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Why there is parity Symmetry ?

  1. Jun 13, 2013 #1
    Greetings,

    Can someone give a detailed explanation of why the expectation value of z coordinate in the ground state of hydrogen atom is zero due to parity symmetry? In addition how do you represent parity inversion in spherical coordinates and how do spherical harmonics behave under this inversion ( I found the formulas but I want the derivation part if applicable)? Any help is appreciated.

    P.S: This came up in first order perturbation theory while discussing the Stark Effect.

    Thanks
     
    Last edited: Jun 13, 2013
  2. jcsd
  3. Jun 14, 2013 #2
    The ground state, or 1s state has parity symmetry because if you jump from one side to the other, it looks the same. Take a look at this nice graphic of the real part of the spherical harmonics:
    http://en.wikipedia.org/wiki/File:Harmoniki.png
    From top to bottom, each row has increasing l (lowercase L), and m goes from left to right. The top is the s (l = 0) state. Notice that the states with even l have parity symmetry since if you go from one side to the opposite side, you stay on the same color. With odd l, you have parity antisymmetry, or odd parity, since you switch colors.

    z is the same thing as r * cos(theta), which is proportional to the [itex]Y_1^0[\itex] spherical harmonic, which has odd parity. A more advanced way of saying it is that z is a spherical tensor of rank 1, so it has an odd parity. Odd parity times even parity = Odd parity. When you calculate an expectation value, you have something like this
    [tex] \left< even | odd | even \right> = \int \text{odd function} d^3\mathbf{r} = 0[\tex]

    Integrating over all space of an odd function always gives 0 since both sides cancel out.
     
    Last edited: Jun 14, 2013
  4. Jun 14, 2013 #3

    Simon Bridge

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    Hmmm ... it is not so much that the expectation value is zero due to parity symmetry in the sense of being a causative agent, but that the parity symmetry and this expectation value are both manifestations of each other. Therefore we can use the symmetry to shortcut a calculation process.

    It can look a lot like waving a magic wand.
    You can always just compute the expectation value the usual way if you like.
     
  5. Jun 14, 2013 #4
    It's pretty obvious that <z> = 0 in this case, but the point is that the expectation value of any odd function or operator is going to be 0 in any odd or even eigenstate. Only some superposition of odd and even could possibly give a non-zero expectation for z.
     
  6. Jun 14, 2013 #5
    I am grateful for the replies; I think I understood a bit also in the later derivations for Stark effect my instructor used squeezed [itex]P^-1P[/itex] ρin from of the bra and before the ket while finding other expectation values. I think this is the same but he also wrote [itex]PzP^-1=-z[/itex] and I was unable to follow this. Also inverse parity operator when acting on a bra from the left does it produce the same result as if the parity operator was acting on a ket?

    Thanks
     
  7. Jun 14, 2013 #6
    The ##P## operates on a ket after it, and the ##P^{-1}## operates on a bra before it. You can imagine ##z## is always operating on something before and after it.
    Applying the transformation to an operator is the same as applying the transformation to all the states around it. As an analogy, if you were somehow mirror flipped, it is the same as if the universe were mirror flipped around you. Well, that's not an exact analogy because of parity violations, but you get the idea.
     
  8. Jun 15, 2013 #7
    Thanks for your answer what I inferred was that P is a unitary operator hence its inverse is its Hermitian conjugate. However, I did not get the second part and why z should operate both on a bra and a ket.
     
  9. Jun 15, 2013 #8

    Simon Bridge

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    What would normally stop it?
    What is different about z?
     
  10. Jun 16, 2013 #9
    I just thought when it acts on a bra or a ket it can no longer act. That is it cannot act on a bra and a ket at the same time.
     
  11. Jun 16, 2013 #10
    Maybe I was a little sloppy with the explanation. In quantum mechanics, we don't observe the state vectors, usually represented by the ket vectors. We only observe some inner product between bra vectors and ket vectors. If we transform both the bra and the ket vector, then nothing changes, sort of as if you took a photograph, and you rotated both the camera and the scene by the same amount. Now, consider putting a filter like a polarizer between the camera and the scene. If you rotate the filter, it's the same as rotating both the camera and the scene in the opposite direction but not moving the filter. You get the same measurement.
     
  12. Jun 16, 2013 #11

    samalkhaiat

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    Parity Invariance in Quantum Mechanics

    In the Schrodinger equation, let [itex]x \rightarrow - x[/itex] and then write [itex]\Psi ( - x , t ) = P \Psi ( x , t )[/itex]:
    [tex]
    i \frac{ \partial }{ \partial t } P \Psi ( x , t ) = - \frac{ \partial^{ 2 } }{ \partial x^{ 2 } } P \Psi ( x , t ) + V ( - x ) P \Psi ( x , t ) .
    [/tex]
    Now, if [itex]V ( x )[/itex] is an even function of [itex]x[/itex], i.e., [itex]V ( - x ) = V ( x )[/itex], then the wave function [itex]P \Psi ( x , t )[/itex] satisfies the same Schrodinger equation
    [tex]
    i \frac{ \partial }{ \partial t } P \Psi ( x , t ) = H P \Psi ( x , t ) . \ \ \ (1)
    [/tex]
    This means that the Schrodinger equation is invariant under the coordinate change [itex]x \rightarrow - x[/itex]. Now, operate with [itex]P[/itex] (from the left) on the original Schroginger equation, [itex]i \partial_{ t } \Psi ( x , t ) = H \Psi ( x , t )[/itex], and insert the identity operator [itex]P^{ 2 } = P^{ - 1 } P = 1[/itex]. You get
    [tex]
    i \partial_{ t } P \Psi ( x , t ) = P H P^{ - 1 } P \Psi ( x , t ) . \ \ \ (2)
    [/tex]
    Comparing Eq(1) with Eq(2), we see that the Hamiltonian is invariant under parity:
    [tex]H = P H P^{ - 1}.[/tex]
    This implies that [itex][ H , P ] = 0[/itex], which implies that the eiginvales of the parity operator are conserved.

    Sam
     
  13. Jun 16, 2013 #12

    Simon Bridge

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    It's just math.
    You can do the math in any order that makes it come out right - so you pick the order that makes the math easy. Sometimes that will mean operating "backwards".

    To see how it's done, try converting the statement into the integration formulation.

    Conceptually it is like Khashishi said.
     
  14. Jun 17, 2013 #13
    Thanks for taking your time, I think now I am able to understand the situation.
     
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