# Lorentz action on creation/annihilation operators

1. Feb 8, 2013

### jackson1

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
$\alpha(k)^\dagger$ is given by

$$U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger = \alpha(\Lambda k)^\dagger .$$
Can someone tell me how to show this, or at least how to get started?

2. Feb 8, 2013

### G01

Momentum eigenstates $|k>$ transform (by definition given how the momentum, k, transforms) according to:

$$U(\Lambda)|k>=|\Lambda k>$$
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. Feb 8, 2013

### jackson1

Thanks for the hint. However, I'm still stuck/puzzled. I have

$$U(\Lambda)\alpha(k)^\dagger |0\rangle = U(\Lambda)|k\rangle = |\Lambda k\rangle = \alpha(\Lambda k)^\dagger |0\rangle ,$$
and so, I would think $U(\Lambda)\alpha(k)^\dagger = \alpha(\Lambda k)^\dagger$. I'm not sure how the other transformation comes into play.

4. Feb 8, 2013

### G01

You want to start from this line:
$$U(\Lambda)|k>=|\Lambda k>$$

Your first instinct is correct. So, you can rewrite this line as:

$$U(\Lambda)\alpha(k)^\dagger |0\rangle=\alpha(\Lambda k)^\dagger|0\rangle$$

Now, remember that $U^{\dagger} U = 1$, and that the vacuum is invariant. (i.e. All observers agree that there are no particles.)

5. Feb 8, 2013

### jackson1

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

6. Feb 8, 2013

### G01

Yes, that's it.

7. Feb 8, 2013

### jackson1

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

8. Feb 8, 2013

### G01

$U(\Lambda)|0>=|0>$ implies $<0|U^{\dagger}(\Lambda)=<0|$ not $U(\Lambda)^\dagger |0\rangle = |0\rangle$

Use unitarity to insert lorentz operators inbetween the creation operator and the vacuum ket. Then make use of the invariance of the vacuum.

9. Feb 8, 2013

### jackson1

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 $U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger U(\Lambda)|0\rangle = \alpha(\Lambda k)^\dagger |0\rangle$ which implies $U(\Lambda)\alpha(k)^\dagger U(\Lambda)^\dagger = \alpha(\Lambda k)^\dagger$. My next question is, what happens if you act on the vacuum ket with $U(\Lambda)^\dagger$. Also, why not stop once you know that $U(\Lambda)\alpha(k)^\dagger |0\rangle = \alpha(\Lambda k)^\dagger |0\rangle$, implying $U(\Lambda)\alpha(k)^\dagger = \alpha(\Lambda k)^\dagger$; that is, why do you want to conjugate (if this is the correct usage of the word) $\alpha(k)^\dagger$. 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. Feb 8, 2013

### G01

Correct.

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.

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

Remember that we want to preserve the normalization of our kets. i.e. :

$$\langle k |k\rangle = \langle \Lambda k |\Lambda k\rangle = 1$$

This will imply operator transformations like $U(\Lambda )a^{\dagger}(k)U(\Lambda )^{\dagger}=a^{\dagger}(\Lambda k)$ if the kets transform like $U( \Lambda )|k\rangle=|\Lambda k\rangle$

It follows directly from the work you just did. Just take the hermitian conjugate of the equation we just derived.

11. Feb 8, 2013

### jackson1

Thanks so much for your help.

12. Feb 8, 2013

### Fredrik

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

13. Feb 8, 2013

### G01

Of course! I should have known that!