# What does well defined parity mean

I'm reading a textbook (Physics of Quantum Mechanics by Binney) and it says that the ground state ket $\left\lvert 1 0 0 \right \rangle$ of the hydrogen atom has well defined (even) parity. What does this mean?

Does it mean that the wave function is even? The wave function for this is an exponential decay. The potential for this is not even either.

• Yes, if you change (x,y,z) to (-x,-y,-z) the wavefunction is the same so it has even parity. If it picked up a minus sign (and it does if L=1) it would have odd parity. – octonion Mar 7 '16 at 1:13
• is your L the 1 in the ket |1 0 0>? the wave function is an exponential decay so it should have odd parity but it says that it has even parity. – physicsnoob1000 Mar 7 '16 at 1:19
• it also says that the expectation value of and component of (x,y,z) vanishes if in a state of well defined parity, I dont see why this is true – physicsnoob1000 Mar 7 '16 at 1:20
• L is the second number, for instance $|2 1 0\rangle$ and $|2 1 -1\rangle$ has well defined odd parity but $|2 0 0\rangle$ has well defined even parity, as you can check. – octonion Mar 7 '16 at 1:21

"Well-defined parity" here means it is an eigenstate of the parity operator that sends $\vec x \mapsto -\vec x$. Even wavefunctions are such eigenstates for the eigenvalue 1, odd wavefunctions are eigenstates for the eigenvalue -1.
Answering your second question in the comments, whether the wavefunction has well defined even or odd parity when you take the expectation value $$\langle \psi|f(x)|\psi\rangle = \int f(x) |\psi(x)|^2 d^3 x$$ the function $|\psi(x)|^2$ is an even function (the product of two odd functions is even).
But the functions x, y, z are odd functions, so $\langle \psi|x|\psi\rangle=0$ for the same reason that for instance $$\int^{\infty}_{-\infty} x e^{-x^2}dx=0$$