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A question about Dirac's quantization

  1. Jul 16, 2005 #1
    In his book Lectures on Quantum Mechanics,
    Dirac gives his way of quantizing system with contraints.

    I do not understand the use of weak inequality in some equations.
    [tex]C \approx 0[/tex]

    What is that for?
  2. jcsd
  3. Jul 16, 2005 #2
    I've been slightly confused by this too, but my understanding of it is that while the [itex]\phi_m[/itex] may equal zero, their derivatives perhaps don't... For this reason then, the quantity [itex]\left[g,\phi_m\right][/itex] (e.g. in equations 1-15, 1-17 to 1-19) would be trivially zero if you plug in [itex]\phi_m=0[/itex], however if you examine the Poisson bracket you actually have
    [tex]\left[g,\phi_m\right] = \frac{\partial g}{\partial q_n}\frac{\partial \phi_m}{\partial p_n} - \frac{\partial g}{\partial p_n}\frac{\partial \phi_m}{\partial q_n}[/tex]
    and this is not necessarily zero. Writing the constraint equations as [itex]\phi_m\approx0[/itex] is just supposed to serve as a reminder not to substitute [itex]\phi_m=0[/itex] into your equations until after you've evaluated any Poisson brackets.
  4. Jul 16, 2005 #3


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    No, weak egaility means that, if 2 phase space functions are weakly equal

    [tex] F\approx G [/tex]
    , it means that they're equal when evaluated on the surface of all constraints. Denoting this surface by [itex] \phi_{\Delta}=0 [/itex] , we have BY DEFINITION

    [tex] F\approx G \Leftrightarrow F|_{\phi_{\Delta}=0}=G|_{\phi_{\Delta}=0} [/tex]

    Last edited: Jul 16, 2005
  5. Jul 16, 2005 #4
    That doesn't seem to be exactly what Dirac is getting at in his book. He says:

    So he actually has the constraints weakly equal to zero, not functions evaluated on surfaces of all constraints. He also explicity says that while the constraints are equal to zero, this substitution must not be made until after the Poisson brackets have been evaluated. I mean, I guess we could say

    [tex]\phi\approx0 \Leftrightarrow \phi|_{\phi=0}=0|_{\phi=0}[/tex]

    but I'm not sure how useful that is. Or am I completely misinterpreting Dirac when he says that the only reason to use [itex]\approx[/itex] is to remind us that we must evaluate Poisson brackets before using the fact that the constraints are in fact equal to zero?
  6. Jul 16, 2005 #5


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    I've given you the definition for the weakly equality

    One writes the finite countable set of all constraints
    [tex] \phi_{\Delta}\approx 0 [/tex]

    , instead of [itex] \phi_{\Delta}=0 [/itex] to specify that the expressions [itex] \phi_{\Delta}\left(q,p\right) [/itex] equal zero NOT on the entire phase space, but only on the submanifold of all constraints.

    As for the definition of weak equality between 2 arbitrary functions defined on ALL phase space, it is given two posts above.

    There's a theorem whose result is used to establish the following connection between strong equality and weak equality

    [tex] F (q,p)\approx G(q,p) \Longleftrightarrow (F-G) (q,p)=c^{\Delta}(q,p) \phi_{\Delta}(q,p) [/tex]

  7. Jul 17, 2005 #6
    Dirac quantization method is not completely rigorous (the deformation quantization is actually considered as the most accomplished one).
    For the ones who are interested on a good compilation of the different quantization methods, I recommend the excellent and impressive paper (which also indicates the problems/restrictions associated with some methods):
    Quantization methods: a guide for physicists and analysts, S. Twareque and M. Englis, math-ph/0405065.


    P.S. I wish I would like to have an analogue paper for the different formulations of QM : )
  8. Jul 17, 2005 #7
    I accept that you've given a definition of a weak equality... I just still am not sure that it is the same as the one Dirac uses in his book. He specifically states that the notation is solely as a reminder, not some mathematical formalism.

    This is the part that confuses me - I thought that setting [itex]\phi_m=0[/itex] was the constraint. That is, we constrain that it must be zero. If you are saying that the phi equal zero only where they equal zero, then I agree, but do not understand why that statement requires a new symbol for equality...

    Don't suppose you want to define what [itex]c^{\Delta}[/itex] is? Or what happens when [itex]F=\phi[/itex] and [itex]G=0[/itex], which is the example we are discussing here? In that case it seems that you just get

    [tex] \phi_{\Delta}(q,p)\approx0 \Longleftrightarrow \phi_{\Delta}(q,p)=c^{\Delta}(q,p)\phi_{\Delta}(q,p)[/tex]

    which seems to imply that [itex]c^{\Delta}\equiv1[/itex] and that everything is weakly equal to zero since everything is equal to itself... But that doesn't tell me anything useful - anything is equal to itself times the identity, so why does this statement require a different symbol for equals than the standard '='? Perhaps I should also confirm that your [itex]\phi_{\Delta}[/itex] is the same as Dirac's [itex]\phi_m[/itex]?

    Perhaps what is confusing me most is why my original interpretation (second post in this thread) of Dirac's weak equality is wrong. His meaning seems fairly straightforward in that passage I quoted from his text. Could you (or anyone) perhaps explain exactly what he means by saying that he rewrites the equations [itex]\phi_m=0 \longrightarrow \phi_m\approx0[/itex], since that is the entire point of this thread and I apparently am incorrect?
  9. Jul 17, 2005 #8


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    (Disclaimer: I'm assuming I know what's going on here)

    In what Dextercioby wrote, I think he's saying that the function cΔ exists. In other words:

    [tex]F \approx G \Leftrightarrow \exists c^\Delta: F - G = c^\Delta \phi_\Delta[/tex]

    You made the freshman's mistake of dividing by zero. :smile: Recall that the proper inference from ab = a is that b = 1 or a = 0.

    The point about weak equality is that you only care about what happens on a particlar set. Maybe an elementary example would help:

    On the set Δ = {1}, which is defined by the constraint φΔ(x) = x - 1 (i.e. it's the solution set to φΔ(x) = 0), we have the following weak equality:

    [tex]\frac{d}{dx} x^2 \approx 2[/tex]

    This is a weak equality because it is true at every point in Δ. It is not a strong equality because it is false on points outside of Δ.

    Yet another way of looking at it is that [itex]F \approx G[/itex] means that φ(P) = 0 → F(P) = G(P).
    Last edited: Jul 17, 2005
  10. Jul 18, 2005 #9


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    My guess: may be he just wants to point out that the total Hamiltonian contains [itex]\phi_m[/itex] that are [itex]\phi_m = 0[/itex] on the constraint surface, i.e. [itex]\phi_m \approx 0[/itex]. The general expression for this Hamiltonian (a Hamiltonian which is extended from the the canonical Hamiltonian on the constraint surface) has to be used to find the unphysical degrees of freedom. Well, I think I did not say anything new to the other previous posts.
  11. Jul 18, 2005 #10
    Oops, I did at that... Although, I did only divide by a weak zero :)

    The example did help, thanks. I think I now know where Daniel was coming from. I'm still not sure about the original question though, as to how/why Dirac used the weak inequality. This is because from your example it seems that to speak of a set of [itex]\phi_\Delta(x)[/itex] as being constraints is to mean that we require [itex]\Delta=\{x:\phi_\Delta(x)=0\}[/itex]...? (Now that I write this it seems awefully recursive as a definition....) So why do we say that [itex]\phi_\Delta\approx0[/itex]? Does it change the definition to [itex]\Delta=\{x:\phi_\Delta(x)\approx0\}[/itex]? If so, what does that mean, is it not further recursion?

    Perhaps this is an irrelevent point, I don't know... It just seems to me that all the examples in this thread, whether of arbitrary or specific functions, are using weak inequalities of functions on a set of constraints - but none of them talks about what it means for the constraint equations themselves to contain weak equalities...
  12. Jul 18, 2005 #11
    This is essentially what I said in my very first post in this thread, without using the language of 'constraint surfaces' - that while the functions [itex]\phi_m[/itex] are required to be zero at the actual constraints, they are not everywhere zero so their derivatives may be non-zero and must therefore be evaluated in any Poisson brackets before making the substitution for [itex]\phi_m[/itex]. Daniel said this was wrong, which is the only reason I am still trying to figure out what the correct interpretation is......
  13. Jul 18, 2005 #12


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    The constraints are just functions, so the definition is exactly the same.

    φ ~ 0 means φ = 0 → φ = 0.

    It's a fairly trivial statement, but sometimes the trivial can be fairly profound!

    (For example, ideals are one of the most important objects of study when studying rings... but an ideal is simply the set of things that "ought" to be equal to zero, when looking at the ring from a particular perspective)

    The point is, I suppose, is that it's not a "reminder" about procedure -- weak equivalence has an honest-to-goodness mathematical meaning.

    When you don't make the distinction, you suffer from a sort of cognative dissonance -- sometimes you treat the objects as equal, and sometimes you don't treat the objects as equal, and that's bad because that's not how equality works.
  14. Jul 19, 2005 #13
    Correct me if I am wrong.

    Weak equality is used in the equations that are true ONLY when the constraints hold.

    Note the word "ONLY", if weak equality had been defined in the following way without the word "ONLY"

    Weak equality is used in the equations that are true when the constraints hold.

    Then all strong equalities are automatically weak equalities. That's not what Dirac wanted.

    Of course, the constraint function itself is zero only when the constraints hold. So [tex]\phi \approx 0[/tex]

    For example, if [tex]\{f, g\} = \phi[/tex]
    [tex]\phi[/tex] is some constraint
    Then we may say
    [tex]\{f, g\} \approx 0[/tex]

    I think they are related. Computing poisson brackets involve computing partial derivatives, partial derivatives are defined assuming all variables are independent, which is only true without any contrainst.

    I mean Poisson Bracket lives in the whole phase space, not only the constrained phase space.
    Last edited: Jul 19, 2005
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