# Two confusing details of Weinberg's exposition of spontaneous symmetry breaking.

+ 4 like - 0 dislike
833 views

I have been confused by two details of Weinberg's delivery of spontaneous symmetry breaking(SSB) in Volume II of his QFT textbook. I'm reading a paperback, so the edition must be after 2005, and both of the confusions come from page 164-166.

(1)On page 164, when Weinberg tries to justify that there are preferred basis among the degenerate vacua for the system to fall into, he starts with a system with a simple symmetry $\phi \to -\phi$, and he writes

For instance, in a theory with a symmetry $\phi \to -\phi$, even if $\Gamma(\phi)$(quantum effective action) has a minimum for some nonzero value $\bar{\phi}$ of $\phi$, how do we know that the true vacuum is one of the states $|\text{VAC},\pm\rangle$ for which $\Phi$ has expectation values $\bar{\phi}$ and $-\bar{\phi}$, and not some linear combination like $|\text{VAC}, +\rangle+|\text{VAC}, -\rangle$ that would respect the symmetry under $\phi \to -\phi$? The assumed symmetry under the transformation $\phi \to -\phi$ tells us that the vacuum matrix elements of the Hamiltonian are

$\langle \text{VAC}, +|H|\text{VAC}, +\rangle=\langle \text{VAC}, -|H|\text{VAC}, -\rangle\equiv a$

(with $a$ real) and

$\langle \text{VAC}, +|H|\text{VAC}, -\rangle=\langle \text{VAC}, -|H|\text{VAC}, +\rangle\equiv b$

(with $b$ real), so the eigenstates of the Hamiltonian are $|\text{VAC}, +\rangle \pm |\text{VAC}, -\rangle$, with energies $a\pm|b|$.

My confusion only concerns the last sentence, shouldn't $a$ be a common eigenvalue of $|\text{VAC}, \pm\rangle$, so $a$ will also be the eigenvalue of $|\text{VAC}, +\rangle \pm |\text{VAC}, -\rangle$? Where does $a\pm|b|$ come from? (In fact, the later discussions seem to remain intact if the eigenvalue is $a$ instead of $a\pm|b|$.)

(2)Starting from the end of page 165 continuing to the beginning of 166, he writes:

For infinite volume, a general vacuum state $|v\rangle$ may be defined as a state with zero momentum

$\mathbf{P}|v\rangle=0$

for which this is a discrete momentum eigenvalue.(This excludes single-particle or multiparticle states, for which the momentum value zero is always part of a continuum of momentum values in a space of infinite volume.) In general there may be a number of such states. They can usually be expanded in a discrete set, ...

This brings up three questions:

A. Why would he bother to require $\mathbf{P}|v\rangle=0$, shouldn't it be an automatic consequence of the more general requirement that vacuum must be Poincare-invariant?

B. And how do we implement the other requirement that "for which this is a discrete momentum eigenvalue"? I mean, if we only look at 3-momentum spectrum, there's no way to tell on the spectrum if a point represents a vacuum or a zero-momentum multipaticle state, and surely the 3-momentum spectrum is always a continuum.

C. Why would one expect "They can usually be expanded in a discrete set"?

asked Dec 27, 2014
@RonMaimon, Could you have a look at this question? Mostly question (2), I suspect (1) is just a careless mistake of Weinberg,
I don't understand question 1). In the basis VAC+, VAC-, $H$ is a 2 by 2 matrix with $a$ on the diagonal and $b$ as antidiagonal entries. The characteristic polynomial of $H$ is $X^2 - 2aX+(a^2-b^2)=0$, of discriminant $(2a)^2-4(a^2-b^2)=4b^2$, and so the eigenvalues are indeed $a \pm |b|$.
@40227, but shouldn't vacua be degenergate eigenstates by definition? And hence VAC+ and VAC- should both be eigenstates with eigenvalue $a$, so is any of their linear combinations.

## 1 Answer

+ 4 like - 0 dislike

(1): The symmetry assumptions let one only conclude that the Hamiltonian $H=cP_0$ satisfies the conditions given by Weinberg, not more. On the other hand, the quantum vacuum state state is defined as the state of minimal energy, hence would be the eigenstate corresponding to the eigenvalue $a-|b|$, which is a superposition of the two classical vacua. For a finite system, nothing more can be said; there is no reason why $b$ should vanish. (Your intuition for this is classical - but $|VAC,\pm\rangle$ are not quantum vacua but classical vacuum states regarded as the quantum states corresponding to a semiclassical model.)

If the volume is large, Weinberg indicates an argument that shows that $b/a$ decays exponentially with the volume, hence $b$ vanishes in the thermodynamic limit of an infinite system. But as long as the volume is huge but finite, and the system is very slightly perturbed to break the symmetry, the ground state is (as Weinberg argues) very close to one of the classical vacuum states. In the thermodynamic limit the broken symmetry persists and the ground state becomes one of the two classical vacuum states.

(2): (A) The requirement that vacuum must be Poincare-invariant is an extra axiom that Weinberg doesn't impose. The requirement that the spatial momentum vanishes can be imposed on an arbitrary eigenstate of the Hamiltonian, as it just removes the degeneracy caused by spatial translation invariance. As a result, the spectrum of the 4-momentum $P$ (which is a union of mass shells) is restricted to the spectrum of the Hamiltonian in the center of mass framework. This is the same trick that removes translation invariance in nonrelativistic systems and produces a Hamiltonian for which the spectrum has the usual form discrete (bound state) spectrum plus continuous (scattering state) spectrum. (B) Thus it makes sense to ask for a discrete energy eigenvalue, which is what Weinberg meant although he talks about a momentum eigenvalue - an obvious nonsense since an operator vector cannot have eigenvalues, only a vector-valued spectrum. (Another item for your list of corrections!) (C) The relativistic situation is not really different in essence from the nonrelativistic version, where the ground state usually has finite multiplicity only. (''Natural'' exceptions would require an infinite-dimensional lie group of symmetries.)

answered Mar 8, 2015 by (15,488 points)

Thanks for the answer, but I'm not quite convinced on a few points.

(1)I would say at the very least $|VAC\pm\rangle$ must be eigenstates of $H$, then the condition

$\langle \text{VAC}, +|H|\text{VAC}, +\rangle=\langle \text{VAC}, -|H|\text{VAC}, -\rangle\equiv a$

is equivalent to that $|VAC\pm\rangle$ both have $a$ as their eigenvalue, so will be any of their linear combination. Sure we don't have to impose that $|VAC\pm\rangle$  are mutually orthogonal(hence a nonzero $b$), but this has no effect to my above argument.

(2)Now I'm thinking maybe Weinberg restricts the discussion only to the context of discrete symmetry? In the case of continuous symmetry: there has to be infinitely many mutually orthogonal ground states, so your (C) can't be true; there are massless excitations(goldstone), so even the energy spectrum is continuous, your (B) can't be true either. (unless by " infinite-dimensional lie group of symmetries" you simply mean continuous symmetries)

What is your evidence for your claim that the classical states (i.e., the basis of the quantum space of potential vacuum states) are eigenstates of $H$? There is none.

You are right, in Section 19.1 Weinberg considers a reflection symmetry, and my explanation above is (as Weinberg's) only for this case.

In case of continuous symmetries, the set of of classical states is infinite, so the quantum space of potential vacua is infinite-dimensional, and the picture is more involved, treated by Weinberg in Section 19.2. In this case, the energy spectrum is indeed continuous, and the ground state is a (still isolated) branch point. This has no nonrelativistic analogue.

By " infinite-dimensional lie group of symmetries" I mean infinitely many independent symmetries of the quantum ground state - without the broken part.

What is your evidence for your claim that the classical states (i.e., the basis of the quantum space of potential vacuum states) are eigenstates of H?

Emm, I don't know what exactly do you mean by classical states. However, following Weinberg's definitions and arguments, when VEV minmizes the effective potential, the corresponding states $|VAC\pm\rangle$ will have the minimal possible expectation value on Hamiltonian operator, hence they must be eigenstates, as can be proven mathematically(easy in finite dimensional case, but I believe it holds for infinite dimensional case with some care).

Ok, I now see what you mean. I was reading the text too locally, trusting that what Weinberg said was correct, and interpolated the text by assuming that the two vacuum states had a classical meaning only (zero loop approximation). But this is not what he assumes; he takes them to be two global minimizers of the effective quantum action, or with translation invariance, of the effective potential. But then his argument is correct only cum grano salis. Indeed, if the effective potential attains its global minimum at two symmetric points then this minimum is the minimal expectation value of the Hamiltonian over all constant fields, hence both vacuum states and their superpositions are eigenstates of this minimal value. He confirms this indirectly close to the end of p.164 where he discusses the ground state of an isolated chair.

But this means that he means to discuss something different when discussing two distinct eigenvalues, as otherwise his argument has no force. The intended $H$ in this argument must be a symmetric perturbation of the exact $H$ of the field theory (e.g., finite size truncations), since for these perturbations his arguments hold, and the eigenstates are symmetric mixtures of the edge vacua. Indeed, in the middle of  p.165 he continues the argument with considering odd perturbations, saying that these affect the diagonal elements only, and much more than the exponentially suppressed off-diagonal part, so that the off-diagonal entries are negligible in comparison and the eigenstate with smallest energy (of the perturbed system) is nearly one of the edge vacua. So the whole story makes sense.

Thus $H$ on p.164 should be read not as the Hamiltonian of the quantum field theory but as the Hamiltonian of any nearby symmetric system.

That makes better sense, I'll check the book again later.

## 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): Email me at this address if my answer is selected or commented on: 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:p$\hbar$ysicsOverflo$\varnothing$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). To avoid this verification in future, please log in or register.