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Math Help - Parity

  1. #1
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    Parity

    With my exam on Friday, I'm trying to get all these practice probs done since some of them will be on the exam.

    If P is the Parity Operator, determine the parity of the functions below:

    P\sin{x} = ?
    Pe^{-x} = ?
    P(e^{x} + e^{-x}) = ?

    Now, if H is the Hamiltonian Operator with V(-x) = V(x), find the parity of

    PHe^{-x} = ?

    ------

    So for the first func, we know that sin(x) is an odd function, and hence that would be -sin(x) if I'm not mistaken since the parity would take sin(x) and make it sin(-x) and thus it'd have parity -1? We only dealt with cos/sin in our notes, so I'm not sure what an extra operator does and the role of exponential functions.
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  2. #2
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    Quote Originally Posted by DiscreteW View Post
    With my exam on Friday, I'm trying to get all these practice probs done since some of them will be on the exam.

    If P is the Parity Operator, determine the parity of the functions below:

    P\sin{x} = ?
    Pe^{-x} = ?
    P(e^{x} + e^{-x}) = ?

    Now, if H is the Hamiltonian Operator with V(-x) = V(x), find the parity of

    PHe^{-x} = ?

    ------

    So for the first func, we know that sin(x) is an odd function, and hence that would be -sin(x) if I'm not mistaken since the parity would take sin(x) and make it sin(-x) and thus it'd have parity -1? We only dealt with cos/sin in our notes, so I'm not sure what an extra operator does and the role of exponential functions.
    Is this related to some Physics concepts?
    I am guessing that we need to define a function called P that will map like this P(f(x)) = f(-x).

    P\sin{x} = -\sin{x} \Rightarrow P(x) = -x

    Pe^{-x} = e^{x}\Rightarrow P(x) = \frac1{x}

    P(e^{x} + e^{-x}) = e^{-x} + e^{x} \Rightarrow P(x) = x

    Hopefully Topsquark, our physics maestro, will help us out with the last one. I have no clue what you are talking about. What does Hy do?
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  3. #3
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    Quote Originally Posted by DiscreteW View Post
    With my exam on Friday, I'm trying to get all these practice probs done since some of them will be on the exam.

    If P is the Parity Operator, determine the parity of the functions below:

    P\sin{x} = ?
    Pe^{-x} = ?
    P(e^{x} + e^{-x}) = ?

    Now, if H is the Hamiltonian Operator with V(-x) = V(x), find the parity of

    PHe^{-x} = ?

    ------

    So for the first func, we know that sin(x) is an odd function, and hence that would be -sin(x) if I'm not mistaken since the parity would take sin(x) and make it sin(-x) and thus it'd have parity -1? We only dealt with cos/sin in our notes, so I'm not sure what an extra operator does and the role of exponential functions.
    Quote Originally Posted by Isomorphism View Post
    Is this related to some Physics concepts?
    I am guessing that we need to define a function called P that will map like this P(f(x)) = f(-x).

    P\sin{x} = -\sin{x} \Rightarrow P(x) = -x

    Pe^{-x} = e^{x}\Rightarrow P(x) = \frac1{x}

    P(e^{x} + e^{-x}) = e^{-x} + e^{x} \Rightarrow P(x) = x

    Hopefully Topsquark, our physics maestro, will help us out with the last one. I have no clue what you are talking about. What does Hy do?
    The parity operator in 1 - D, as Isomorphism suggests, maps f(x) to f(-x). In 3-D it maps the function f(x, y, z) to f(-x, -y, -z).

    I think all that is being asked for here is what the parity of the function is. So
    Psin(x) = sin(-x) = -sin(x), so sin(x) has an odd parity, or -1.

    Pe^{-x} = e^x \neq \pm e^{-x} so e^{-x} is not a parity eigenstate.

    P(e^x + e^{-x}) = e^{-x} + e^x = +(e^x + e^{-x}) so e^x + e^{-x} has an even parity, or +1.

    H is the Hamiltonian operator: H = -\frac{\hbar ^2}{2m} \frac{d^2}{dx^2} + V(x)

    We can do PHe^{-x} in two ways: Apply H to e^{-x} then apply P, or we can apply P to He^{-x} directly as P is a linear operator. I think at this level it is more instructive to take the former route, so
    PHe^{-x} = P \left ( -\frac{\hbar ^2}{2m}e^{-x} + V(x)e^{-x} \right ) = -\frac{\hbar ^2}{2m}e^x + V(-x)e^x

    We know that V(x) has an even parity so this is
    PHe^{-x} = -\frac{\hbar ^2}{2m}e^x + V(x)e^x = He^x \neq \pm He^{-x}
    so this is not a parity eigenstate.

    -Dan
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