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Compatibility between normal ordering and Lorentz covariance.

+ 4 like - 0 dislike

For simplicity let's consider a current operator $J^{\mu}$ that is bilinear in elementary fields(e.g. $J^{\mu}=\bar{\psi}\gamma^{\mu}\psi$), then $J^{\mu}$ and its normally ordered form only differ by a c-number, that is, $:J^{\mu}:= J^{\mu}+c^\mu$. We know $J^{\mu}$ is formally Lorentz covariant in the sense that 

$$ U(\Lambda)^{-1}J^\mu U(\Lambda)=\Lambda^\mu_{\ \ \nu}J^{\nu},$$

but then $:J^{\mu}:$ doesn't seem to be Lorentz covariant in the same sense:

\begin{split} U(\Lambda)^{-1}:J^{\mu}:U(\Lambda)&=U(\Lambda)^{-1}J^{\mu}U(\Lambda)+U(\Lambda)^{-1}c^\mu U(\Lambda)\\ &=\Lambda^\mu_{\ \ \nu}J^{\nu}+c^\mu\\ &=\Lambda^\mu_{\ \ \nu}:J^{\nu}:+ (\delta^\mu_{\ \ \nu}-\Lambda^\mu_{\ \ \nu})c^\nu \end{split} Covariance would only be achieved if $c^\nu=0$, but we know this is not the case: $c^\nu$ are divergent numbers rather than 0. It seems I have to reach the conclusion that $:J^{\mu}:$ is not covariant. However, the covariant perturbation theory is derived using $:J^{\mu}:$, so how is the covariance restored in the perturbation theory?

asked Jun 27, 2015 in Theoretical Physics by Jia Yiyang (2,465 points) [ revision history ]
edited Jun 27, 2015 by Jia Yiyang
Most voted comments show all comments

I believe $c$ also transforms like a vector, no?

@RyanThorngren, not under the sandwiching of $U^{-1}\cdots U$, also see my comment below your answer.

It is a vexing question, as your computation looks superficially correct. But you are not allowed to treat a component of $c^\mu$ as a scalar - classically it doesn't transform as such, but as a component of a vector. One is transforming the quantum computation in a tensor product representation, and only the first factor trivializes upon taking the expectation value. I must think about how to make it plain, before posting it is an answer.

@ArnoldNeumaier, I'm glad that I can finally get my point across to you. Looking forward to your answer.

@ArnoldNeumaier, well my instinctual diagnosis for the problem is that it again has something to do with the ill-definedness of the local product that forms $J^\mu$, and $J^\mu$ doesn't even satisfy the necessary condition for Lorentz covariance $\langle 0|J^\mu|0\rangle=0$. So my guess would be somehow we make $:J^\mu:$ covariant but not $J^\mu$ itself. However I need a more concrete analysis than this sort of handwaving.

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@ArnoldNeumaier, @drake

I have a crude thought, maybe I can prove the covariance of a general $:J:$, compatibly with the formal covariance of $J$, as long as we interpret 

$U(\Lambda)^{-1}J^\mu U(\Lambda)$ in exactly the very way we formally prove it's covariant, in particular, no manipulations through commutation relations shall happen. I'll try to write it down tomorrow.

Formal computations are just heuristics - they may or may not have a trustworthy version. Physicists typically work on the heuristic level until they encounter manifest problems, at which time they either take more care or decide upon the right form based on knowing the result. 

I trust to some extent formal power series with well-defined terms (which at least have a rigorous meaning as asymptotic series), But I never trust anything formal if it involves infinities.

The $J$'s arising in quantum field theory are typically for a free field, where its action on momentum states is easy to compute (using for fermions Gordon identities), and is a useful exercise anyway. There are also currents for interacting fields (see old work by Brandt and by Zimmermann), but these are far more elusive, and I have seen them used only in case of exactly solvable problems in 2 dimensions.

3 Answers

+ 1 like - 0 dislike

You are  right in your observation, namely, both \(J\)and  \(:J:\) —as collections of operators acting on a Fock space—cannot transform covariantly.  It is, however, the normal ordered operator that transforming in a covariant way. Note that this already happens to any Hamiltonian: the normal-ordered Hamiltonian does transform as zero component of a 4-vector, but not the one giving a vacuum energy. 

Therefore, one can see normal ordering as the prescription to define the product of two local quantum fields at the same point that ensures Lorentz covariance. As I said above, one has to use normal ordering to define a Poincare algebra of operators, including the Hamiltonian, acting on a Fock space, rather than a trivial central extension of itself. 

answered Jun 29, 2015 by drake (885 points) [ revision history ]
edited Jun 30, 2015 by drake
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$J$ and $:J:$ are quantizations of the same classical observable. The spin of the classical observable equals the spin of its quantization. Do you disagree?

If the quantization is to preserve the covariance of the observable under the symmetry group of the theory (Lorentz), then the normal-ordered  operator is a quantization in Fock space, but the non-normal-ordered is not. 

If by the spin of an object, you mean the representation with which the object transforms, then I think that preserving the spin is a nice criterion to select a quantum representation (J with no normal-ordering does not transform with any representation). Note, however, that in general it is impossible to implement this without adding fictitious degrees of freedom, namely, there is no helicity-2 massless causal fields transforming as a 4-vector in a Fock space. And it seems there are photons in nature. So we welcome gauge redundancies. 

I think I agree with your conclusion, and I like your example of Hamiltonian. Yet I'd like to see a more elaborate argument on how this works in general. Note even in the case of a free particle Hamiltonian, formal manipulations already give weird contradictions(at least in a formal sense): for example, when we calculate the Hamiltonian of free scalar field as an integral of Hamiltonian density, we first reach the expression $H=\int dp E_{p}\frac{1}{2}(a_p a_p^\dagger+a_p^\dagger a_p)$, and this is already formally covariant. However, once we apply the commutation relation and get $H=\int dp E_{p}(a_p^\dagger a_p+ \delta(0))$, this is no longer covariant, but then throwing the c-number away makes it covariant again. The only moral of the story I can make out of it is that, the first form is formally covariant but since the first form is not a well-defined operator, we can't take much face value of the formal covariance, so let's manually check the covariance on its normally ordered form. But this is all like a happy coincidence to me that the normal ordering keeps the covariance, because such covariance doesn't clearly follow from the covariance of elementary quantum fields. Does this mean every time I see a formally convariant density, I really have to manually check the covariance of the normal ordering?(or maybe do the reverse engineering: require the normally ordered density to be covariant, and possibly revise the original density?)   

Hi @JiaYiyang With the conventional expressions (normal-ordered) for the other generators, your first Hamiltonian is not covariant. What expressions for boost and momentum generators do you have in mind? 

My high-level view here is that in QFT, as opposed to quantum mechanics with countable degrees of freedom, there are inequivalent representations (not related by unitary transformations) of the canonical commutation relations. This ambiguity is potentially problematic because different representations give rise to different physics. However, if the classical theory is Poincare invariant, one can make use of this symmetry to fix the ambiguity in the representation, namely, picking that representation that preserves the Lorentz covariance of the theory. 

Following Weinberg's approach, I define $U(\Lambda)$ such that $U(\Lambda) a_{p} U^{-1}(\Lambda)=\sqrt{\frac{E_{\Lambda p}}{E_p}}a_{\Lambda p}$(the normalization convention is following Peskin's, however.) 

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@JiaYiyang: Actually, your formula $U(\Lambda)|p\rangle=|\Lambda p\rangle$ (and my Fock version of it) is good only in the scalar case, as it ignores what happens on the spin/vector indices.

@ArnoldNeumaier, I ignored other indices for the sake of simplicity. If you want to include spin indices, you need to go through the little group analysis, which takes quite some extra effort. Or we can just quote the results presented by Weinberg in Chap2 of his textbook.

+ 1 like - 1 dislike

One way to see $c$ is a vector is to notice $c = -\langle 0 | J | 0 \rangle$. Since $|0\rangle$ is Lorentz-invariant, $c$ transforms like $J$ does.

answered Jun 27, 2015 by Ryan Thorngren (1,605 points) [ no revision ]

But note how the $U^{-1}\cdots U$ is applied: it's implemented as $U^{-1}\langle 0 | J | 0 \rangle U$ instead of $\langle 0 | U^{-1}J U | 0 \rangle$.

I find your statement very confusing. What is $U$ exactly? I thought it was an operator on the Hilbert space.

Yes it is, and you take $:J^{\mu}:= J^{\mu}+c^\mu$ as an operator identity, $c^\mu$ is really short for $c^\mu I$, with $I$ being the identity operator.

So what do you mean by $U^{-1} \langle 0 | J |0 \rangle U$?

It means $U^{-1} c I U$ where $c = -\langle 0 | J | 0 \rangle$.

Isn't $\langle 0|J|0\rangle$ a number (or scalar times the identity operator)? Then $U^{-1}I U=U^{-1}U=I$.

@Meng: yes, that's why he gets the apparent contradiction.

+ 0 like - 0 dislike

I think I've figured this out. The point is that, the rigorous meaning one can draw from the formal covariance of $J^\mu$ is that the momentum-space coefficient functions of $J^\mu$ (i.e. the functions in front of monomials of $a_p$ and $a^\dagger_p$) transform covariantly under the change of variable $p\to \Lambda p$. The covariance of the coefficient functions is unaffected by normal ordering, and is sufficient to give rise to the covariance of $:J^\mu:$. The rest of this answer will be an elaboration of the first paragraph. 

Let me first clarify the notations used and the meaning of the formal covariance of the ill-defined current $J^\mu$. I'm going to ignore the spin degrees of freedom in this discussion, but one should see the generalization to include spin only involves a straightforward (but perhaps cumbersome) change of notations. I'm also ignoring the spacetime dependence, that is to say I'm only considering the covariance of $J^\mu(0)$, and the generalization to $J^\mu(x)$ is straightforward and easy.

In the context of my question, $U(\Lambda)$ is defined as such that

$$U(\Lambda) a_{p} U^{-1}(\Lambda)=\sqrt{\frac{E_{\Lambda p}}{E_p}}a_{\Lambda p}.$$

The covariance of $J^\mu$ must be understood in a very formal and specific sense, the sense in which the covariance is formally proved. For example, in the case of a fermionic bilinear:

$$U(\Lambda)J^{\mu}U(\Lambda)^{-1}=U\bar{\psi}\gamma^{\mu}\psi U^{-1}\\ =U\bar{\psi}_iU^{-1}(\gamma^{\mu})_{ij}U \psi_j U^{-1}=\bar{\psi}D(\Lambda)\gamma^{\mu}D(\Lambda)^{-1}\psi= \Lambda^{\mu}_{\ \ \nu}\bar{\psi}\gamma^{\nu}\psi, $$

where $D(\Lambda)$ is the spinor representation of Lorentz group, typically constructed via Clifford algebra. Note in this formal proof, what's important is that, under the change $a_{p}\to \sqrt{\frac{E_{\Lambda p}}{E_p}}a_{\Lambda p}$ (ignoring spin indices of course) the elementary field transforms as $\psi \to D(\Lambda)\psi$. In the proof, no manipulation of operator ordering and commutation relations ever occurs: all we do is to do a change of integration variable, and let the algebraic properties of the coefficient functions take care of the rest. In fact, we'd better not mess with the operator ordering, as it can easily spoil the formal covariance (example: $H=\int \text{d}p\frac{1}{2}E_{p}(a_p a_p^\dagger+a_p^\dagger a_p)=\int \text{d}p E_{p}(a_p^\dagger a_p+\delta(0))$, see my longest comment under drake's answer).

To explain what's going on in more details without getting tangled with notational nuisances, let me remind you again I'll omit the spin degrees of freedom, but it should be transparent enough by the end of the argument that it's readily generalizable to spinor case, since all that matters is that we know the coefficient functions(even with spin indices) transform covariantly. The mathematical gist is, after multiplying the elementary fields and grouping c/a operators (during the grouping no operator ordering procedure should be performed at all, e.g. $a^\dagger(p_1)a(p_2)$ and $a(p_2)a^\dagger(p_1)$ should be treated as two independent terms), a typical monomial term in $J^\mu(0)$ has the form

$$ \int \left(\prod\limits_{i=1}^{n}\text{d}p_i\right)M(\{a^\dagger(p_i), a(p_i)\})f^\mu(\{p_i\}),$$

where $M$ is a monomial of c/a operators not necessarily normally ordered, but has an ordering directly from the multiplication of elementary fields.

The formal covariance of $J^\mu$ means

$$\Lambda^\mu_{\ \ \nu}\int \left(\prod\limits_{i=1}^{n}\text{d}p_i\right)M(\{a^\dagger(p_i), a(p_i)\})f^\nu(\{p_i\})\\=\int \left(\prod\limits_{i=1}^{n}\text{d}p_i\right)M(\{a^\dagger(\Lambda p_i), a(\Lambda p_i)\})f^\mu(\{p_i\})\\=\int \left(\prod\limits_{i=1}^{n}\text{d}q_i\right)\left(\prod\limits_{i=1}^n \frac{E_{\Lambda^{-1} q_i}}{E_{q_i}}\right) \left(\prod\limits_{i=1}^{m}\sqrt{\frac{E_{q_i}}{E_{\Lambda^{-1} q_i}}}\right) M(\{a^\dagger(q_i), a(q_i)\})f^\mu(\{\Lambda^{-1}q_i\}) ,$$

where $\prod\limits_{i=1}^n {E_{\Lambda^{-1} q_i}}/{E_{q_i}}$ comes from the transformation of measure and $\prod\limits_{i=1}^{m}\sqrt{{E_{q_i}}/{E_{\Lambda^{-1} q_i}}}$ from the transformation of c/a operators in $M$. This is equivalent to 

$$f^\mu(\{\Lambda^{-1}q_i\})\left(\prod\limits_{i=1}^n \frac{E_{\Lambda^{-1} q_i}}{E_{q_i}}\right) \left(\prod\limits_{i=1}^{m}\sqrt{\frac{E_{q_i}}{E_{\Lambda^{-1} q_i}}}\right)=\Lambda^\mu_{\ \ \nu}f^\nu(\{q_i\}).$$

The above equation makes completely rigorous sense since it's a statement about c-number functions. Obviously, this equation is sufficient to prove the covariance of the normal ordering 

$$ \int \left(\prod\limits_{i=1}^{n}\text{d}p_i\right):M(\{a^\dagger(p_i), a(p_i)\}):f^\mu(\{p_i\}),$$

since on the operator part only a change of integration variable is needed for the proof.

So let's recapitulate the logic of this answer:

1. The current is only covariant when written in a certain way, but not in all ways. (recall the free scalar field Hamiltonian example: $H=\int \text{d}p\frac{1}{2}E_{p}(a_p a_p^\dagger+a_p^\dagger a_p)=\int \text{d} pE_{p}(a_p^\dagger a_p+\delta(0))$, which is formally covariant in the first form but not in the second form.)

2. In that certain way where the current is formally covariant, the formal covariance really means a genuine covariance of the coefficient functions.

3. The covariance of the coefficient functions is sufficient to establish the covariance of the normally ordered current. 

answered Jul 2, 2015 by Jia Yiyang (2,465 points) [ revision history ]
edited Jul 2, 2015 by Jia Yiyang
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You U(Λ) is defined as such that

This is not a definition but a property satisfied by $U(\Lambda)$. In fact, this property fixes the convention used for scaling $a_p$. The formula cannot be said to define $U(\Lambda)$ - it defines only what it means that $U(\Lambda)$ is representation of the Lorentz group. A definition of $U(\Lambda)$ itself would instead say how to compute $U(\Lambda)$ (applied to a state vector). Please change your text accordingly!

@ArnoldNeumaier, well, it implies how it will act on the Fock space in momentum basis, i.e. 

$U(\Lambda)|p_1, p_2,\ldots \rangle= U(\Lambda)a^\dagger_{p_1}a^\dagger_{p_2}\ldots U(\Lambda)^{-1}|0\rangle$, so it is unambiguously defined.

it implies how it will act on the Fock space in momentum basis

No it doesn't, since we do not know how to evaluate the right hand side without knowing already how $U(\Lambda)$ acts.

 I think you forgot in your transformation formula a factor $D(\Lambda)$, else your computation with the $\psi$'s does not make sense.

Also, you can save a lot of notation and hence have a cleaner argument if you change the normalization of the $a$'s so that the $\sqrt{E_p}a_p$ (now) becomes the new $a_p$. 


No it doesn't, since we do not know how to evaluate the right hand side without knowing already how U(Λ) acts.

Why not? For example $U(\Lambda)|p_1, p_2 \rangle=U(\Lambda)a^\dagger_{p_1}a^\dagger_{p_2}U(\Lambda)^{-1}|0\rangle=U(\Lambda)a^\dagger_{p_1}U(\Lambda)^{-1}U(\Lambda)a^\dagger_{p_2}U(\Lambda)^{-1}$

$=\sqrt{\frac{E_{\Lambda p_1}}{E_{p_1}}}\sqrt{\frac{E_{\Lambda p_2}}{E_{p_2}}}a^\dagger_{\Lambda p_1}a^\dagger_{\Lambda p_2}|0\rangle=\sqrt{\frac{E_{\Lambda p_1}}{E_{p_1}}}\sqrt{\frac{E_{\Lambda p_2}}{E_{p_2}}}|\Lambda p_1, \Lambda p_2\rangle$.

 I think you forgot in your transformation formula a factor $D(Λ)$, else your computation with the ψ's does not make sense.

As I said I was ignoring the spin indices, so the latter derivation is really for a scalar case, but it should be transparent enough that it's readily generalizable to spinor case, since all that matters is that we know the coefficient functions(even including spin) transform covariantly. I'll make it clearer in my answer.

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I propose to use $U(\lambda):|p\rangle:=D(\Lambda,p)|\Lambda p\rangle\>$ as definition of $U$ on an arbitrary single-particle state with momentum $p$, to be extended canonically to multiparticle space (by a sum of terms acting on each factor of a tensor product separately). This gives a real definition independent of spin (whereas yours is an implicit definition that needs work to show why this ''defines'' a unitary transformation). As a consequence one has $U(\Lambda)a(p)=D(\Lambda,p) a(\Lambda p) U(\Lambda)$, which is an any-spin version of your abbreviated relation, but with a different normalization of the annihilators. Then you can do everything you did without any sloppiness, with all your square roots replaced by appropriate $D$'s, and your derivation is valid without any handwaving regarding spin indices.

@ArnoldNeumaier, thanks, I'll improve it when I get some time, but I assume you get what I'm conveying by this answer in its present form?

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