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Exponential operators

  1. Dec 31, 2015 #1
    Why is (5.302) an approximation instead of an equality?

    Let ##T## be the operator ##\frac{p_x^2}{2m}##.

    By the law of indices, we should have ##e^{-\frac{i}{\hbar}(T+V)\Delta t}=e^{-\frac{i}{\hbar}T\Delta t} e^{-\frac{i}{\hbar}V\Delta t}## exactly. Is it because ##T## and ##V## do not commute? So the correct equation should be ##e^{-\frac{i}{\hbar}(T+V)\Delta t}=\frac{1}{2}[e^{-\frac{i}{\hbar}T\Delta t} e^{-\frac{i}{\hbar}V\Delta t}+e^{-\frac{i}{\hbar}V\Delta t} e^{-\frac{i}{\hbar}T\Delta t}]##?

    But if this is so, shouldn't (5.302) be correct up to terms of order ##\Delta t## instead of order ##(\Delta t)^2## as claimed by the book?

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    Last edited: Dec 31, 2015
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  3. Dec 31, 2015 #2

    atyy

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  4. Dec 31, 2015 #3

    strangerep

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    Yes.

    Not in general. Check out the Baker-Campbell-Hausdorff formula(s).
     
  5. Dec 31, 2015 #4
    When I work out the term of order ##(\Delta t)^2##, the left hand side gives ##\frac{1}{2}(-\frac{i}{\hbar})^2(T^2+TV+VT+V^2)(\Delta t)^2## but the right hand side gives ##\frac{1}{2}(-\frac{i}{\hbar})^2(T^2+2TV+V^2)(\Delta t)^2##. They are not equal. So is the book wrong?
     
  6. Jan 1, 2016 #5

    strangerep

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    Which book is it?
     
  7. Jan 1, 2016 #6
    It's Quantum Mechanics 2nd ed. by Bransden & Joachain, page 243.
     
  8. Jan 1, 2016 #7

    samalkhaiat

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    To 2nd order, the BCH identity reads
    [tex]e^{\lambda (T+V)/N} = e^{\lambda T/N}\ e^{\lambda V/N} - \frac{1}{2}(\frac{\lambda}{N})^{2} [ T , V ][/tex]
    However, one can prove that, as [itex]N \to \infty[/itex],
    [tex]\left( e^{\lambda (T+V)/N} \right)^{N} = \left( e^{\lambda T/N}\ e^{\lambda V/N} \right)^{N}[/tex]
    This helps you to rewrite the Green's function as
    [tex]\langle x | e^{\lambda H} | y \rangle = \langle x | ( e^{\lambda H /N} )^{N}| y \rangle = \langle x | ( e^{\lambda T /N} e^{\lambda V /N} )^{N}| y \rangle ,[/tex]
    which leads to the Path Integral when you insert the identity operators
    [tex]\int dx_{j} |x_{j}\rangle \langle x_{j}| = I, \ \ \ \int dp |p \rangle \langle p| = I.[/tex]
     
  9. Jan 1, 2016 #8

    atyy

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    I think you are right and the book is wrong.
     
  10. Jan 1, 2016 #9
    But this equation can be simplified to one independent of ##N##:

    [tex] e^{\lambda (T+V)} = e^{\lambda T}\ e^{\lambda V} [/tex]

    So if it is true as [itex]N \to \infty[/itex], then shouldn't it be true for all values of ##N##? But this is not true, so the equation should not be true for any value of ##N##?
     
  11. Jan 1, 2016 #10

    blue_leaf77

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    If ##T## and ##V## do not commute, you can't interchange ##e^{\lambda T}## and ##\ e^{\lambda V}##.
     
  12. Jan 1, 2016 #11
    So that means
    [tex]\left( e^{\lambda (T+V)/N} \right)^{N} = \left( e^{\lambda T/N}\ e^{\lambda V/N} \right)^{N}[/tex]
    is false even as [itex]N \to \infty[/itex]?

    I think the correct statement should instead be as [itex]N \to \infty[/itex],

    [tex]e^{\lambda (T+V)/N} = e^{\lambda T/N}\ e^{\lambda V/N} [/tex]

    Is this correct?
     
    Last edited: Jan 1, 2016
  13. Jan 1, 2016 #12

    blue_leaf77

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    What I implied is that, I don't think you can reduce the RHS of ##\left( e^{\lambda (T+V)/N} \right)^{N} = \left( e^{\lambda T/N}\ e^{\lambda V/N} \right)^{N}## to ##\left( e^{\lambda T}\ e^{\lambda V} \right)## because in doing so, you must have brought ##N## in the outer most to inside, which means you are interchanging some pairs of the two exponential operators in the process.
     
  14. Jan 1, 2016 #13
    Since (https://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula)
    Screen Shot 2016-01-02 at 3.12.33 am.png
    then to the second order, ##\ e^Xe^Y=e^{X+Y+\frac{1}{2}[ X , Y ]}##.

    So shouldn't it be

    [itex]e^{\lambda (T+V)/N} = e^{\lambda T/N}\ e^{\lambda V/N}\ e^{\frac{1}{2}(\frac{\lambda}{N})^{2} [ T , V ]}[/itex]?
     
  15. Jan 1, 2016 #14

    strangerep

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    @Happiness,

    Sam has (almost) given you the crucial missing piece in post #7. I'm reasonably sure Sam meant "in the limit as ##N\to\infty##". In general, some properties might be true only in the sense of limits, though untrue for finite values of ##N##.

    I should have also given you the Wiki link to the Lie-Trotter product formula. I.e., $$e^{A+B} ~=~ \lim_{N\to\infty} \Big( e^{A/N} e^{B/N} \Big)^N ~.$$The way your textbook expresses it seems a bit misleading, imho, but the proper Lie-Trotter formula can be applied more successfully in the path integral material.
     
  16. Jan 2, 2016 #15

    samalkhaiat

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    No. You can write
    [tex]e^{H} = (e^{H/N})^{N} ,[/tex]
    But
    [tex](e^{T/N}e^{V/N})^{N} = (e^{T/N}e^{V/N}) \cdots N \mbox{-factors} \cdots (e^{T/N}e^{V/N}) .[/tex]
     
  17. Jan 2, 2016 #16

    samalkhaiat

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    No. It is true and has a name. It is called the Lie-Trotter product formula. You should not jump to a wrong conclusion just because you can not prove it. Do you think I lied to you when I said “we can prove that”?

    Yes, it is correct because as I said
    [tex]e^{T/N}e^{V/N} - e^{(T+V)/N} = \frac{1}{2N^{2}} [ T , V ] \sim \mathcal{O}(\frac{1}{N^{2}}) . \ \ \ \ (1)[/tex]
    So,
    [tex]\lim_{N \to \infty} \left( e^{T/N}e^{V/N} - e^{(T+V)/N} \right) = 0 .[/tex]
    But this does not help you in setting up the Path integral representation of the propagator. In fact, we can use this result to prove the Lie-Trotter formula which is essential in deriving the path integral. Let me show you how. Let [itex]A[/itex] and [itex]B[/itex] be two operators, and consider the difference [itex]A^{N} - B^{N}[/itex]. By adding and subtracting equal terms, I can write the following identity
    [tex]
    \begin{equation*}
    \begin{split}
    A^{N} - B^{N} =& (AB^{N-1} - B^{N}) + (A^{2}B^{N-2} - AB^{N-1}) \\
    & + (A^{3}B^{N-3} - A^{2}B^{N-2}) + (A^{4}B^{N-4} - A^{3}B^{N-3}) \\
    & + \cdots + (A^{N} – A^{N-1}B) .
    \end{split}
    \end{equation*}
    [/tex]
    This can be rewritten as
    [tex]
    \begin{equation*}
    \begin{split}
    A^{N} - B^{N} =& (A - B ) B^{N-1} + A (A - B ) B^{N-2} \\
    & + A^{2} (A - B ) B^{N-3} + A^{3} (A - B ) B^{N-4} \\
    & + \cdots + A^{N-1} ( A - B ) .
    \end{split}
    \end{equation*}
    [/tex]
    Okay, now take
    [tex]A = e^{T/N}e^{V/N} , \ \ B = e^{(T+V)/N} .[/tex]
    But, we know that
    [tex]A - B = \frac{1}{2N^{2}} [ T , V ] \sim \mathcal{O}(N^{-2}) .[/tex]
    So, [itex](A-B) \to 0[/itex] as [itex]N \to \infty[/itex]. Now, the above identity consists of [itex]N[/itex] terms each of which has the factor [itex](A-B)[/itex] which is of order [itex](1/N^{2})[/itex]. Hence, in the limit [itex]N \to \infty[/itex], the difference [itex]A^{N} - B^{N}[/itex] is zero. Thus, we obtain the Lie-Trotter formula
    [tex]\lim_{N \to \infty} \left( e^{T/N}e^{V/N} \right)^{N} = \lim_{N \to \infty} \left( e^{(T+V)/N} \right)^{N} = e^{(T+V)} .[/tex]
     
  18. Jan 2, 2016 #17

    samalkhaiat

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    This equation is correct to all orders provided that
    [tex][ X , [ X , Y ] ] = [ Y , [ X , Y ] ] = 0 .[/tex]
    We speak of “orders” when we expand the exponentials.



    No. Again, if [itex][ T , V ][/itex] commutes with both [itex]T[/itex] and [itex]V[/itex], you can write
    [tex]e^{\lambda (T + V)/N } = e^{\lambda T/N} e^{\lambda V/N} e^{- \frac{1}{2}(\frac{\lambda}{N})^{2} [ T , V ] } .[/tex]
    Notice the minus sign on the right hand side.
    But, when I wrote
    [tex]e^{X+Y} = e^{X}e^{Y} - \frac{1}{2} [ X , Y ], \ \ \ \ (2)[/tex]
    I did not assume that [itex][ X , Y ][/itex] commutes with both [itex]X[/itex] and [itex]Y[/itex]. Equation (2) is an identity up to quadratic terms, i.e., when you expand and keep only the quadratic terms [itex]X^{2},Y^{2},XY[/itex] and [itex]YX[/itex].
     
  19. Jan 2, 2016 #18
    Yes, I realized my mistake when @blue_leaf77 pointed it out earlier in post #12, sorry!
     
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