Confusion over parity violation

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I'm trying to get my head around the cobalt-60 beta decay experiment that apparently was used to show the weak decay did not conserve parity. It basically has a bunch of cobalt nuclei at low temperature in a magnetic field so that their spins are all aligned parallel to the field. The experiment was apparently looking to see whether there was "spatial asymmetry", which I take it means that more beta-electrons would be emitted in one direction relative to the field direction that another.

I get that spin, being an axial vector, won't change direction under the parity operation, r transforming to -r, and that momentum vectors will. I also get that if you saw particles emitted predominantly in one direction, you would see the mirror image as having them being emitted in the opposite direction, but the spin vector direction would be the same in both the mirror and the real world.

what I don't get is what this says about parity violation? I mean, for an eigenstate x, for even parity we have Px = +x, and for odd parity we have Px = -x. Where is the connection between seeing the difference in momentum vector direction in a mirror reflection, and having the parity eigenvalue go from +1 to -1?

The textbooks talk abotu the weak force being able to "distinguish right and left" and therefore this means parity is not conserved, but I just do not understand this.
 

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  • #2
Meir Achuz
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The decay rate depends on the amplitude squared.
If the amplitude is is either pure even or pure odd under parity, there will be no asymmetry. In beta decay, the amplitude goes like
(1+p.spin), which is a mixture of even and odd parity.
Its square shows a p.spin asymmetry.
 
  • #3
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Mathamatically, something is not conserved means that it does not commute with the hamiltonian. You first parity transform a given state then allow it to evolve(action of hamiltonian). Or you first allow the state to evolve and then parity transform it. For the beta decay expt, you end up having two different states when you do the above stated transformations. so we say beta decay violates parity.

Lets say the particles are spinning along z axis and decay happens only along spin direction only.

First we parity transform. but spin is even parity, so in the transformed system, again the spin is along z.
Now we evolve the system, the law(weak interaction) means decay happens along spin direction(z axis). so final state is decay products moving in z axis.

Now lets reverse the operations.

First evolve the system, that means the particles are emitted in spin direction(z axis)
Now you parity transform, that means we have decay particles moving in -z direction.

That is we obtain two different final states when we reverse the order of parity transformation and the hamiltonian. so decay along spin direction leads to parity non conservation.
 
  • #4
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Mathamatically, something is not conserved means that it does not commute with the hamiltonian. You first parity transform a given state then allow it to evolve(action of hamiltonian). Or you first allow the state to evolve and then parity transform it. For the beta decay expt, you end up having two different states when you do the above stated transformations. so we say beta decay violates parity.

Lets say the particles are spinning along z axis and decay happens only along spin direction only.

First we parity transform. but spin is even parity, so in the transformed system, again the spin is along z.
Now we evolve the system, the law(weak interaction) means decay happens along spin direction(z axis). so final state is decay products moving in z axis.

Now lets reverse the operations.

First evolve the system, that means the particles are emitted in spin direction(z axis)
Now you parity transform, that means we have decay particles moving in -z direction.

That is we obtain two different final states when we reverse the order of parity transformation and the hamiltonian. so decay along spin direction leads to parity non conservation.
So then if we've got our parity-hamiltonian commutator [P,H] = X, X would be non-zero if the two operators do not commute.
Is this saying that eigenstates of the parity operator are not eigenstates of the Hamiltonian?

and is the interpretation of this that, if we can measure the outcome of the final state (ie. what the energy is?), it means that we know what Hamiltonian eigenstate we have, therefore we do not know what parity eigenstate we have, so...

...linking back to the first response, apparently if we do not have a definite parity state then we have not conserved parity?

my question now is, assuming what I just said makes sense, where does the energy come in? the experiment was measuring the number of electrons going in a given direction, trying to see if more went one way than another. what has this got to do with their energy, ie. the hamiltonian eigenstate?
 
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Yes, if X is non zero, then it means that eigen states of parity operator are not eigensatets of hamiltonian. You can interpret non commuting of two operators as you said(not measuring both simultaneously). But hamiltonian is much more than just energy operator. Hamiltonian also corresponds to time evolution operation(e^-iH t). I will explain with an example.

Hamiltonian commutes with momentum operator P.(I am using P for momentum here) Let psi be an eigenstates with definite momentum p. ie P psi = p psi
HP=PH therefore e(-iH t) P=P e(-iH t)
Lets take the state psi and evolve it. e(-iH t) psi
Now lets find its momentum P e(-iH t) psi = e(-iH t) P psi = p e(-iH t) psi
That is we started with a state with momentum p. After time evolution we still have a state with momentum p. So commutaion of P with H implies momentum is conserved. If P were parity, then since parity does not commute with H, it means if we start with a parity eigenstate what we will end up will not be a parity eigenstate.
Nuclei spinning along z axis is an even parity state, but the decay products moving along z axis(the time evolved state) is not. So parity is not conserved. This is just another way of seeing the problem(as I did in the last post).
 
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thanks man I think I get it now, cheers.
 
  • #7
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you are welcome, cheers.
 
  • #8
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Hi guys,

In Co60 experiment, the electron emission is taken with respect to the spin direction, i.e., <spin.p> != 0 shows parity violation. What if there is decay of zero spin particle? How one can measure the parity violation in decay of spin zero particle like K+, K0 by measuring angular distribution of the decay products like in Co60 experiment?

[tex] K^{+} \rightarrow \pi^{+} + \pi^{0} [/tex] and is parity violating decay. [tex] K^{+} [/tex] has zero spin. If we measure the relative angular distribution of the two pions, the <[tex] p_{\pi^{+}}.p_{\pi^{0}} [/tex]> with be +ve under parity operation. So, it does not give asymmetry and what the parity of [tex] K^{+} [/tex] system is.

PS: [tex] \pi^{-} [/tex] should be [tex] \pi^{0} [/tex], I tried to update but some how it is not updating.

Thanks.
 
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Hello, any body home?
 
  • #10
Meir Achuz
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All charge states of the pion have negative intrinsic parity as does the k meson. A spin zero pseudoscalar cannot decay into two spin zero pseudoscalars if parity is conserved. This was shown in the 1950's.
 

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