Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Commutator of operators

  1. Jul 29, 2014 #1

    dyn

    User Avatar

    When performing matrix multiplication with 2 matrices A and B ;AB might exist but BA might not even exist. Hermitian operators can be thought of as matrices but in everything I have seen so far AB and BA always exist even though they can be different depending on the value of the commutator. Do AB and BA always exist ? And if so , why ? Thanks
     
  2. jcsd
  3. Jul 29, 2014 #2

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    You can always multiply any two square matrices so I don't know where you get that from.

    Thanks
    Bill
     
  4. Jul 29, 2014 #3

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    As bhobba already suggested, maybe you're thinking of matrices that aren't square. Operators are more like square matrices, at least when they're defined on the entire Hilbert space H. But there are operators in QM that are only partially defined, i.e. defined only on a subspace. If the subspace is dense in H, the operator is said to be densely defined. Position and momentum are the best examples of operators that are only densely defined.

    If A and B are two densely defined operators, then the domain of AB consists of those x in H such that Bx is in the domain of A. Similarly, the domain of BA consists of those x in H such that Ax is in the domain of B. The domain of the commutator [A,B] is the intersection of the domains of AB and BA. So when we write [x,p]=i, the right-hand side shouldn't be interpreted as ##iI##, where ##I## is the identity operator on H. It should be interpreted as ##iI_D##, where ##I_D## is the identity operator on the domain of [x,p].
     
  5. Jul 29, 2014 #4

    dyn

    User Avatar

    Thanks. I had forgot that Hermitian matrices must be square. But why must they be of the same size , ie the same number of rows and columns ?
     
  6. Jul 29, 2014 #5

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    If the two matrices are not the same size, then they don't act on the same space. In that case neither AB nor BA are defined.
     
  7. Jul 29, 2014 #6
    I hope it's ok for me to step in with my own question. Which are the (dense) subsets of the Hilbert space on which the position and momentum operators are defined?

    How would an operator like x fail to be defined on a subset? Is it by failing to be bounded on that subset?

    I would very much appreciate it if you could suggest to me some (introductory) reference where I can read about this kind of thing. Thanks.
     
  8. Jul 29, 2014 #7

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    To ensure operators are always defined you have to go to RHS formalism. The position operator is always defined in the position representation ie regardless what F(x) is, Lebesgue square integrable, a distribution, anything x F(x) is always defined. The issue is the momentum operator. That's best done in the RHS formalism. You take some kind of space that is differentiable so df(x)/dx always exists - they can be test functions, fairly good functions (they are nice because their Fourier transform is a fairly good function) then as per distribution theory - given any bra F you can define its derivative as minus <F|df/dx>. Thus it's defined on the dual of whatever space you based it on - which for fairly good functions are the well tempered distributions, which is a superset of square integrable functions - so virtually anything can have the momentum operator defined on it.

    Thanks
    Bill
     
  9. Jul 30, 2014 #8
    Now I know what I should read about next: rigged Hilbert spaces...any suggestions for a good reference? (preferably something written for quantum mechanists as opposed to mathematicians). Thank you!
     
  10. Jul 30, 2014 #9
    but what about ##F(x)=\frac{1}{x}##, then F(x) is square integrable, and so is in the Hilbert space L2, whereas ##xF(x)=1## is not square integrable. So the operator x takes us outside the Hilbert space - and is therefore not defined on that F(x)
     
  11. Jul 30, 2014 #10

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    1/x is not normally square integrable because of the pole at zero. But one can use a trick so in distribution theory it is hence 1/x or even F(x)/x is legit if F(x) is a distribution.

    I wont detail the trick (OK you got it out of me - Cauchy principle value) - best you study the whole thing as part of Distribution theory. Here is my go to text:
    https://www.amazon.com/The-Theory-Distributions-Nontechnical-Introduction/dp/0521558905

    I highly recommend it to anyone into physics or applied math.

    That's just a build up to the full treatment:
    http://physics.lamar.edu/rafa/webdis.pdf

    Terry Tao has also written an introduction:
    http://www.math.ucla.edu/~tao/preprints/distribution.pdf

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  12. Jul 30, 2014 #11

    strangerep

    User Avatar
    Science Advisor

    Yes. See below.

    Umm, not sure what you (really) meant there,...

    If ##F(x)## is square integrable, it might be that ##G(x) := x F(x)## is not. And if the position operator is well-defined, than ##x G(x) = x^2 F(x)## must be square integrable too, and so on, for arbitrarily high ##x^n##. Similarly for the momentum operator.

    That leads to Schwartz space as the relevant domain on which the position and momentum operators are well-defined.

    :biggrin: See the item which is in both my signature line, and Bhobba's. (I.e., Ch1 of Ballentine.)
     
  13. Jul 30, 2014 #12

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    Sorry I wasn't clear - I meant exists as a function - of course it may not be square integrable. One can only take derivatives of some functions - but xF(x) is always a function.

    Thanks
    Bill
     
    Last edited: Jul 30, 2014
  14. Jul 30, 2014 #13

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In the quantum theory of a spin-0 particle moving in 1 dimension, the Hilbert space is ##L^2(\mathbb R)##, i.e. the set of (equivalence classes of) square-integrable functions from ##\mathbb R## to ##\mathbb C##. I will denote that Hilbert space by ##\mathcal H##. We'd like to define the operators Q (position) and P (momentum) through the formulas
    \begin{align}
    (Qf)(x)=xf(x)\\
    (Pf)(x)=f'(x)
    \end{align} for "all f and all x". But "all f" can't mean "all f in ##\mathcal H##", because there are functions f in ##\mathcal H## such that one or more of the following statements are true.

    1. The g defined by g(x)=xf(x) for all x isn't square integrable.
    2. The g defined by g(x)=f'(x) isn't square integrable.
    3. f isn't differentiable.

    We have to keep that in mind when we define Q and P. Let E be the set of all ##f\in\mathcal H## such that the g defined by g(x)=xf(x) is square integrable. Define ##Q:E\to\mathcal H## by ##(Qf)(x)=xf(x)## for all ##f\in E## (not all ##f\in\mathcal H##) and all ##x\in\mathbb R##. Let F be the set of all ##f\in\mathcal H## such that f is differentiable and ##f'\in\mathcal H##. Define ##P:F\to\mathcal H## by ##(Pf)(x)=f'(x)## for all ##f\in F## and all ##x\in\mathbb R##.

    In the case of Q and P (or x and p if you prefer that notation), the reason is what I said above. But I think (not sure) that there's also a theorem that says that only bounded operators can be defined on the whole space.

    I've heard good things about https://www.amazon.com/Introductory-Functional-Analysis-Applications-Kreyszig/dp/0471504599/. But I think even the easiest books are quite difficult. There's no way to really learn the mathematics of QM without studying at least 500 pages (probably more like 1000 pages) of topology, integration theory, Hilbert spaces and operator algebras.
     
    Last edited by a moderator: May 6, 2017
  15. Jul 30, 2014 #14
    Ok...following what you guys have explained - I did some reading and the whole thing is much clearer now in my mind. I don't want to spend too much time studying RHS in detail... I will instead jump into the theory of distributions, as it is what seems to be lacking in my background. (After all, one must understand distributions well to completely understand Gelfand's triplet).

    Thanks to everybody.
     
  16. Jul 31, 2014 #15

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    Good idea.

    It should be part of any physicists or applied mathematicians arsenal.

    Thanks
    Bill
     
  17. Jul 31, 2014 #16

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I see that forgot a factor of -i in my definition of P in post #13. The defining equation should be ##(Pf)(x)=-if'(x)##, not ##(Pf)(x)=f'(x)##. The operator defined by the latter equation should be denoted by D, not P (i.e. Df=f'). The relationship between them is P=-iD.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Commutator of operators
  1. Commuting operators (Replies: 1)

  2. Commuting of operators (Replies: 4)

  3. Operators Commutation (Replies: 6)

Loading...