• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,079 questions , 2,229 unanswered
5,348 answers , 22,758 comments
1,470 users with positive rep
819 active unimported users
More ...

  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?
This post has been migrated from (A51.SE)
asked Oct 15, 2011 in Theoretical Physics by Anirbit (585 points) [ no revision ]
Most voted comments show all comments
@Moshe I guess I have good idea of what homeworks are, given that I have done hundreds of them! It seems that the websites of MathOverflow and this one are raising the levels to even beyond beginning graduate level stuff. This is going to probably cut out a very large clientele base. Of course from the beginning both the websites started as being for "research level" but graduate level questions were not shunned - like I could learn a hell lot about Riemannian geometry and space-time topology through discussions on that site.

This post has been migrated from (A51.SE)
@Moshe But these days MathOverflow is working such that questions on ramification divisors of algebraic curves are considered too elementary and here it seems that basic QFT questions are looked down upon! I am increasingly feeling that there is an increasing assumption that the people asking the questions are necessarily from large institutes which have adequately capable peer group and TAs and other such resources which will help deal with anything "elementary" and a website like this will not be needed for such questions which can be dealt locally.

This post has been migrated from (A51.SE)
@Moshe Till even recently I wasn't so fortunate to have so many resources at my disposal! (..now may be some..) So these websites basically seem to be closing its doors to people from small places or for self-learners beyond the institutionalized system. I am not sure that this should be the way to go.

This post has been migrated from (A51.SE)
@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.

This post has been migrated from (A51.SE)
Yes, please, we'd love to have this sort of question at physics.SE.

This post has been migrated from (A51.SE)
Most recent comments show all comments
Also, the purpose of homework is for the student to practice sitting down quietly, organize their thoughts and figure things out for themselves. Getting the answers on the internet defies that purpose.

This post has been migrated from (A51.SE)
Dear @Anirbit, $\phi^*$ is linearly independent of $\phi$ simply because you can't write the former as a linear combination of the latter. Complex conjugation is not a linear operation: it is antilinear and in the context of complex calculus, it is a very unnatural operation. Sorry to mention this general point but if you haven't understood why complex conjugate quantities are independent of the original ones, you couldn't have possibly understood the background for a single among 100 questions you previously asked (about representations of exotic SUSies in exotic dimensions etc.).

This post has been migrated from (A51.SE)

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.

This post has been migrated from (A51.SE)
answered Oct 15, 2011 by Squark (1,725 points) [ no revision ]
Most voted comments show all comments
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.

This post has been migrated from (A51.SE)
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)

This post has been migrated from (A51.SE)
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!

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
About the counting - I guess the $4$ conserved quantities from $T^{\mu \nu}$ are okay and hence you would count the full symmetry size as either $3+3+4$ or as $6+4$ depending on taste!

This post has been migrated from (A51.SE)
Most recent comments show all comments
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.

This post has been migrated from (A51.SE)
+ 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 $\phi$.

This post has been migrated from (A51.SE)
answered Oct 15, 2011 by Qmechanic (3,120 points) [ no revision ]
+ 1 like - 0 dislike

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 $\phi^\ast=b+d^\ast$.

This post has been migrated from (A51.SE)
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?

This post has been migrated from (A51.SE)
Like Moshe wrote above, you avoid problems by writing everything in terms of 4 independent real fields.

This post has been migrated from (A51.SE)
@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..)

This post has been migrated from (A51.SE)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights