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Why transition rate independent of time in perturbation theory?

  1. Feb 12, 2014 #1
    After time t, the probability of monochromatic absorption of the ground state |1> to the energy state |n> is given by:

    [tex]|<n|1>|^2=4|U_{n1}|^2\frac{\sin^2((E_n-E_1-\hbar\omega)t/2\hbar)}{(E_n-E_1-\hbar\omega)^2} [/tex]

    where U is the transition matrix. The claim is that as t goes to infinity, the fraction becomes (up to factors of pi and stuff):

    [tex]t*\delta(E_n-E_1-\hbar\omega) [/tex]

    So if you take the probability per time by dividing by the time, it is independent of t!

    But what's really going on here? It looks to me like if you divide the first equation I have by t, then as t goes to ∞, then the numerator just oscillates, and t in the denominator blows up, so the transition rate is zero.

    If the claim is that [itex]E_n-E_1-\hbar\omega [/itex] is really small so that the sine term can keep growing as t increases, can't you expand the sine term in power series, and get a [itex]t^2[/itex] dependence, so that when you divide by time to get the transition probability, it depends linearly on time? So you would get that transition rate is proportional to t rather than [itex]\delta(E_n-E_1-\hbar\omega) [/itex]?
     
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  3. Feb 13, 2014 #2

    DrDu

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    The point is rather that for large t, the sin^2 cannot get larger than 1, so that dividing by t, this term goes to 0. The only exception occurs very near the point of energy conservation.
     
  4. Feb 13, 2014 #3
    If the probability of the state to make a transition is of the form:

    |<n|1>|2=at+b

    Then the transition rate is a+b/t ≈ a

    In the formula above, a is a delta function.

    However, doesn't the probability at+b grow unbounded with t?

    The maximum of the probability should be 1?

    If you have a bounded function f(t), then f(t)/t as t → ∞ should equal zero?
     
  5. Feb 13, 2014 #4

    DrDu

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    It is a first order result, only. Depletion of the initial state is a higher order effect and thus not included.
     
  6. Feb 13, 2014 #5
    So would it be correct to say that you can divide the expression [tex]|<n|1>|^2=4|U_{n1}|^2 \frac{\sin^2((E_n-E_1-\hbar\omega)t/2\hbar)}{(E_n-E_1-\hbar\omega)^2} [/tex]

    into two regions: t small and t big.

    t small: power expanding the sine, you get (up to constants), that the probability for transition is proportional to t2, or that the average transition rate from 0 to t is proportional to t.

    Note that it doesn't matter if [itex]E_n=E_1+\hbar\omega [/itex] or not, at small t the probability of transition is independent of En (it does depend on Un1, but not on the value of En).

    t big: presumably you can no longer power expand in sine even if [itex]E_n=E_1+\hbar\omega [/itex] since then you get the t small behavior. This must imply that t>[itex](E_n-E_1-\hbar\omega)^{-1} [/itex]. Because the sine is oscillating, when you divide by t and take t large, you get zero. Unless [itex](E_n-E_1-\hbar\omega)=0 [/itex], in which case the denominator blows up. The numerator is still oscillating (otherwise it's quadratically increasing like the t small case), but the denominator is going to zero due to energy conservation, and this cancels the the division by t when t goes to ∞? Not only do they cancel to get a nonzero result, but the whole oscillatory nature of the sine gets cancelled out even though you can't power expand the sine since it's oscillating.
     
  7. Feb 13, 2014 #6

    ChrisVer

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    In general, you should have a look at it first from the other way around.
    If you write (one out of many defs of Dirac's Delta function):
    [itex] δ(x-x_{0})= \frac{1}{2π}\int_{-∞}^{+∞} dt e^{i(x-x_{0})t}[/itex]
    And now instead of infinites as limits, just take [itex]-T/2, +T/2[/itex] and sent [itex]T[/itex] to infinity.... then you have:

    [itex] δ(x-x_{0})= \frac{1}{2π}lim_{T→∞}\int_{-T/2}^{+T/2} dt e^{i(x-x_{0})t}[/itex]

    The solution of that integral, is the sin (due to symmetric limits the ∫sin=cos die out). Thus you have:

    [itex]δ(x-x_{0})= \frac{2}{2π}lim_{T→∞} \frac{sin((T/2)(x-x_{0}))}{(x-x_{0})}[/itex]

    Up to now it's middle way there... I'll forget the lim but keep in mind that T is going to infinity.
    Takeing the square on both sides:
    [itex](2πδ(x-x_{0}))^{2}= 4 \frac{sin^{2}((T/2)(x-x_{0}))}{(x-x_{0})^{2}}[/itex]

    being almost there only now you have to use that:
    [itex](2πδ(x-x_{0}))^{2}= (2π δ(0)) (2π δ(χ-χ_{0}))= 2π T δ(χ-χ_{0})[/itex]
    there is a quick way to check it from the definition of delta function again that:

    [itex]2πδ(0)=\int_{-T/2}^{+T/2} dt e^{0}= T[/itex]
    Of course that's a tricky way...

    Nevertheless, substituting the squared expression back, the result then reads:
    [itex](2πδ(x-x_{0}))^{2}=2π T δ(χ-χ_{0})=4 \frac{sin^{2}((T/2)(x-x_{0}))}{(x-x_{0})^{2}}[/itex]

    So indeed playing around with the numbers,
    [itex](T/2) δ(x-x_{0})=\frac{sin^{2}((T/2)(x-x_{0}))}{(x-x_{0})^{2}}[/itex]
    Well nobody told me that I should have used the expression T/2, I just used it because I wanted to be sure I didn't make a mistake...just rename it [itex]t[/itex] deal it as a time (don't forget the = symbol for delta function holds only for t→∞ as you also said), change the [itex]x,x_{0}[/itex] accordingly, and you will get the result you asked for....

    (Source: Bjorken and Drell, relativistic quanum mechanics, Chap7 par1)

    I think that this kind of game must have been a Feynman's trick :cool: dealing infinities like that
     
    Last edited: Feb 13, 2014
  8. Feb 13, 2014 #7
    When you squared the delta function like that it reminded me of how transition rates are calculated in quantum field theory.

    If I recall correctly, what happens is that the scattering amplitude is proportional to [itex]\delta^4(k_i-k_f) [/itex] which is conservation of 4-momenta k. You square that to get the probability: [itex]\delta^4(k_i-k_f)\delta^4(k_i-k_f)=\delta^4(k_i-k_f)\delta^4(0)= 2\pi^4\delta^4(k_i-k_f)VT[/itex].

    Then when you divide by T to get the transition rate, you get that the transition rate is independent of time.

    But it seems in quantum mechanics the fact that the transition rate is independent of time is due to assumption that t--->infinity, that you are really calculating the the average transition rate between t=0 and t=infinity. And it also seems that we are assuming that the perturbation is so small that it takes an infinite amount of time to deplete the original state.
     
  9. Feb 13, 2014 #8

    ChrisVer

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    I am not sure for the below, I'm just using instinct.....


    you also have to think how:
    [itex]\frac{sin^{2}x}{x}[/itex] behaves for large x... it of course goes to zero...

    Now what would happen in the case you had something else with which you could play with?
    For example an [itex]a[/itex], whose values you could just adjust in any way you'd like, and it being like the below:

    [itex]\frac{sin^{2}ax}{a^{2}x}[/itex]

    Sending x to infinity, you would still have that:
    [itex]0≤\frac{sin^{2}ax}{a^{2}x}≤\frac{1}{a^{2}x}[/itex]
    But in this case you can also send [itex]a[/itex] to zero... if [itex]a≠0[/itex] then the sandwich will tell you that by sending [itex]x→∞[/itex] you will get 0...

    What would happen if [itex]a→0[/itex]? First of all you can choose by yourself in what rate the [itex]a[/itex] will approach zero.
    The sandwich can't work out nicely now, because [itex]a[/itex] will rule out [itex]x[/itex]'s grows (it's squared) and the result won't be readable.
    You will have to go again back in the first expression case...
    [itex]\frac{sin^{2}ax}{a^{2}x}[/itex]
    and deal that [itex]a[/itex] will rule out any [itex]x[/itex]'s sending to infinity, in such a way that [itex]ax=const[/itex]... the [itex]sin^{2}(ax)[/itex] will just tend to go to a fixed value since in [itex]ax[/itex] one will grow and the other will become less in the same rate... and you will have that [itex]a^{2}x[/itex] will be ruled out by [itex]a[/itex]'s going to zero rate... then you will have a [itex]1/a[/itex] behavior, that for [itex]a[/itex]→0 will explode in infinity....

    So what is the result? for [itex]a[/itex] not approaching zero, you have the expression going to 0, and when [itex]a[/itex] approaches zero, you get infinity... That's a nice spot to put [itex]δ(a)[/itex]
     
    Last edited: Feb 13, 2014
  10. Feb 13, 2014 #9

    ChrisVer

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    I don't really understand what you mean by this paragraph. I think that we take the time to approach infinity, because we consider that we stand really far away from where the interaction happens...
    So you send something that interacts with (a perturbative yes) potential in some region, and you measure the results far away from that region.
     
  11. Feb 13, 2014 #10
    That's a really nice way to explain why the sine stops oscillating when you take t (or x) goes to infinity, that [itex]ax=const[/itex].

    It just seems shocking, if you have [itex]\frac{sin^{2}ax}{a^{2}x}[/itex], then sin doesn't blow up no matter what ax is, so any potential problems must be in a^2x. So what's wrong with saying that [itex]a\sqrt{x}=const [/itex] instead of ax=const?

    For QFT you're right, we take time to infinity because we stand far away. However, for QM, the way that time-dependent perturbation theory is set up, in principle you can make the calculations for any time. In this way it's different from scattering where we're only interested in the far past and far present.
     
  12. Feb 13, 2014 #11

    ChrisVer

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    The reason why I wanted to use [itex]ax[/itex] as a constant, was because it was handy for the problem (it would fix the sin's argument and I wouldn't have to care about oscillations- not because of the denominator). In fact the [itex]a[/itex] as I inserted it, was just an arbitrary parameter whose value you can adjust anyway you like. I mean there is no loss by assuming a handy rate (it doesn't affect anything) since what you want in the end is that [itex]a=0[/itex] instead of just "approaching zero". That's why I was also able to write [itex]=δ(a)[/itex] which defines the infinity appearing at [itex]a=0[/itex].
    It wouldn't be the same I guess, if you couldn't arbitrary adjust [itex]a[/itex] in general... if for example you had an a priori set rate when you defined [itex]a[/itex]. But at least that's fine for the energy differences and time... :wink: So in this case if you tell me that you want to send it faster I will answer I want to send it slower, and if you tell me that you want to send it slower I will say send it faster so that [itex]ax=const[/itex] will hold... it's not self-running like time...

    PS. A mathematician might kill me though...but he would also do that seeing me squaring the delta function and treating it like that.
    (i also corrected the sandwich rule in my previous post - had typed it wrongly with sin^2 to play from -1 to 1...)
     
    Last edited: Feb 13, 2014
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