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Quantum Harmonic oscillator

  1. Jan 30, 2016 #1
    upload_2016-1-31_3-3-25.png upload_2016-1-31_3-33-26.png upload_2016-1-31_3-3-43.png upload_2016-1-31_3-4-23.png
    When I work out $$b^+b$$, I get

    $$\widehat{b^+} \widehat{b} = \frac{1}{2} (ξ - \frac{d}{dξ})(ξ + \frac{d}{dξ}) = \frac{1}{2} (ξ^2 - \frac{d^2}{dξ^2}) = \frac{mωπx^2}{h} - \frac{h}{4mωπ} \frac{d^2}{dx^2}$$

    So base on what I have about, (9) should be
    $$(9) = \frac{hω}{2π} (\frac{1}{2} \widehat{b^+} \widehat{b} + \frac{1}{2} ξ^2)$$
    instead of (10).

    Am I missing out something?
     

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    Last edited: Jan 30, 2016
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  3. Jan 30, 2016 #2

    blue_leaf77

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    ##\frac{1}{2}(ξ-\frac{d}{dξ})(ξ+\frac{d}{dξ})=\frac{1}{2}(ξ^2-\frac{d^2}{dξ^2})##
    The above step is not correct. Instead, it should be
    $$
    \frac{1}{2}(\xi-\frac{d}{d\xi})(\xi+\frac{d}{d\xi}) = \frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2}).
    $$
    Then simplify the second and third terms.
     
  4. Jan 30, 2016 #3
    Based on yours,
    $$
    \frac{1}{2}(\xi-\frac{d}{d\xi})(\xi+\frac{d}{d\xi}) = \frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2}) = \frac{mωπx^2}{h} - \frac{h}{4mωπ} \frac{d^2}{dx^2} -\frac{1}{2} + x\frac{1}{2} \frac{d}{dx}
    $$
    This is where i am stuck again as to what do I do with $$x\frac{1}{2} \frac{d}{dx}$$ and this one( $$\frac{h}{4mωπ} \frac{d^2}{dx^2}$$) has the term 4 instead of 8 as shown in equation (9).
     
  5. Jan 30, 2016 #4

    blue_leaf77

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    That's still incorrect.
    The term ##\frac{d}{d\xi}\xi## should expand into two terms, what are they?
     
  6. Jan 30, 2016 #5
    ##\frac{d}{d\xi}\xi = \frac{d}{d\xi}(\xi * 1) = 1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}##
    Is this correct?
     
  7. Jan 30, 2016 #6

    blue_leaf77

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    Yes, that one is correct.
     
  8. Jan 30, 2016 #7
    What about this one( $$\frac{h}{4mωπ} \frac{d^2}{dx^2}$$) which is supposed to be in equation (9)?

    The only way I can get to $$\frac{h^2}{8mωπ^2} \frac{d^2}{dx^2}$$ is by multiplying $$\frac{h}{4mωπ} \frac{d^2}{dx^2}$$ with $$\frac{h}{2π}$$, which I don't know where to get from.






    How I get $$\frac{h}{4mωπ} \frac{d^2}{dx^2}$$,
    First, $$dξ = \sqrt[] {\frac{2mπω}{h}} * dx$$
    Squaring it, $$dξ^2 = \frac{2mπω}{h} * dx^2$$
    Subbing it into $$ \frac{d^2}{dξ^2} $$ from b+*b, I get $$\frac{h}{2mωπ} \frac{d^2}{dx^2}$$ before multiplying 0.5 from the bracket outside $$\frac{1}{2}(\xi^2 +\xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi - \frac{d^2}{d\xi^2})=b^+*b$$.
     
    Last edited: Jan 30, 2016
  9. Jan 31, 2016 #8

    blue_leaf77

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    The coefficient in front of the second order derivative in the above equation is not correct, as you can check its dimensionality. It should be
    $$
    \frac{h}{4m\omega \pi}
    $$
    which is the result you obtained from your own calculation. This is the correct one as you can check by multiplying it with the constant outside the square bracket and it should lead to ##p^2/(2m)##.
     
  10. Jan 31, 2016 #9
    Thanks a lot!
     
  11. Jan 31, 2016 #10
    Sorry to trouble you again.
    Talking about the second and third terms, $$ \xi\frac{d}{d\xi} - \frac{d}{d\xi}\xi = \xi\frac{d}{d\xi} - (1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}) $$
    Initially I thought I can cancel out the $$\xi\frac{d}{d\xi}$$ by subtracting it with $$\xi * \frac{d1}{d\xi}$$, but in my another thread, you told me that
    $$∅*(\frac{d}{dξ})\neq \frac{d∅}{dξ}$$, thats why I am stuck at this part.
     
  12. Jan 31, 2016 #11

    blue_leaf77

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    I should probably have corrected your post #5, instead of equal sign it might have been more appropriate to use arrow instead:
    $$
    \frac{d}{d\xi}\xi \rightarrow \frac{d}{d\xi}(\xi * 1) = 1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}.
    $$
    with the arrow sign is intended to mean that the expression to the left is to be operated on a function, in the above case a number ##1##. The insertion of a ##1## there is supposed to be thought as an arbitrary object because remember that Hamiltonian is actually an operator, it's not yet a physical quantity. Therefore, writing ##H## only with nothing following it does not have mathematical significance. Only when you make it operate on an arbitrary state ##\psi## will the expression ##H\psi## be meaningful.
    In the above equation, if you make it as if it operates on a ##1## (or an arbitrary state ##\psi##), it should become
    $$
    \xi\frac{d}{d\xi}1 - (1 * \frac{d\xi}{d\xi} + \xi * \frac{d1}{d\xi}).
    $$
     
  13. Jan 31, 2016 #12
    Thanks a lot again! One last thing if you wouldn't mind,
    After converting equation (9) to equation (10),
    $$\hat{b}*∅_0 = 0$$
    Thus, $$\hat{b}^+\hat{b}*∅_0 = 0$$,
    So equation (10) left with $$\frac{hω}{2π}(0+\frac{1}{2})∅_0=E_0∅_0$$
    Shouldn't $$E_0=\frac{hω}{4π}$$ instead of $$E_0=\frac{hω}{2π}$$?
     
  14. Jan 31, 2016 #13

    blue_leaf77

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    Doesn't the equation ##\frac{hω}{2π}(0+\frac{1}{2})∅_0=E_0∅_0## correctly imply that ##E_0=\frac{hω}{4π}##?
     
  15. Jan 31, 2016 #14
    Yes, but from the source ##E_0=\frac{hω}{2π}##.
    Source: http://vixra.org/pdf/1307.0007v1.pdf
     
  16. Jan 31, 2016 #15

    blue_leaf77

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    I think you should look for another source to study this subject, the author seems to be rather sloppy with his derivation.
     
  17. Jan 31, 2016 #16
    Ok thanks!
     
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