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Momentum states normalization

  1. Apr 28, 2015 #1
    What is the value of ##\left\langle {{\bf{p}}|{\bf{p}}} \right\rangle ## when ##a_{\bf{k}}^ + k\left| 0 \right\rangle = \sqrt {2{E_{\bf{p}}}} \left| 0 \right\rangle ##? (like in Peskin)

    I suppose that ##\left\langle {\bf{k}} \right|{a_{\bf{k}}}{\bf{P}}a_{\bf{k}}^ + \left| {\bf{k}} \right\rangle = {\bf{k}}##.


    & \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\bf{p}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{p}}^ + a_{\bf{p}}^{}a_{\bf{k}}^ + \left| 0 \right\rangle } = \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\bf{p}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{p}}^ + \left\{ {\left[ {a_{\bf{p}}^{},a_{\bf{k}}^ + } \right] + \overbrace {a_{\bf{k}}^ + a_{\bf{p}}^{}}^{ = 0}} \right\}\left| 0 \right\rangle } = \cr
    & = \int {{d^3}p{\bf{p}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{p}}^ + \left| 0 \right\rangle \delta \left( {{\bf{k}} - {\bf{p}}} \right)} = {\bf{k}}\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{k}}^ + \left| 0 \right\rangle = {\bf{k}} \cr} ##.

    This implies ##\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{k}}^ + \left| 0 \right\rangle = 1##


    ##\left\langle {{\bf{p}}|{\bf{p}}} \right\rangle = {1 \over {2{E_{\bf{p}}}}}##

    am I right?
  2. jcsd
  3. Apr 29, 2015 #2


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    No you are wrong. Try using the |k> I write below to calculate <q|k> and then see what happens for q=k.

    I think you did a typo, because in Peskin it is:

    [itex] |k> = \sqrt{2E_k} a^\dagger_k |0> [/itex]

    So try working on [itex]<q|k>=?[/itex]

    Also I'd like you to justify your "suppose that [itex]<k| a_k P a_k^\dagger |k> =k [/itex] " .
    Last edited: Apr 29, 2015
  4. Apr 29, 2015 #3
    In Peskin; page 22 just under equation (2.23) it says "So the operator ##a_{\bf{k}}^ + ## creates momentum ##{\bf{p}}## ..."
    and becuse ##{\bf{P}}## meusure momentum i thout that:

    ##\left\langle 0 \right|{a_{\bf{k}}}{\bf{P}}a_{\bf{k}}^ + \left| 0 \right\rangle = {\bf{k}}##.

    Now, Peskin say ##\left\langle {{\bf{p}}|{\bf{q}}} \right\rangle = 2{E_{\bf{p}}}{\left( {2\pi } \right)^3}{\delta ^{\left( 3 \right)}}\left( {{\bf{p}} - {\bf{k}}} \right)##

    but how do you interpret ##\left\langle {{\bf{p}}|{\bf{p}}} \right\rangle = 2{E_{\bf{p}}}{\left( {2\pi } \right)^3}{\delta ^{\left( 3 \right)}}\left( 0 \right)##?
  5. Apr 29, 2015 #4


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    Well you can work this out and get the <p|q> ...

    As for the momentum, I am not sure... shouldn't the a-operators bring you some energy factor? Maybe I'm wrong...

    You have to keep in mind that [itex] \delta^3 (p -p ) [/itex] represents the volume of the space. What do you know about plane wave states from QM?
  6. Apr 29, 2015 #5
    Acording to the interpretation that the delta functio is the volume of space we get ##\left\langle {{\bf{p}}|{\bf{p}}} \right\rangle = \infty ##,
    which mean that ##\left| {\bf{p}} \right\rangle ## is un-normelizable and thuse not representing anything real.
  7. Apr 29, 2015 #6


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    For a plane wave that's normal, since the plane wave is extended in all space..
    If on the other hand you choose a wavepucket state [itex]|f> =\int \frac{d^3p}{(2 \pi)^3} \frac{f(p)}{2E_p} |p>[/itex] then the formula for [itex]<f|f>[/itex] will not vanish.
  8. Apr 29, 2015 #7
    Why dose a wave spread on all space has infinite momentum? and how do you know it's extended in all space?
    besides, it's not what Peskin says:
    page 22 just under equation (2.23) it says "So the operator ##a_{\bf{p}}^ + ## creates momentum ##{\bf{p}}## ..."
  9. Apr 29, 2015 #8


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    It does not have infinite momentum. The infinity you get is the problem to normalize the corresponding wave function.
    A wave function that spreads on all space is in all space by definition. I don't understand that question.
  10. Apr 29, 2015 #9
    So how do you normalize it?
  11. Apr 29, 2015 #10


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    You don't, it is not a physical state.
  12. Apr 29, 2015 #11


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    Where was that referring to?
    A plane wave can have definite momentum (if you like to see it from HUP, the fact that the plane wave has definite momentum means that its position is completely uncertain and so it extends in the whole space)...
    Let's put that in a different way, from QM: are plane waves normalizable?
    Last edited: Apr 29, 2015
  13. Apr 30, 2015 #12


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    Well, I guess, if answered in this style, we should push the whole thread to the homework section ;-).

    A little tip: First think in terms of "first quantization" in non-relativistic quantum theory. The momentum "eigenstates" in position representation are plane waves
    $$u_{\vec{p}}(\vec{x})=N \exp(\mathrm{i} \vec{x} \cdot \vec{p}),$$
    where ##N## is an arbitrary normalization constant. The scalar product is given by
    $$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi_1^*(\vec{x}) \psi_2(\vec{x}).$$
    Now try to answer ChrisVer's questions, particularly whether you can calculate the norm of the momentum eigenfunction. Then try to evaluate
    $$\langle u_{\vec{p}_1}|u_{\vec{p}_2} \rangle.$$
    This should help to understand, what's going on here.

    For a pragmatic but better introduction than usually found in QM textbooks, see
    L. Ballentine, Quantum Mechanics, A Modern Development, Addison Wesley
  14. Apr 30, 2015 #13


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    Sorry (from you? and from the one who posted the OP) for my style, I chose for once to try and put someone do the calculation by himself and figure it out...When it came to explaining things I guess I and mfb gave a normal "answer". I mean it ends up being a QM question from a QFT one.

    To help the OP though understand how the [itex]u_p[/itex] is associated with [itex]|p>[/itex] it's exactly the fact that they are the same thing. Both are the eigenstates of a given momentum [itex]p[/itex]. The [itex]u_p(x)[/itex] vanhees wrote is [itex]<x|p>[/itex] and that's how the integral [itex]\int d^3x[/itex] appears when taking their inner product (by the insertion of unity [itex]I= \int d^3 x |x><x|[/itex] between the inner product ).
  15. Apr 30, 2015 #14


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    It was not meant as a criticism, because indeed, to figure this out once oneself, is crucial for an understanding of the continuous spectra of operators in quantum theory. For that reason, I think the thread is better moved to the homework section, because there it might get more attention than here.
  16. Apr 30, 2015 #15
    The problem is that it is not a homework assigment.
  17. Apr 30, 2015 #16
    Let me see if i got it right:


    answer to my question

    i understand that ##a_{\bf{k}}^ + \left| 0 \right\rangle ## is not a physical state,

    so it is O.K. that ##\left\langle 0 \right|{a_{\bf{k}}}a_{\bf{k}}^+ \left| 0 \right\rangle ## have no meaning.

    Is it right to say that ##\left\langle {{\bf{p}}|{\bf{q}}} \right\rangle = 2{E_{\bf{p}}}{\left( {2\pi } \right)^3}{\delta ^{\left( 3 \right)}}\left( {{\bf{p}} - {\bf{q}}} \right)## have no physical meaning either and it's just an instrument for calculation?

    Now, i had a mistake in the first post, i wrote ##\left\langle {\bf{p}} \right|{a_{\bf{k}}}{\bf{P}}a_{\bf{k}}^+ \left| {\bf{p}} \right\rangle ## instead of ##\left\langle 0 \right|{a_{\bf{k}}}{\bf{P}}a_{\bf{k}}^+ \left| 0 \right\rangle ##.

    setting this straight, the question still remain:

    dose ##\left\langle 0 \right|{a_{\bf{k}}}{\bf{P}}a_{\bf{k}}^+ \left| 0 \right\rangle ## have a meaning and if so, is it true that ##\left\langle 0 \right|{a_{\bf{k}}}{\bf{P}}a_{\bf{k}}^+ \left| 0 \right\rangle = {\bf{k}}##?
  18. Apr 30, 2015 #17


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    Great! Now you got the correct relation,
    $$\left\langle {{\bf{p}}|{\bf{q}}} \right\rangle = 2{E_{\bf{p}}}{\left( {2\pi } \right)^3}{\delta ^{\left( 3 \right)}}\left( {{\bf{p}} - {\bf{q}}} \right),$$
    where the prefactor is conventional. It's the usual (modern) choice in HEP, because it leads to the definition of S-matrix elements that are manifestly Lorentz covariant. Have a look in the section on the definition of the invariant cross section in Peskin and Schroeder, because there it's very well described in which sense, the here discussed "momentum eigenstates" have a physical sense.

    To answer your final question, now think first, what is
    $${\bf P} a^{\dagger}_{{\bf k}} |0 \rangle,$$
    and then think about, wheter it makes sense to multiply this from the left with ##\langle {\bf k}|##!
  19. Apr 30, 2015 #18
    O. K. I THINK I GOT IT!!!

    Following Peskin equation (4.65)

    ##\left| \phi \right\rangle = \int {{{{d^3}k} \over {{{\left( {2\pi } \right)}^3}}}{1 \over {\sqrt {2{E_{\bf{k}}}} }}\phi \left( {\bf{k}} \right)\left| {\bf{k}} \right\rangle } = \int {{{{d^3}k} \over {{{\left( {2\pi } \right)}^3}}}\phi \left( {\bf{k}} \right)a_{\bf{k}}^ + \left| 0 \right\rangle } ##

    taking ##\phi \left( {\bf{k}} \right) = \exp \{ - {{\left| {{{\bf{k}}^2} - {{\bf{p}}^2}} \right|} \over a}\} /b##

    and b such that

    ##\int {{{{d^3}k} \over {{{\left( {2\pi } \right)}^3}}}{{\left| {\phi \left( {\bf{k}} \right)} \right|}^2} = 1} ##

    we finally have:

    & \left\langle {\phi |{\bf{P}}|\phi } \right\rangle = \int {{{{d^3}p{d^3}k{d^3}k'} \over {{{\left( {2\pi } \right)}^9}}}} {\bf{p}}\phi \left( {\bf{k}} \right)\phi \left( {{\bf{k'}}} \right)\left\langle {0|{a_{\bf{k}}}a_{\bf{p}}^ + {a_{\bf{p}}}a_{{\bf{k'}}}^ + |0} \right\rangle = \cr
    & = \int {{{{d^3}k{d^3}k'} \over {{{\left( {2\pi } \right)}^6}}}} {\bf{k'}}\phi \left( {\bf{k}} \right)\phi \left( {{\bf{k'}}} \right)\left\langle {0|{a_{\bf{k}}}a_{{\bf{k'}}}^ + |0} \right\rangle = \cr
    & = \int {{{{d^3}k} \over {{{\left( {2\pi } \right)}^3}}}} {\bf{k}}{\left| {\phi \left( {\bf{k}} \right)} \right|^2} = {\bf{k}} \cr} ##.

    So ##\left| \phi \right\rangle ## is a realistic eigenstate of ##{\bf{P}}## and ##{\left| {\bf{k}} \right\rangle }## is just a mathematical tool.

  20. Apr 30, 2015 #19

    Isn't ##\mathop {\lim }\limits_{a \to \infty } \exp \{ - {{\left| {{{\bf{k}}^2} - {{\bf{p}}^2}} \right|} \over a}\} /b = \delta \left( {\bf{p}} \right)##


    & \mathop {\lim }\limits_{a \to \infty } \left| \phi \right\rangle = {1 \over {\sqrt {{{\left( {2\pi } \right)}^3}2{E_{\bf{p}}}} }}\left| {\bf{p}} \right\rangle \cr
    & \cr
    & \left\langle {\phi |\phi } \right\rangle = 1 \cr
    & \Downarrow \cr
    & \mathop {\lim }\limits_{a \to \infty } \left\langle {\phi |\phi } \right\rangle = {1 \over {{{\left( {2\pi } \right)}^3}2{E_{\bf{p}}}}}\left\langle {{\bf{p}}|{\bf{p}}} \right\rangle = 1 \cr
    & \Downarrow \cr
    & \left\langle {{\bf{p}}|{\bf{p}}} \right\rangle = 2{E_{\bf{p}}}{\left( {2\pi } \right)^3} \cr
    & \Downarrow \cr
    & \left\langle {0|{a_{\bf{p}}}a_{\bf{p}}^ + |0} \right\rangle = {\left( {2\pi } \right)^3} \cr
    & \cr
    & \delta \left( {{\bf{p}} - {\bf{p}}} \right) = 1 \cr
    & \cr
    & H = \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\omega _{\bf{p}}}\left( {a_{\bf{p}}^ + {a_{\bf{p}}} + {1 \over 2}\left[ {{a_{\bf{p}}},a_{\bf{p}}^ + } \right]} \right)} \cr
    & \cr
    & H = \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\omega _{\bf{p}}}a_{\bf{p}}^ + {a_{\bf{p}}} + } {1 \over 2}\int {{d^3}p{\omega _{\bf{p}}}} \cr} ##.

    We can define ##\delta \left( {{\bf{p}} - {\bf{p}}} \right) = 1## which will extend the kroneker delta.

    Now ##H = \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\omega _{\bf{p}}}\left( {a_{\bf{p}}^ + {a_{\bf{p}}} + {1 \over 2}\left[ {{a_{\bf{p}}},a_{\bf{p}}^ + } \right]} \right)} ## (Peskin e. (2.31)

    will be ##H = \int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{\omega _{\bf{p}}}a_{\bf{p}}^ + {a_{\bf{p}}} + } {1 \over 2}\int {{d^3}p{\omega _{\bf{p}}}} ## which is still infinity.

    and ##\left\langle {0|a_{\bf{p}}^ + {a_{\bf{p}}}|0} \right\rangle = \left\langle {0|\left[ {a_{\bf{p}}^ + ,{a_{\bf{p}}}} \right] + {a_{\bf{p}}}a_{\bf{p}}^ + |0} \right\rangle = - {\left( {2\pi } \right)^3} + {\left( {2\pi } \right)^3} = 0##

    so everything works fine.

    (I played a little with the ##{2\pi }## so it will fit)
  21. Apr 30, 2015 #20


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    Point 1: your delta function definition, as a limit of the exponential is wrong in form. The delta function should have the limit going to 0, and not of that expression I think.

    Is it logical to say [itex]\delta(0)=1[/itex]?

    Also for [itex]<0|a_p^\dagger a_p |0> =0[/itex] you can see that immediately for you are taking the "average" number of quanta in the vacuum (which is zero).
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