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16

Good question! First you need to know that parity refers to the behavior of a physical system, or one of the mathematical functions that describe such a system, under reflection. There are two "kinds" of parity: If $f(x) = f(-x)$, we say the function $f$ has even parity If $f(x) = -f(-x)$, we say the function $f$ has odd parity Of course, for most ...


9

The thing is that the operation "exchange of two particles" has to be defined properly. What is the meaning of $P$ ? We can imagine the operator $P$ is not physical (in the sense that is does not correspond to a physically possible operation). For instance, $P\psi(x_1,x_2)=\lambda\psi(x_2,x_1)$ in the sense that it only exchange the argument of the ...


9

No, it really is arbitrary. The reason we use the right hand rule today (although it may have been chosen for different reasons of convenience in the past) is simply that our coordinate system of choice is right-handed. Mathematically, this means that we define the directions of the axes so that you have to use the right-hand rule to evaluate this cross ...


8

Our spacetime cannot be unorientable. That's because the laws of physics describing our spacetime are not left-right symmetric. We say that they break the P-symmetry (parity) or that they are "chiral" (derived from a Greek word for the hand which is either left or right.) For example, a left-handed neutrino would turn into a right-handed neutrino if you ...


7

Presumably you are asking about the communication ambiguity in physics: can we unambiguously specify what we mean by "a right handed coordinate system" to a correspondent far away without a pre-arrnage communications channel (i.e. using SETI)? For a long time the answer seemed to be "no", but the discovery of parity violation in 1957 changed the answer to ...


6

First, an assignment of the parities. The parity of fermions is a bit ambiguous because one may always redefine parity by $$ P \to P (-1)^{2J}, P(-1)^L, P(-1)^{3B}, P(-1)^{3Q} $$ or one may add the product of several factors of this kind because the second factor is a multiplicatively conserved sign. By this definition, one gets another parity that is still ...


5

There is no mathematical difficulty of a non-time-orientable spacetime in GTR, and they can be generated by taking a quotient space rather easily, as you suggest. On the other hand, the same kind of action can be in some sense be undone as well. For any spacetime $M$, let $M' = (p,o)$, where $p\in M$ and $o$ is a time orientation. Then this double ...


4

I) Consider an arbitrary coordinate transformation $$x^{\mu}\longrightarrow x^{\prime \nu}~=~f^{\nu}(x).$$ Let $$J ~:=~\det(\frac{\partial x^{\prime \nu}}{\partial x^{\mu}})$$ denote the corresponding Jacobian. Traditionally in physics, a scalar $\sigma$ transforms as $$ \sigma ~\longrightarrow~ \sigma^{\prime}~=~\sigma, $$ a pseudo-scalar ...


4

If you have two coordinate systems with the same origin, you can represent a (linear) transformation of coordinates from one to another as a matrix. This matrix has either positive or negative determinant. This sign of the determinant is what gives the transformation its parity. (All this applies to any number of dimensions, not just 3.) If you compose ...


4

I have a partial answer to my question, which I'm posting down here because the question was getting too long. After a good look online at a bunch of confusing (to me) papers from the nuclear physics literature, I came upon this review: Intrinsic reflection asymmetry in atomic nuclei. P. A. Butler and W. Nazarewicz. Rev. Mod. Phys. 68 no. 2, pp. 349-421 ...


3

$PT-, T-, P-$ transformations refer to subgroup of discrete transformations of the Lorentz group. They transform connected components of the Lorentz group between each other ($PT$ transformation transforms $L^{\uparrow}_{+}$ representation to $L^{\downarrow}_{+}$). In general, they can't be represented as the special case of rotation, which refer to subgroup ...


3

No, actually it implies that the original particle has positive parity. That's because parity is a multiplicative quantum number, which means that when you want to find the overall parity of a system that consists of multiple parts, you multiply the parities of the individual parts. If you have two decay products each with parity -1, then you multiply them ...


3

No, it's not true. Suppose I'm floating in outer space (presumably in a space suit or something else to keep me alive). I'm still me, and I still know that, for example, my left hand is the one on the left, and my right hand is the one I can write with. Even on Earth, we don't need environmental clues to distinguish left from right; it's more a matter of ...


3

You prove the equality of operators by applying them to a function, we have $$ H = - \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x) $$ Ergo: $$ HP f(x) = H f(-x) = (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(-x) = - \frac{\hbar^2}{2 m} f''(-x) + V(x) f(-x) $$ and $$ PH f(x) = P (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(x) = P (- ...


3

Dear Pie86, the emission of particles in a weak decay is a complicated reaction, and Gell-Mann's totalitarian principle applies: every process or effect that is not prohibited by a symmetry will occur at a nonzero probability. The asymmetry or the spin-momentum correlation for the electrons is such an effect. In this case, it is infinitely unlikely that the ...


3

Dear lurscher, the quote is the kind of C-physics described by the C-word which is a favorite word of mine but is discouraged on this server, so I won't use it - but you have used it. You don't misunderstand anything - quite on the contrary, you're right on the money. These comments about a non-existent test of parity in the equivalence principle are due to ...


3

The statement you cited does not imply that a complex representation of a gauge group implies a chiral gauge theory in general. This only holds true if the gauge group corresponds to a chiral symmetry in the first place. A chirally symmetric theory contains massless fermions. Regarding your counterexample: it is true that QCD contains fermions in the ...


3

I slightly deviate from your notation and use $\phi $ to denote the scalar field as its more standard. Also I should point out that quantum fields are operators and thus under a transformation they get acted on from both the left and the right. The complex scalar field is given by, \begin{equation} \phi (x) = \int \frac{ \,d^3p }{ (2\pi)^3 } \frac{1}{ ...


3

spin 1/2 fermions (electron, proton, neutron, muon, tau, quarks) have +1 parity (by convention as pointed out in Anna's comment). The corresponding anti-fermions have -1 parity. Bosons and their anti-particles have the same parity. See this and this lecture for more information on parity.


2

I haven't read that book, but I did read Feynman's discussion of (sounds like) exactly the same thing. Easy: Tell the aliens how to build a telescope, then describe the configuration of some galaxies near them. OK OK, but suppose we rule that out: We can't see any objects in common. Easy: Send them circularly-polarized radio waves (thanks @Anonymous Coward). ...


2

Well you'll have the typical conservation laws: Conservation of energy Conservation of momentum Conservation of (generalized) angular momentum Conservation of electric charge And some more particle-specific: Conservation of lepton number Conservation of baryon number Conservation of flavour (note that anti-particles have opposit flavour) For the ...


2

I think it's a matter of choice. If you look through several books you'll see all the possible combination $C\Psi(x)C$, $C\Psi(x)C^{-1}$, $C\Psi(x)C^{\dagger}$ (and the same for $P$ and $T$). I think it all comes down to the representation you are using. Like it is said in the book of Sterman (page 524) :"The precise nature of $T$ depends on the ...


2

generally under symmetry transformation $S$, $$ O \to S O S^{-1} $$ if $S O S^{-1}=O$ then $O$ is invariant under the symmetry transformation $S$, so $S$ commutes with $O$: $$ [S,O]=0 $$ This is correct as you said. $$ C(\hat{O}| v \rangle)=(C\hat{O}C^{-1})(C| v \rangle)\\ P(\hat{O}| v \rangle)=(P\hat{O}P^{-1})(P| v \rangle)\\ T(\hat{O}| v ...


2

I assume OP means an even potential $V(x)~=~V(-x)$, e.g., a finite square well potential $V(x) ~\propto~ \theta(|x|-a) $. Then the answer to the question(v1) is No. Sketched proof: Under the assumption that $V$ is even, the Hamiltonian $$H= \frac{p^2}{2m}+V(x)$$ then commutes with the parity operator $P$. So the operators $H$ and $P$ can be ...


2

The charged current part of the Lagrangian of the electoweak interaction, for the first generation of leptons, is : $$L_c = \frac{g}{\sqrt{2}}(\bar \nu_L \gamma^\mu e_L W^+_\mu + \bar e_L \gamma^\mu \nu_L W^-_\mu )$$ The first part corresponds to different versions of the same vertex : $e_L + W^+ \leftrightarrow \nu_L \tag{1a}$ $(\bar\nu)_R + W^+ ...


2

Now when we operate parity operator, does that mean we are taking any physical entity at x to −x. Or we are just reverting axes of the co-ordinate system? Well, either operation should adhere to the same rules, and you mention the correct term: it depends on whether we see the operation as active or passive. Either view has the same end result: we move ...


2

The process could in general take place. A sample diagram is: $\hspace{4cm}$ With regards to Parity: Assuming the electron-positron pair don't have any angular momentum, the initial Parity is $-1$. Assuming the $\eta_C\eta_C$ pair don't have any angular momentum, their Parity is $+1$. Thus in this case the reaction cannot occur. If we assume the ...


1

1) because the potential energy function has even symmetry, the solutions to Schrodinger's Equation results in purely even or odd solutions. Consider the Time Independent Schrodinger Equation (TISE) below $$-\frac{\hbar^2}{2m}\nabla^2 \psi(x) + U(x)\psi(x) = E\psi(x)$$ But the potential contains the symmetry that $U(-x) = U(x)$. What this implies is that ...


1

Because $V(x)$ is even, one can show that the eigenfunctions of $H$ are either even or odd. Let's call these eigenfunctions $\psi_n(x)$, which are odd if $n$ is odd, even if $n$ is even. One also shows that $\psi_n$ is real. Now, $|\psi_n(x)|^2=\psi^*_n (x)\psi_n (x) $ is obviously even. It's trivial if $n$ is even, and is easily shown if $n$ is odd. One ...


1

Have a look at the paper Evidence for the spin-0 nature of the Higgs boson using ATLAS data. We don't have enough data to rule out the possibility the particle recently discovered might be spin 2, but it looks very unlikely. The data we have gives us about 98% confidence that the new particle is spin 0 with positive parity - i.e. it's the Higgs boson.



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