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A Parity and time reversal on derivatives and electric field?

  1. Oct 27, 2016 #1
    I am trying to learn how parity and time reversal transform the electric field, ##A_\mu## and ##\partial_\mu##. In other words what: what are ##P \partial_\mu P##, ##T \partial_\mu T##, ##T A_\mu T## and ##P A_\mu P##?

    My first guess was that ##P A_\mu(t,\vec{x}) P = A_\mu(t,-\vec{x})##, ##T A_\mu(t,\vec{x}) T = A_\mu(-t,\vec{x})##, ##P \partial_\mu(t,\vec{x}) P = \partial_\mu(t,-\vec{x})##, ##T \partial_\mu(t,\vec{x}) T = \partial_\mu(-t,\vec{x})##, but this gives the wrong answer when I try to do exercises. So there's probably something that I dont understand correctly.

    I've looked in Peskin, and searched the internet, and I cannot find the the answer to this. These would be useful to anyone starting out with QED, so it would be good if someone would give me the answer I'd be very greateful :)

    Thanks!
     
  2. jcsd
  3. Oct 27, 2016 #2

    PeterDonis

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    The 4-vector ##A_{\mu}## is the electromagnetic potential, not the electric field. The electromagnetic field (which is the proper term) is ##F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}##. But for this discussion we should probably stick with the potential ##A_{\mu}##.

    [Edit: removed incorrect statement about how the operators are applied.]

    In addition to the above, yes, there is. You are thinking of ##P## and ##T## as "flipping spacetime" (so to speak) along a spatial or time axis. But that's not really the right way to think of them. You need to think of them as transforming vectors at a point into different vectors at the same point.

    It might help to take a step back and see where these things ##P## and ##T## come from in the spacetime picture. Consider the entire set of 4-vectors at some point in Minkowski spacetime; call this set ##V##. There is a group ##L## of transformations, which is usually called the "Lorentz group" (though that term is also used for one of its subgroups, see below), that consists of all possible transformations that take every vector in ##V## to some other vector in ##V##. But it turns out that these transformations fall into four classes.

    The first class, which is a subgroup of the full group ##L##, is correctly called the "proper orthochronous Lorentz group", but is sometimes just called the "Lorentz group", with its elements being called "Lorentz transformations" (instead of "proper orthochronous Lorentz transformations"). This class includes the identity transformation, and consists of all transformations in ##V## that are continuously connected to the identity. It is often broken up further into "rotations" (more precisely spatial rotations) and "boosts" (though some transformations in this class are combinations of the two). This class takes timelike vectors into other timelike vectors in the same light cone (future or past), and takes right- (left-) handed triplets of spacelike vectors into other right- (left-) handed triplets of spacelike vectors. In other words, these transformations "preserve orientation" in both time and space.

    The three other classes of transformations in ##L## are just those that are obtained by combining some transformation in the first class with one of three discrete transformations: ##P##, ##T##, or ##PT##. The first two are usually called "parity reversal" and "time reversal"; the third is just a combination of both. Given a transformation ##\Lambda## in ##L##, the transformation ##P \Lambda## takes timelike vectors into other timelike vectors in the same light cone, but takes right- (left-) handed triplets of spacelike vectors into left- (right-) handed triplets. So it reverses spatial orientation but not time orientation. The transformation ##T \Lambda##, by contrast, takes timelike vectors in the future (past) light cone into timelike vectors in the past (future) light cone, so it reverses time orientation but not spatial orientation. The transformation ##P T \Lambda## reverses both orientations.

    Now, from the above, it should be evident that there is no single answer to the questions of what is ##P A_{\mu}##, ##T A_{\mu}##, ##P \partial_{\mu}##, ##T \partial_{\mu}##. You need to know what kinds of vectors these are (timelike or spacelike--or null, but we can leave those out for the time being). For example, if we are talking about the vector ##\partial_t##, which is timelike, then it will be unaffected by ##P## but will change sign under ##T##; but if we are talking about the vector ##\partial_x##, which is spacelike, it will change sign under ##P## (as will ##\partial_y## and ##\partial_z##), but will be unaffected by ##T##.
     
    Last edited: Oct 28, 2016
  4. Oct 28, 2016 #3

    vanhees71

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    I'm not sure about the Op's notation. Are you dealing with quantum fields (i.e., field operators)? Then you have unitary parity and charge conjugation operators and a antiunitary time-reversal operator ##\hat{P}##, ##\hat{C}##, and ##\hat{T}##, defined on Fock space. From these you have to distinguish the corresponding matrices transforming the space-time arguments. Let's concentrate on parity (space reflection). The corresponding Lorentz matrix is ##{\mathcal{P}^\mu}_{\nu}=\mathrm{diag}(1,-1,-1,-1)##. QED admits a parity transformation, i.e., it is space-reflection co-variant. Thus you have (for the field operators),
    $$A^{\prime \mu}(x)=\hat{P} \hat{A}^{\mu}(x) \hat{P}={\mathcal{P}^{\mu}}_{\nu} \hat{A}(\mathcal{P} x),$$
    where I've used ##\mathcal{P}^{-1}=\mathcal{P}##.
     
  5. Oct 28, 2016 #4
    Yes exactly.

    My professor hasn't taught this, yet we are expected to know it. o_O The whole class is stuck :smile:

    So how do the rest transform? I am just making a wild guess, please let me know if it's right or wrong:

    Does there exist a matrix ##{\mathcal{T}^\mu}_\nu = \text{diag}(-1,1,1,1)## such that:
    $$
    T A^\mu(x) T = {\mathcal{T}^\mu}_\nu A^\mu ({\mathcal{T}^\mu}_\nu x), \qquad T \partial^\mu(x) T = {\mathcal{T}^\mu}_\nu \partial^\mu ({\mathcal{T}^\mu}_\nu x) ?
    $$

    And
    $$
    P \partial^\mu(x) P = {\mathcal{P}^\mu}_\nu \partial^\mu ({\mathcal{P}^\mu}_\nu x)?
    $$

    Or are these not completely correct?

    Thank you for your time :)
     
  6. Oct 28, 2016 #5

    vanhees71

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    I don't undestand your notation. ##\hat{P}## commutes with c-numbers and derivatives with respect to c-numbers. It's an operator in Hilbert space (Fock space of photons in this case). Forget time reversal for a moment. It's more complicated since ##\hat{T}## is antiunitary.
     
  7. Oct 28, 2016 #6

    PeterDonis

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    Ah, ok, in this case ##P A P## (which should actually be ##P A P^{-1}##, but as vanhees71 says, ##P = P^{-1}##) is correct, since you're applying ##P## to an operator, not a state. (But there is still an additional wrinkle with ##T##, as vanhees71 says.)
     
  8. Oct 28, 2016 #7
    Ok, thank you for your replies :)

    (1) Did i get guess the transformation rule for parity acting on the derivative correctly?

    (2) Clearly if T is antiunitary, then ## {\mathcal{T}^\mu}_\nu \neq \mathrm{diag}(-1,1,1,1) ##

    Will you please correct me and show me what the transformation rule for ##A_\mu## and ##\partial_\mu## is under time reversal? :)
     
  9. Oct 28, 2016 #8

    PeterDonis

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    Your notation still doesn't make sense. First of all, you have the same indexes ##\mu## and ##\nu## appearing too many times in your expressions; that doesn't make sense. Second, ##x## is a 4-vector (spacetime position), so it has an index too; you can't properly express the result of applying the matrix ##\mathcal{P}## to ##x## without the index.

    For example, vanhees71's expression

    would be written in index notation as

    $$
    \hat{A^{\prime \mu}} (x^{\alpha}) = \mathcal{P}^{\mu}{}_{\nu} \hat{A^{\nu}} (\mathcal{P}^{\alpha}{}_{\beta} x^{\beta})
    $$

    Notice how the free indexes match on both sides of the equation, and how indexes that are repeated appear once "upstairs" and once "downstairs", indicating summation.
     
  10. Oct 29, 2016 #9
    It should have said ##T A(x) T = {\mathcal{T}^\mu}_\nu A^\nu ({\mathcal{T}^\sigma}_\rho x^\rho), \ T \partial(x) T = {\mathcal{T}^\mu}_\nu \partial^\nu ({\mathcal{T}^\sigma}_\rho x^\rho)## and ## P \partial(x) P = {\mathcal{P}^\mu}_\nu \partial^\nu ({\mathcal{P}^\sigma}_\rho x^\rho)##. But I think these are incorrect.

    Are you kindly able to tell me the correct transformation rules for ##A_\nu## and ##\partial_\gamma##?
     
  11. Oct 29, 2016 #10

    samalkhaiat

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    In field theory, you have to distinguish between the parity [itex]P[/itex] and time reversal [itex]T[/itex] transformations of coordinates from the corresponding quantum unitary [itex]\mathscr{P}[/itex] and anti-unitary [itex]\mathscr{T}[/itex] operators on the states and field operators:
    [tex]P: \ (t , \vec{x}) \to (t , - \vec{x}) , \ \mbox{or} \ \ Px^{\mu} = x_{\mu} ,[/tex] [tex]T: \ (t , \vec{x}) \to (-t , \vec{x} ) , \ \mbox{or} \ \ Tx^{\mu} = - x_{\mu} ,[/tex]
    [tex]\mathscr{P}: A_{\mu}(x) \to \mathscr{P} A_{\mu}(x) \mathscr{P}^{\dagger} = A^{\mu} (Px) , \ \ \ (1)[/tex] [tex]\mathscr{T}: A_{\mu}(x) \to \mathscr{T}A_{\mu}(x) \mathscr{T}^{\dagger} = A^{\mu}(Tx) . \ \ \ \ (2)[/tex]
    These transformation laws of the photon field operator can be deduced from the invariance of the interaction term in the action integral [tex]S(t_{1},t_{2}) = \int_{\mathbb{R}^{3}} \int_{t_{1}}^{t_{2}} \ dt \ d^{3} \vec{x} \ A_{\mu}(x) J^{\mu}(x) , \ \ \ \ (3)[/tex] and the known behaviour of the charge and current densities [itex](J^{0} , \vec{J})[/itex] under parity and time reversal: The electric charge is invariant under parity (looking at the charge from left or from the right should not change its value), i.e., [tex]\mathscr{P} Q \mathscr{P}^{\dagger} = Q = \int d^{3}\vec{x} \ J^{0}(x) .[/tex] This implies [tex]\mathscr{P} J^{0}(t , \vec{x}) \mathscr{P}^{\dagger} = J^{0}(t , - \vec{x}) .[/tex] However, parity reverses the direction of the current [itex]\vec{J}[/itex], i.e., [tex]\mathscr{P} \vec{J}(t , \vec{x}) \mathscr{P}^{\dagger} = -\vec{J} (t , -\vec{x}) .[/tex] So, for the 4-vector current, we have [tex]\mathscr{P} J^{\mu}(x) \mathscr{P}^{\dagger} = J_{\mu}(Px) . \ \ \ (4)[/tex] Thus, from (4) and the invariance of the action (3) we can read off the transformation law (1).
    By the same reasoning, we deduce the transformation of [itex]J^{\mu}[/itex] under time reversal: [tex]\mathscr{T}J^{\mu}(x) \mathscr{T}^{\dagger} = J_{\mu}(Tx) \equiv J_{\mu} (-t , \vec{x}) . \ \ \ (5)[/tex] Now, you can easily show that (2) and (5) leave the action (3) invariant (up to time translation): Since now we have, under time reversal, [tex]A_{\mu}(x) J^{\mu} (x) \to A^{\mu}(-t , \vec{x}) J_{\mu}( -t , \vec{x}) ,[/tex] then
    [tex]\mathscr{T}S(t_{1},t_{2}) \mathscr{T}^{\dagger} = \int_{\mathbb{R}^{3}} \int_{t_{1}}^{t_{2}} d^{3}\vec{x} \ dt \ \mathcal{L}_{int}(-t , \vec{x}) .[/tex] Changing the integration variable [itex]t \to –t[/itex], we get
    [tex]\mathscr{T}S(t_{1},t_{2}) \mathscr{T}^{\dagger} = \int_{\mathbb{R}^{3}} \int_{-t_{2}}^{-t_{1}} d^{3}\vec{x} \ dt \ \mathcal{L}_{int}(t , \vec{x}) = S(-t_{2} , -t_{1}).[/tex]
    However, [itex]S(-t_{2},-t_{1})[/itex] can be related to the original action [itex]S(t_{1},t_{2})[/itex] by a time-translation which is also an assumed symmetry.
    As for [itex]\partial_{\mu}[/itex], you need to know that, just like [itex]x^{\mu}[/itex], [itex]\partial_{\mu}[/itex] is not an operator in field theory. It transforms just like the coordinates: [tex]P \partial_{\mu} = \partial^{\mu}, \ \ \ T\partial_{\mu} = - \partial^{\mu} .[/tex]
     
  12. Oct 29, 2016 #11
    Thanks man! :)
     
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