When a lambda particle decays into proton and a pion, I am told it does not conserve parity. Why?
[Physics] Why does lambda decay violate parity
parityparticle-physicsquantum mechanics
Related Solutions
Action of the parity operator on a state is not determined only by intrinsic parities of the particles at hand, but also by the state itself. For example, if you have a one-particle state $\Psi_p$ with impulse $p$, then $P\Psi_p=\pm\Psi_{-p}$, and such a state does not even have a definite parity (it is not an eigenstate).
You probably know that if there is an orbital angular momentum $l$, you get additional partity $(-1)^l$, but it is for momentum eigenstates. It is not the case for your sitation, but your state happens to be an eigenstate of $P$ with value $-1$. You can explicitly write your state down (You might consider using Dirac spinors in order to be able to see the difference for, say, antiproton, I guess.) to see it.
The magnetic field is a "pseudovector" (more properly, a 2-form), as opposed to the electric field, which is a vector (or a 1-form). That is, under parity, $\mathbf B$ is left unchanged. You can see this from the Lorentz force, $$\mathbf F = q(\mathbf E + \mathbf v\times \mathbf B)$$ where since force is a vector, $\mathbf E$ must also be a vector. Since $\mathbf v \times \mathbf B$ is a product, if $\mathbf B$ were a vector, this term would not transform properly under parity. Thus $\mathbf B$ does not change when we perform a parity transformation.
However I think the more correct way to see this is from the relativistic formulation of electrodynamics. Introduce the electromagnetic field tensor $$F_{\mu\nu} =\begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix}$$ and note that the purely spatial part $F_{ij}$, $1 \le i, j \le 3$ is equivalent to the magnetic field. Since there are two indices, the components are invariant under parity transformations. The electric field is given by $E_i = F_{0i}$, so it changes sign under parity, it is a vector.
The more sophisticated yet way to see this decomposition is that if there is timelike 1-form $dt$ we can decompose the field strength 2-form as $$F = E\wedge dt + B.$$ We see that $E$ is a 1-form (equivalent to a vector after raising the index), but $B$ is a 2-form (often called pseudovector, because not enough people know about the wonders of differential forms).
Now your transformation is not quite $P : (x,y,z) \mapsto (-x,-y,-z).$ It is $RP : (x,y,z) \mapsto (-x,y,z)$, or $P$, then a rotation by $\pi$ in the $yz$-plane. Here the $x$-axis is along $\mathbf v$ and the $z$-axis along $\mathbf B$. Since $\mathbf v$ is perpendicular to the plane of the rotation, it is just affected by the reflection, and is to the left. $\mathbf B$, lying in the plane of rotation, is rotated half a revolution, and is now out of the page, so the force is upward. The force being a vector, this is precisely what we expect. I hope this answers your question.
Best Answer
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 conserved (at least in low-energy processes that also conserve $B,L$).
However, there's a convention that assigns a particular parity to fermions. Note that Weinberg has proved that $$ P^2 = (-1)^{2J} $$ so parity behaves much like the rotation by 180 degrees: its square changes the sign of states with an odd number of fermions. In the standard convention, electron, proton, and neutron are set to have $P=+1$: one is allowed to make three choices like that. The parity of the pion is then determined to be negative, $P=-1$, because a deuteron-pion ground state may decay into two neutrons with $L=1$ - the odd orbital momentum changes the sign of the parity, too.
Strong and electromagnetic interactions preserve parity, so one may assign parity to all hadrons, as determined from various strong and electromagnetic interactions. Lambda then turns out to have a positive parity $P=+1$ much like protons and neutrons because it's just another bound state of three quarks and the change of the quarks' identity doesn't change parity.
The neutral $\Lambda^0$ decays to a nucleon and a pion, $p+\pi^-$ or $n+\pi^0$, so a $P=+1$ state decays into one particle (nucleon) with $P=+1$ and one (pion) with $P=-1$. That violates the parity because $(+1)\neq (+1)(-1)$. It's because the decay is due to the weak interactions. Weak interactions don't preserve the parity because even the spectrum doesn't: for example, a parity-transformed partner of a left-handed neutrino - the right-handed neutrino - doesn't even exist. This asymmetry is confirmed by the weak interactions that contain two-component (fundamentally left-right asymmetric) spinors or, in the four-component spinor formalism, combinations $(1\pm \gamma_5)$ - scalars and pseudoscalars - that maximally violate the parity.
Because of these interactions, any parity-violating process that respects all the other laws is allowed, although it may be slow because it must rely on the weak interactions which are weak. The decays of the neutral Lambda baryon - and similar decays of the other Lambda particles - belong among the parity-violating ones.
The violations of parity conservation, discovered half a century ago, was a shock. But once you appreciate the simple fact that fields may be described by a 2-component spinor that prefers one chirality over the other - may describe left-handed particles without the right-handed ones - it's not so shocking. The two-component spinors work because $SL(2,C)=SO(3,1)$, locally. The fundamental representation of $SL(2,C)$ has a spin-1/2, and it also has a small enough number of components that there's no oppositely spinning particle (with the same other charges). The theory is still Lorentz-invariant.
Not only parity is violated in some processes: at very high energies, those processes become so common that it isn't even possible to define parity accurately. After all, the parity transformation of the neutrino states is ill-defined. That's why it didn't hurt that I could have redefined parity by the signs coming from the lepton and baryon numbers (which are ultimately violated at very high energies, perhaps with the exception of $B-L$) at the beginning: none of the parity operators is actually fully well-defined on the spectrum of particle states.