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

Commutators and traces

  1. Mar 31, 2015 #1

    dyn

    User Avatar

    I read the following as a model solution to a question but I don't understand it - " there is no possible finite dimensional representation of the operators x and p that can reproduce the commutator [x,p] = I(hbar)(identity matrix) since the LHS has zero trace and the RHS has finite trace. My questions are what has the trace got to do with it ? Also an infinite dimensional identity matrix will not have zero trace either so how is the equation satisfied in that case ?
    Thanks
     
  2. jcsd
  3. Apr 1, 2015 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    In finite-dimensional vector spaces the trace of an operator always exists and for any two operators ##\mathrm{Tr} (A B)=\mathrm{Tr} (B A)## or, equivalently, ##\mathrm{Tr}([A,B]=0##. This immediately leads to a contradiction, when applied to the Heisenberg algebra, ##[x,p]=\mathrm{i} \mathbb{1}##, if you assume that it is realized on a finite-dimensional Hilbert space.

    In the separable Hilbert space of infinite dimension, there's no such contradiction, because the unit operator has no trace.
     
  4. Apr 1, 2015 #3

    dyn

    User Avatar

    so for the equation to be valid the trace must be the same on both sides ? The commutator has zero trace but wouldn't an infinite dimensional identity matrix have an infinite trace ?
     
  5. Apr 1, 2015 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    That's the point. The trace operation has a limited domain, and if the commutator is not in that domain, it's simply not defined. In infinite-dim. vector spaces you thus can not prove that the trace of any commutator necessarily vanishes. That's precisely what happens in the case of the Heisenberg commutation relation. Note that also position and momentum operators have a limited domain, i.e., they are only defined on a dense subspace and not on the entire Hilbert space. However, you can extend the definition of the pre-unitary operators induced by the self-adjoint operators (e.g., for momentum you get a spatial translation operator) to unitary operators on the entire Hilbert space.
     
  6. Apr 1, 2015 #5

    dyn

    User Avatar

    In infinite dim space doesn't the commutator still have zero trace as Tr(AB)=Tr(BA) ? I thought the model solution implied that HUP is not valid for finite dim spaces but valid for infinite dim spaces ?
     
  7. Apr 1, 2015 #6

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    The trace of the unit operator doesn't exist in infinite-dimensional Hilbert space. The trace of the commutator ##[x,p]## is not defined at all and that's why it cannot be 0.
     
  8. Apr 1, 2015 #7

    dyn

    User Avatar

    I'm getting confused now. Is the x,p commutator relation satisfied anywhere ? It seems it isn't satisfied with finite dim or infinite dim ?
     
  9. Apr 2, 2015 #8

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Well, look for the concrete relization of ##x## and ##p## on the Hilbert space ##L^2(\mathbb{R},\mathbb{C})## ("wave mechanics"). You have
    $$\hat{x} \psi(x)=x \psi(x), \quad \hat{p} \psi(x)=-\mathrm{i} \partial_x \psi(x).$$
    Both operators are self-adjoint with a domain that is for sure not the entire Hilbert space but a dense subspace (like Schwartz's space of rapidly falling ##C^{\infty}##. Their spectrum is entirely continuous and the whole real line ##\mathbb{R}##. The "eigenvectors" are in the dual of their domain, which is much larger than ##L^2##. They are distributions rather than real functions,
    $$u_{x_0}(x)=\delta(x-x_0), \quad u_p(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).$$
    The commutator is
    ##[\hat{x},\hat{p}]=\mathrm{i} \mathbb{1},##
    and neither the trace of ##\hat{x}##, ##\hat{p}##, ## \hat{x} \hat{p}##, and ##\hat{p} \hat{x}## nor that of ##[\hat{x},\hat{p}]## exists.

    For a more mathematically rigorous treatment, see, e.g., the books by Galindo and Pascual.
     
  10. Apr 2, 2015 #9

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

  11. Apr 2, 2015 #10

    dyn

    User Avatar

    Some of this is going over my head. I know that the commutation relation is not satisfied for finite dim spaces as I did an example which showed that and I know in finite dim the commutator has zero trace and the identity matrix has finite trace. I know x and p are infinite dim in reality and in infinite dim traces do not exist. Am I right so far ? If yes then is the commutator satisfied exactly in infinite dim ?
     
  12. Apr 2, 2015 #11
    It is not exactly that the traces don't exist, they are either undefined or their value depends on the basis chosen, so that's all that can be said of their commutator. The first option is useless and QM is a pragmatic theory.
     
  13. Apr 3, 2015 #12

    strangerep

    User Avatar
    Science Advisor

    You're confronting (one of) the problems that make a framework larger than ordinary Hilbert space desirable. That framework is called "Rigged Hilbert Space".

    Ch1 of Ballentine gives a gentle introduction.
     
  14. Apr 3, 2015 #13

    dyn

    User Avatar

    I came across the following question which requires a true or false answer. " for any integer N ≥ 2 there exists N x N matrices X and P such that [ X , P ] = I(hbar)(identity matrix)
    From what has been said above I hope the correct answer is FALSE. Can somebody confirm for me that I am correct ? Thanks
     
  15. Apr 3, 2015 #14

    samalkhaiat

    User Avatar
    Science Advisor

    For finite N, it is false. See below
    The set of all [itex]n \times n[/itex] matrices with entries in [itex]\mathbb{C}[/itex] forms a (finite-dimensional) vector space, [itex]M_{n} ( \mathbb{C} )[/itex], of dimension [itex]n^{2}[/itex]. The trace operation is a functional over [itex]M_{n} ( \mathbb{C} )[/itex] which satisfies:
    i) Linearity: for all [itex](A , B , C) \in M_{n} ( \mathbb{C} )[/itex] and [itex]( \alpha , \beta ) \in \mathbb{C}[/itex], [tex]\mbox{Tr} ( \alpha A + \beta B ) = \alpha \mbox{Tr} A + \beta \mbox{Tr} B ,[/tex]
    ii) Cyclicity: [tex]\mbox{Tr} (A B C ) = \mbox{Tr} ( C A B ) = \mbox{Tr} ( B C A ) .[/tex] Now, let [itex]A_{1}[/itex] and [itex]A_{2}[/itex] be any matrices such that [itex]A_{1} A_{2} \in M_{n} ( \mathbb{C} )[/itex] and [itex]A_{2} A_{1} \in M_{n} ( \mathbb{C} )[/itex], then (i) and (ii) imply that [tex]\mbox{Tr} ( A_{1} A_{2} - A_{2} A_{1} ) = \mbox{Tr} ( A_{1} A_{2} ) - \mbox{Tr} ( A_{2} A_{1} ) = 0 .[/tex] This means that the equation [tex]A B - B A = \mathbb{I}_{n} ,[/tex] has no solution: [tex]\mbox{Tr} ([ A , B ]) = 0 < \mbox{Tr} ( \mathbb{I}_{n} ) = n .[/tex] Now, let us consider infinite-dimensional matrices, i.e. matrices with infinite but countable number of rows and columns. Let us also assume that multiplication of such matrices is sensible, i.e. for any two such matrices [itex](A , B)[/itex], the series [itex](A B)_{i j} = \sum_{k}^{\infty} A_{i k} B_{k j}[/itex] and [itex](B A)_{i j} = \sum_{k}^{\infty} B_{i k} A_{k j}[/itex] converge for all [itex](i,j) \in \mathbb{Z}[/itex]. In this sense we can speak of the infinite-dimensional space [itex]M_{\infty} ( \mathbb{C} )[/itex]. Let us now ask similar question as in the finite-dimensional case, namely, is the equation [tex]A B - B A = \mathbb{I}_{\infty} , \ \ \ \ \ \ (1)[/tex] solvable in [itex]M_{\infty} (\mathbb{C})[/itex]? The answer is yes. As you can easily check, the following (infinite-dimensional) matrices do satisfy Eq(1):
    [tex]
    A = \sqrt{n} \ \delta_{m , n - 1} = \begin{pmatrix}
    0 & \sqrt{1} & 0 & 0 & \cdots \\
    0 & 0 & \sqrt{2} & 0 & \cdots \\
    0 & 0 & 0 & \sqrt{3} & \cdots \\
    \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} ,
    [/tex]
    [tex]
    B = A^{\dagger} = \sqrt{n + 1} \ \delta_{m , n + 1} = \begin{pmatrix}
    0 & 0 & 0 & \cdots \\
    \sqrt{1} & 0 & 0 & \cdots \\
    0 & \sqrt{2} & 0 & \cdots \\
    0 & 0 & \sqrt{3} & \cdots \\
    \vdots & \vdots & \vdots & \ddots \end{pmatrix} .
    [/tex] These, as you might know, are infinite-dimensional matrix representation of the operator algebra [itex][ a , a^{\dagger} ] = 1[/itex] of the SHO in the energy eigen-states. Furthermore, you can also check that the following Hermitian combinations [tex]X = \frac{1}{\sqrt{2}} ( A^{\dagger} + A ) , \ \ \ P = \frac{i}{\sqrt{2}} ( A^{\dagger} - A ) ,[/tex] satisfy [tex][X , P] = i [A , A^{\dagger}] = i \mathbb{I}_{\infty} .[/tex] These are the well-known matrix representation of position and momentum operators in the energy-eigen states of the simple harmonic oscillator.
    The definition of the trace operation in [itex]M_{\infty} ( \mathbb{C} )[/itex] is a bit technical. However, if the infinite sum of “diagonal elements” converge, we can define the trace by [itex]\mbox{Tr} ( M ) = \sum^{\infty} M_{i i}[/itex]. Now, we can make the following observation: the matrix [itex]A^{\dagger} A[/itex], which is in [itex]M_{\infty} ( \mathbb{C} )[/itex], is non-negative and Hermitian. Therefore, using the expressions above for [itex](A , A^{\dagger})[/itex] and our definition for the trace, we find (according to Euler) [tex]\mbox{Tr} ( A^{\dagger} A) = \mbox{Tr} ( A A^{\dagger} ) = \sum_{k = 1}^{\infty} k = - \frac{1}{12}.[/tex] Thus, on one hand [tex]\mbox{Tr} ( A A^{\dagger} ) - \mbox{Tr} ( A^{\dagger} A ) = 0 ,[/tex] on the other hand [tex]\mbox{Tr} ( A A^{\dagger} - A^{\dagger} A ) = \mbox{Tr} ( \mathbb{I}_{\infty}) = 1 + 1 + 1 + \cdots .[/tex] Since everybody (including Euler) agree that the infinite sum [itex]( 1 + 1 + 1 + \cdots )[/itex] diverges, we conclude that [tex]\mbox{Tr} ( A A^{\dagger} - A^{\dagger} A ) \neq \mbox{Tr}( A A^{\dagger} ) - \mbox{Tr}( A^{\dagger} A ) .[/tex] This means that “our definition of the trace” can not be linear functional on [itex]M_{\infty}( \mathbb{C} )[/itex], which seems very bizarre.

    Sam
     
    Last edited: Apr 4, 2015
  16. Apr 3, 2015 #15

    dyn

    User Avatar

    Thanks for confirming me correct. As for infinite dim spaces I give up !
     
  17. Apr 4, 2015 #16

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Of course, you should explain, how you come to the formula
    $$\sum_{k=1}^{\infty} k=-\frac{1}{12}.$$
    As it stands, it's of course wrong. The usual definition of the limit of a series is the limit of the sequence of partial sums, and that give ##\infty##.
     
  18. Apr 4, 2015 #17
    Do not. That's the meaningful bit.
     
  19. Apr 4, 2015 #18
    Hopefully Sam will answer but in the meantime this is my understanding.
    You have to use analytic continuation, that is, impose dependence on Riemann zeta function, in this particular case for s=1, and the particular series 1+2+3+.... , in other bases you get other convergences. This comes at the cost of losing basis-independence of the trace i.e. incompatibility with linearity that Sam found bizarre above.
     
  20. Apr 4, 2015 #19

    dyn

    User Avatar

    Thanks. I won't give up but at this moment in time I know nothing about Riemmann zeta functions and the like so I will have to come back to it when I know more but even some of you advanced guys seem to have different interpretations of it.
     
  21. Apr 4, 2015 #20
    Actually that was meant to be -1, 1 is the pole of the function.

    I'm sure the advanced guys(I'm certainly not included in that group) will agree on the math.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Commutators and traces
  1. Trace of a matrix (Replies: 9)

Loading...