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  Some questions about the theory of complex scalar fields

+ 0 like - 3 dislike

In a theory with two complex scalar fields say $\phi_1$ and $\phi_2$, one often takes the independent degrees of freedom to be $\phi_1$, $\phi_1^*$, $\phi_2$, $\phi_2^*$. This confuses me at times to be able to clearly discern as to what is independent.

  • Should there be a non-trivial commutation relation between say $\phi_1$ and the canonical momenta conjugate to $\phi_1^*$?

I would think that there should not be any since they are by definition independent degrees of freedom but then it is also true that the canonical momenta conjugate to $\phi_1^*$ is the partial time-derivative of $\phi_1$ - so I am not very sure.

But one after all has to explicitly use this identification of the canonical momenta conjugate to $\phi_1^*$ with $\frac{\partial \phi_1}{\partial t}$ when rewriting the Hamiltonian in terms of only $\phi_1$, $\phi_2$, $\phi_1^*$, $\phi_2^*$ and their conjugate momenta. (..removing all time derivatives..). So what is "independence".

  • Since these four are independent fields, I guess one should use a different pair of creation and annihilation operators for each of $\phi_1$, $\phi_1^*$, $\phi_2$, $\phi_2^*$ and let them be $a$s, $b$s ,$c$s and $d$s. (By $a$s I would mean its creation partner also)

Since the canonical momentum conjugate to $\phi_1$ is the time-derivative of $\phi_1^*$ and similarly for $\phi_2$, there will be non-trivial commutation relations induced only between the $a$s and $b$s and between $cs$ and $ds$ rather than the ``usual" situation of having non-trivial commutation relations between $a$ and $a^*$s (which would be zero here). Is this right?

  • How many conserved quantities does this theory have? Since this theory has a global $SU(2)$ symmetry that gives $3$ conserved quantities. Now the rank $3$, Belinfante-Rosenfeld tensor (..anti-symmetric in its last two indices..) should give $6$ conserved quantities (the same as the dimension of the homogeneous part of the Lorentz group) and the conservation of the rank $2$ stress-energy tensor should give $4$ more conserved quantities (the 3 momenta and total energy). So I would have thought that there are $3+6+4$ conserved quantities. Is that right? (or am I massively over counting!?) Now for a similar theory with $n$ complex scalar fields I should get $6+4+(n^2-1)$ conserved quantities. (The last addition coming from the dimension of the $SU(n)$ group). Is this argument right?
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asked Oct 15, 2011 in Theoretical Physics by Anirbit (585 points) [ no revision ]
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This might be an opportunity for us to figure out how we feel about homework problems. This question is more or less identical to a problem from Peakin and Schroeder. In my mind it is too elementary for this site, more about accounting than any concepts, but I can see how reasonable people may have a different opinion.

This post has been migrated from (A51.SE)
The website will fail to become the physicist's MathOverflow if non-research level questions like this one are allowed. As simple as that.

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@Moshe Its very difficult for me to have a judgement as to what is a "homework" problem and what is not. I don't understand what is the cut-off at these sites. I generally would have thought that anything in a graduate level text-book should be acceptable - at least that used to be the case with MathOverflow earlier. Now it seems that both MathOverflow and this site are trying to cut out even beginning graduate level texts. Then it gets very tricky. Instead of Peskin and Schroeder if it was from say Weinberg's book then would it become any more acceptable?

This post has been migrated from (A51.SE)
@Moshe Though its a different matter that I didn't know that Peskin and Schroeder talked of this question until very recently. If graduate level questions are also pushed to the stackexchange site then that would cause an immense problem there given that the other site is so full of undergraduate, high-school and even completely non-sensical or bizarrely amateurish questions! Hence I am not sure that trying to draw a line between "graduate level" and "research level" is a very productive idea and very soon that line gets very hard to draw.

This post has been migrated from (A51.SE)
This is not my site, I am just initiating a discussion and expressing my opinion. In my opinion allowing specific homework problems will overwhelm the site, there are too many of them and they are easier to ask and answer. It would be however reasonable to allow graduate level questions, though they are not research level, as long as they are about general issues (and not minutiae of specific examples) and therefore have a chance of being of general interest. Again, reasonable people may have a different opinions, which is why I initiated the discussion (on meta) to begin with.

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@Anirbit, there is a perfectly good place for such questions, which is physics.SE. If they are not selective enough in their choice of questions or answers, you can take it up with them. This site has a different purpose, and transforming it into another physics.SE site to compensate for its shortcomings is not the way to go.

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Yes, please, we'd love to have this sort of question at physics.SE.

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3 Answers

+ 2 like - 0 dislike

The 1st part of the question doesn't really need QFT, it can be asked in mechanics too. Consider a harmonic oscillator of mass m with two coordinates x, y and a symmetric potential

V(x, y) = k(x^2 + y^2) / 2

We can define the variables

z = (x + iy) / sqrt(2)

z* = (x - iy) / sqrt(2)

The conjugate momenta of x and y are

p = m dx / dt

q = m dy / dt

Straightforward computation shows

{x + iy, p + iq} = 0

{x + iy, p - iq} = 2

{x - iy, p + iq} = 2

{x - iy, p - iq} = 0

So the conjugate momentum of z is

w = m dz* / dt

and the conjugate momentum of z* is

w* = m dz / dt

However, {z, w*} = {z*, w} = 0, so this commutation relation is trivial.

Regarding conserved quantities it depends on the potential which you haven't specified. For the free theory with equal masses for phi1 and phi2, the full internal symmetry group is O(4) since we can rewrite the theory in terms of 4 real fields. This yields dim O(4) = 6 conserved quantities.

The Belifante-Rosenfeld tensor actually yields only 3 conserved quantities in the usual sense: the angular momenta. These correspond to rotation symmetry. The other 3 quantities correspond to generators of the Lorentz group which don't commute with time translation: the boosts. So these quantities are not conserved unless you introduce explicit time dependence into their definition.

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answered Oct 15, 2011 by Squark (1,725 points) [ no revision ]
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Thanks for the detailed answer. I would think that the quantum mechanics analogy is somewhat trivial because there the [z,w^*] commutator is coming out as a consequence of something we know about the x and p commutators and things can be reconstructed back. But in field theory we have nothing more basic than the z, z*, w and w*. So at the level of z and w itself one would have to make a decision of what the commutators are - then how does one decide that only [z,w] and [z*,w*] commutators are non-trivial?

This post has been migrated from (A51.SE)
Can you kindly explain why the "spatial" part of the Belinfante-Rosenfeld tensor is not conserved? If I think of the conserved tensor as being $M^{\mu \nu \lambda}$ then the conservation equation is $\partial _\mu M^{\mu \nu \lambda} = 0$ Now I would think that $\int M^{0 \nu \lambda} d^3x$ is conserved for any value of $\nu$ and $\lambda$. What is wrong here? Though the $O(4)$ argument will produce $6$ conserved charges, one still can't match the count to the total conserved charges coming from stress-energy tensor (4), Belinfante-Tensor (?) and global $SU(2)$ (3). I am a bit confused here.

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The quantum mechanics analogy is precise since we can define the real fields Re phi_i, Im phi_i as analogues of x, y. Of course you can always compute the Poisson brackets directly starting from the Lagrangian formalism and it is doable in the field theory case as well and yields the same answer. The computation is closely analgous to what happens in the harmonic oscillator case. In fact a free field is nothing but a collection of an infinite number of harmonic oscillators (Fourier modes)

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The Belifante-Rosenfeld tensor depends explicitely on the spacetime coordinates hence doesn't yield conserved quantities in the usual sense (because of explicit dependence on time). The explicit time dependence only drops out for angular momenta. There is a perfect analogy in classical Newtonian mechanics. The "conserved quantities" corresponding to Galilean boosts are the coordinates of the center-of-mass at t = 0 i.e. X_i - V_i t where X_i are the coordinates of the center of mass and V_i is the velocity of the center-of-mass.

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The global SU(2) is contained in the global O(4). Just consider the 2-dimensional complex vector space on which SU(2) acts as a 4-dimensional real vector space and SU(2) becomes a subgroup of O(4)

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I guess you are alluding to the fact that what the conservation equation implies is that $\frac{\partial}{\partial t} \int M^{0\mu \nu}d^3x = 0$ and this does not imply that $\frac{d}{d t} (\int M^{0\mu \nu}d^3x) = 0$ .. right? But then I am wondering how one can be careful about this - since the Noether procedure only gives us a "conserved current" in the sense that some $j^\mu$ such that $\partial _\mu j^\mu = 0$. Now given this how does one go about deciding whether $\frac{d}{dt} (\int j^0 d^3x) $ is zero. I haven't seen the books being clear about this point!

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The problem with the quantum mechanics analogy is that here there is a clear notion of a $p$ to begin with and one knows the x-p commutation. But in field theory everything has to be defined between $\phi$ and $\pi$ to start off. I guess the only clean way out is to follow a postulate that commutation has to be trivial between any field and any other field's canonical momenta. I can't see any other way out of this but to imagine that as a part of the quantization principles.

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+ 1 like - 0 dislike

I will only answer the first parts of the OP's question(v1). The fields $\phi^*_a$ are not independent, but the complex conjugate of the $\phi_a$ fields, so there are 2 complex degree of freedom, or 4 real degree of freedom in total. Concerning the canonical momenta and canonical commutation relations, I recently answered a similar question on physics.SE for the case of one complex scalar field $\phi$.

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answered Oct 15, 2011 by Qmechanic (3,120 points) [ no revision ]
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For the first part. A complex field $\phi$ is equivalent to two independent real fields $\phi_1,\phi_2$, i.e. $\phi=\phi_1+\mathrm i\phi_2$. By real field I mean a hermitian field in the sense that $\phi(f)^\ast = \phi(\bar f)$.

To use different pairs creation and annihilation operators for $\phi$ and $\phi^\ast$ would be wrong, because $\phi$ "annihilates" what $\phi^\ast$ "creates" and viceversa. If $a_1$ and $a_2$ are the creation operators for $\phi_1,\phi_2$, respectively, then $b=a_1+\mathrm i a_2$ and $d=a_1 - \mathrm i a_2$ are the creation operators of $\phi$ and $\phi^\ast$, in other words formally $\phi=d+b^\ast$ and $\phi^\ast=b+d^\ast$.

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answered Oct 15, 2011 by Marcel (300 points) [ no revision ]
If while writing the canonical momenta and their commutation relations we are treating $\phi$ and $\phi^*$ as independent fields then why shouldn't their mode decomposition also respect that and use independent creation and annihilation operators? Otherwise where would that independence translate to?

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Like Moshe wrote above, you avoid problems by writing everything in terms of 4 independent real fields.

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@Marcek Thanks for the comments. Its a bit queer to come to terms with the fact that the fields $\phi$ and $\phi^*$ are complex conjugates of each other but will be treated independently for quantization purposes. Its not completely intuitively obvious that the quantum commutation rules that are imposed should be such that $[\phi, \frac{\partial \phi^*}{\partial t}]$ is non-zero but $[\phi, \frac{\partial \phi}{\partial t}]=0$. (...one so rarely ever runs into complex scalar fields that intuition is weak here..)

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