Gauss formula of limit

Discussion in 'Physics & Math' started by neelakash, Nov 10, 2009.

  1. neelakash Registered Senior Member

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    The following limit result is well known:

    \(e^x = \lim_{n\to \infty} \left(1 + x/n \right) ^n\)

    Now,with matrix valued argument,the above equation can be written as

    \(\displaystyle\lim_{k\to\infty}\ [\ I_{\ n\times\ n}\ + \frac{\ i \vec{\phi}\ . \mathbb{D} }{k}] ^k\) \( = \ exp\ [ \ {\ i \ {\vec{\phi}\ . \mathbb{D} ] \)

    My question is will the relation still hold if \(\mathbb{D}=\mathbb{D}_{1}\ +\mathbb{D}_{2}\) and the two operators D1 and D2 do not commute...?

    I think it will(as for angular momentum matrices,which are generators of rotation),but how to see it?
     
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  3. temur man of no words Registered Senior Member

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  5. D H Some other guy Valued Senior Member

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    Matrix addition commutes, always. It is matrix multiplication that in general does not commute. However, given any real square matrix A and a non-negative integer n, A[sup]n[/sup]A=AA[sup]n[/sup]. In other words, A[sup]n[/sup] is unambiguous.
     
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  7. neelakash Registered Senior Member

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    My question was if the result will hold good for the case where D1 and D2 do not commute (in the sense of matrix multiplication).Say,[D1,D2]=cD3 where c is in general a complex number.

    In the proof of the Gaussformula, we binomially expnad the RHS and taking limit,show that the formula conforms to exp x.

    But when I am using matrix argument,I am not sure how to write the higher powers of i*phi*D...
     
  8. AlphaNumeric Fully ionized Registered Senior Member

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    Of course the formula holds if you just replace \(D_{1}\) with \(D_{1}+D_{2}\). It's like asking if a formula is still value if you replace x by y+z. Substituting \(D_{1}+D_{2}\) in the place of \(D_{1}\) is just a relabelling, you don't need to know anything about \(D_{1}\)'s relationship with \(D_{2}\) to do that.

    It's only when you then start manipulating that expression, such as wondering if \(e^{D_{1}+D_{2}} = e^{D_{1}}e^{D_{2}}\), which is only true if \([D_{1},D_{2}]=0\).
     
  9. neelakash Registered Senior Member

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    ya...that's what I found...I was confusing with something else....
     

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