I would be grateful if somebody could help me brining the parity operator and the particle exchange operator together.

Suppose, there is a two-proton system, where one proton is sitting at $+r$ on the $x$-axis, one is sitting at $-r$, as shown in the figure.

Two-proton system

In this situation, the parity operator $\hat P_{\text{pa}}$ should be the same as the particle exchange $\hat P_{\text{ex}}$ operator, right?

Therefore, I will denote both operators just by $\hat P$.

$$ \hat P :\equiv \hat P_{\text{pa}} = \hat P_{\text{ex}} \tag{1} $$

Particle Exchange

Since this is a two-fermion system, the combined state should be anti-symmetric under the exchange of the two protons:

$$ \hat P \, |\Psi_{12}\rangle = p_{12}\, |\Psi_{12}\rangle, \ \ p_{12} = -1 \tag{2} $$


The intrinsic parity of each proton is $+1$.

$$ \hat P \, |\Psi_i\rangle = p_i \, |\Psi_i\rangle, \ \ p_i = +1, \ \ i \in \{1,2\} \tag{3} $$

Since parity is multiplicative, the parity of the two-proton system should be the product of the single parities:

$$ p_{12} = p_1 \cdot p_2 = (+1) \cdot (+1) = +1 \tag{4} $$

Why the different sign?

Apparently, the parity of the two-proton system is $p_{12}=+1$, whereas the particle exchange considerations lead me to $p_{12}=-1$.

I could not resolve this by considering the spin as well.

Could someone give me a pointer what I am doing wrong here?

Thank you!


1 Answer 1


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.

  • $\begingroup$ Thanks @peter. As you've said, for $l=0$, $(-1)^l$ just gives me a factor of $1$. Assuming, $p_{12}=p_1\cdot p_2$ is correct, how is the parity $p_1$ of one of the single-proton states to be calculated in my scenario? $\endgroup$
    – fiedl
    Apr 28, 2013 at 22:25
  • $\begingroup$ @fiedl: But you dont have $l=0$, your state is a sum of products of delta-functions, which is not rotationally-invariant ($l=0$ means rotation invariance for the coordinate dependence). In fact, your state just dont have a definite $l$. I used it just as an illustration of the fact that intrisic parities are not the only information you need. However, you can say that your state is a superposition of states with odd $l$'s. $\endgroup$ Apr 28, 2013 at 22:46
  • $\begingroup$ @fiedl: As for one-proton states, you indeed have your state to be equal to (antisymmetrized) product of one-particle states. But these states do not have definite parity -- $P$ takes a state with one proton at $r$ to a state with one proton at $-r$, which is a different state. However, the full two-particle state turns out to have a definite parity, which is $-1$. $\endgroup$ Apr 28, 2013 at 22:48
  • $\begingroup$ @fiedl: If you undestand this: a general state of some number $n$ of fermions can be represented as a wavefunction of $r_1,r_2,...r_n$ vectors, and this wavefunction would carry $n$ Dirac spinor indices. Intrinsic parity tells you how to deal with the spinor indices, but not how to deal with the coordinates when you apply $P$. $\endgroup$ Apr 28, 2013 at 22:56
  • $\begingroup$ Thanks @peter, that helped a lot. If I understand you correctly, I can assume that my statements (1), (2) and (3) hold, but (4) has to be refined to $p_{12}=p_1 \cdot p_2 \cdot (-1)^l$, where $l$ is the quantum number of the angular momentum of the relative motion of the two protons [Stöcker, chap. 27.3.1]. Thus, in order to satisfy $(2)$, $l$ has to be $l\in \{1,3,5,\dots \}$, as you have pointed out. $\endgroup$
    – fiedl
    Apr 29, 2013 at 8:16

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