quantum field theory - Physical meaning of partition function in QFT

When we have the generating functional $Z$ for a scalar field

\begin{equation} Z(J,J^{\dagger}) = \int{D\phi^{\dagger}D\phi \; \exp\left[{\int L+\phi^{\dagger}J(x)+J^{\dagger}(x)}\phi\right]}, \end{equation}

the partition function is $Z(0,0)$. We know that the derivatives of the generating functional give the propagator for the system, and it is often said that $Z(0,0)$ relates to the vacuum energy, and it is formally given by

\begin{equation} Z(0,0) = \langle 0,t_f|0,t_i \rangle. \end{equation}

How does this matrix element represent the vacuum energy of the system? Is it to do with the size of the fluctuations between the times $t_i$ and $t_f$? Or what is another interpretation of $Z(0,0)$?

Answer

The partition function $Z[J]$, both in QM and in CM, is underdetermined: any multiple of $Z[J]$ gives rise to the same dynamics. This means that $Z[0]$ is arbitrary, and is usually set to one: $$ Z[0]\equiv 1 \tag{1} $$ effectively getting rid of vacuum diagrams, that is, we set $H|\Omega\rangle=0$. In other words: the energy of the vacuum is not measurable and can be set to any number we want. We can only measure differences in energies (except in GR), which means that a constant offset of energies is irrelevant.

The matrix element $$ \langle 0,t_f|0,t_i\rangle \tag{2} $$ can be interpreted as the amplitude of ending up with a vacuum state at the time $t_f$ if you start with vacuum at a time $t_i$. Or put it another way, it is the amplitude to get nothing if you initially have nothing. This number is, naturally, one: $$ \langle 0,t_f|0,t_i\rangle\equiv 1 \tag{3} $$ in agreement with $(1)$.

condensed matter - How do Dirac fermions arise in graphene, and, what significance (if any) does this have for high-energy physics?

Graphene has a honeycomb lattice (in the absence of defects and impurities). By considering the low-energy limit of the half-filled Hubbard model used to model the strongly interacting electron gas we find that the low-energy quasiparticles obey the dispersion relation for massless fermions. These details are all covered very nicely in a paper by Gonzalez, Guniea and Vozmediano (reference) among others.

It might seem like I'm answering the question. I'm following this line of exposition because I don't want to assume that this is a topic something commonly known or understood outside the condensed matter community. Any answers which elaborated on these basics would be very useful as they would help make the discussion more broadly accessible.

My primary question is more about the implications this fact has for high-energy physics, in particular the question of emergent-matter in theories of quantum gravity. The case of graphene is a canonical example in that regard where one obtains relativistic, massless excitations in the low-energy corner of an otherwise non-relativistic system - the 2D electron gas (2DEG).

Obviously I have my own beliefs in this regard and I will try to outline them in an answer. But I also want to solicit the communities views in this regard.

Answer

The answer you'll get from most high-energy physicists is that there are no implications whatsoever. Lorentz invariance is extraordinarily well-tested: see, e.g., http://arxiv.org/abs/0801.0287. In particular, there are many relevant operators in the Standard Model that one would expect to be generated if physics at a high scale is not Lorentz-invariant. Even some irrelevant operators that one might naively expect to appear with order-one Planck-suppressed coefficients are constrained to have smaller coefficients. Adding in gravity only makes the problem worse. For instance, most attempts to generate emergent GR from nonrelativistic theories will have an extra scalar mode and run into massive phenomenological difficulties, because they aren't really gauging the full diffeomorphism group.

To be slightly more clear: there are cases (and the free relativistic fermion emerging in the long-distance limit of graphene is one of them) where lattice symmetries can forbid dangerous relevant operators. This shouldn't happen for the full Standard Model (I assume someone has written down a careful argument for this somewhere, but I don't know a reference offhand). Still, even for the graphene case there are irrelevant operators, and we have bounds on those too. Furthermore, once you start thinking about gravity you're more or less forced to give up the hope of an underlying highly symmetric lattice that forbids all the dangerous operators.

One more half-joking comment: this argument also tells you the correct answer to the FQXi essay contest "Is Reality Digital or Analog?," so if someone fleshes it out carefully they could possibly make up to $10k from it.

(It is a good question, by the way; there's an obvious conventional wisdom from effective field theory that explains why you don't see high-energy theorists pursuing this sort of thing much, but from the outside it might not be so clear why such ideas don't generate much interest.)

electrostatics - Realistic vs Idealistic capacitance

I am doing an investigation into the differences of calculating capacitance using the well know formula for an idealistic parallel plate capacitor, based on the assumption of a uniformly distributed electric field:$C=\epsilon_0 \frac{A}{d}$ vs using numerical methods for calculating the capacitance of a realistic model i.e. with fringe fields (see here for a more detailed explanation of my methods).

I am investigating how the capacitance changes as i move the plates further away from each other, and what i have found is that, based on the results I'm getting, as the plates are moved further away from each other the ratio of realistic/idealistic increases, implying that the realistic electric field can store more energy.

My question is, is this correct and if not why?

Thanks

[Edit] I originally thought that this may be because the realistic formula is only valid for small d, however the ratio seems to increase linearly, rather than converging to a value and then dropping off leading me to think that this is not the problem?

[Info for comments] $$C=\frac{\epsilon_0 L}{V} \sum_{bound} |\phi_{outer}-\phi_{plate}|$$

$$C_{\infty}=\frac{\epsilon_0 A}{d}$$

Dividing them, where $A=lL$ i.e the area of the plate, gives:

$$\frac{C}{C_{\infty}}=\frac{d}{V l} \sum_{bound} |\phi_{outer}-\phi_{plate}|$$

Quantum state where uncertainty in kinetic energy is zero?

While reading Shankar's book on Quantum Mechanics, I encountered an interesting problem:

Compute $\Delta T\cdot\Delta X$, where $T = P^2/2m$.

I found several solutions online which arrive at the result $\Delta T\cdot\Delta X \ge 0$.

My question is: does there exist a state $|{\psi}\rangle$ which saturates this inequality, i.e. for which $\Delta T\cdot\Delta X = 0$? We know $\Delta X\ne 0$ (from the uncertainty relation between $X$ and $P$), so then we must surely have that $\Delta T = 0$. But I'm struggling to imagine a physical state with a well-defined kinetic energy! If it is indeed possible, please provide an example of such a state. Thanks!

Capillary action and conservation of energy

When I dip a paper towel in a cup of water the water gets drawn up due to capillary action. How is this reconciled with conservation of energy, as it would seem on the surface that the potential energy of the system has gone up with no work being done?

Answer

as the Wikipedia article you linked explains, capillary action is due to the "surface tension" (cohesion within the liquid) as well as "adhesion" (attraction between the water and the paper).

In particular, surface tension

http://en.wikipedia.org/wiki/Surface_tension

is the energy per unit area of the surface of the liquid. The molecules of water inside the liquid are nicely sitting in a potential "hole" induced by the adjacent molecules.

However, the molecules of water that are on the surface lack the attractive force to the now non-existent neighbors on the side where the liquid no longer exists. Because attractive forces are related to a negative potential energy, the lack of them leads to a positive potential energy.

Consequently, water will contain some extra positive energy called the "surface tension". It's a form of potential energy but it can only be interpreted in this way at the molecular level. As hinted in the previous paragraph, the surface tension is a contribution to the energy that is proportional to the surface. That's why the surface tension causes water to take the shape of spherical droplets, among many other things, because the sphere minimizes the surface among shapes of the same volume.

Similarly, adhesion - between the paper and water in this case - adds a negative energy proportional to the surface of contact. Again, it may be explained by the (now added, not lacking) negative potential energy between the molecules of the water and the molecules of the paper that attract. To reduce the energy, water tries to touch the paper as much as it can.

Only the total energy is conserved. The energy needed to lift the water is obtained from the surface tension and adhesion. Microscopically, you may literally imagine that the molecules of water at the top were attracted by the nearby (and so far higher) molecules of paper.

experimental physics - Is it possible to use "negative sound waves" to "cancel out" a sound to create silence?

I saw youtube videos that claimed to do this, although I'm quite certain the videos just excluded sound and lied.

However, I am wondering if the physics of this is actually possible - to create a completely negative sound wave that cancels the sound out to a flat line i.e. complete silence.

p.s. I am not referring to white noise, which prevents the sound from being heard by overriding it with a different sound (white noise). Rather, I am referring to making a sound be completely silent by generating the inverse, opposite, or "negative" sound wave.

Answer

Yes it can be done, and indeed it's a well established technology called active noise control.

The idea is based on destructive interference. If at some point two sound waves have the same amplitude and frequency and they're 180º out of phase then they will sum to zero and the sound intensity at that point will be zero. Your phrase negative sound just means sound that is 180º out of phase with the sound you're trying to cancel.

However it's rarely possible to cancel the sound over more than a very small region. The cancellation requires the amplitude of the cancelling sound to be precisely matched to amplitude of the noise. The trouble is that the amplitude of sound typically decreases as the inverse square of distance from its source. As a result it's hard to get the sound amplitudes to match over more than a restricted area. However noise cancellation is used in special cases like noise cancelling headphones.

experimental physics - What differs string theory from philosophy or religion?

Possible Duplicate:
What experiment would disprove string theory?

A hypothesis without hard evidence sounds very much like philosophy or religion to me. All of them tries to establish a functional model explaining how the world works. In my understanding, string theory is an unprovable theory. What differs string theory from philosophy or religion?

Answer

How does string theory differ from philosophy or religion?

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  As the field of mathematics, string theory differs from philosophy and religion not by its experimental verifiability but by its methods.

  
  


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  It is mostly based on mathematical reasoning.

  
  


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  It uses basically the same tools that are used in other areas of quantum physics.

  
  


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  It has many points of contact with other areas of physics and mathematics.

  
  


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  It is studied in physics and math departments.

  
  


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  Some of those working on string theories are also highly regarded in other areas of theoretical physics or mathematics. (E.g. Witten got the Fields medal, the highest distinction in mathematics.)

  
  


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  Some concepts first discussed in string theory found later use in particle physics. (e.g., AdS/CFT, hep-th/9905111; hep-ph/0702210)

  
  
  


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  It may make one day testable predictions of previously unknown effects, and can then be checked for its validity.