, so ##U(\Lambda)^\dagger=U(\Lambda)^{-1}=U(\Lambda^{-1})## is another Lorentz transformation operator.

Of course! I should have known that! :redface:
 

1. What is the Lorentz action on creation/annihilation operators?

The Lorentz action on creation/annihilation operators is a mathematical description of how these operators transform under Lorentz transformations, which are transformations in space and time. It is a way to incorporate the principles of special relativity into quantum field theory.

2. How does the Lorentz action affect the creation and annihilation of particles?

The Lorentz action on creation/annihilation operators affects the transformation of these operators when particles are created or annihilated. It allows us to understand how these operators behave under different reference frames and how they are related to each other.

3. Why is the Lorentz action important in quantum field theory?

The Lorentz action is important because it allows us to describe the behavior of particles in a way that is consistent with the principles of special relativity. It helps us understand how particles interact and how they transform under different reference frames.

4. How is the Lorentz action related to Lorentz invariance?

The Lorentz action is related to Lorentz invariance because it describes how the creation/annihilation operators transform under Lorentz transformations, which are transformations that preserve the laws of physics in different inertial reference frames. This is a fundamental principle in special relativity and is essential for understanding the behavior of particles.

5. Are there any practical applications of the Lorentz action on creation/annihilation operators?

Yes, there are practical applications of the Lorentz action on creation/annihilation operators in various fields such as particle physics, quantum mechanics, and cosmology. It is used to study the behavior of particles in high-energy collisions, to calculate scattering amplitudes, and to understand the expansion of the universe. It is also important for developing new theories and models in these fields.

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