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Adjoint of an operator for particles

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Let x be the position coordinate for a particle that moves in one dimension and let [tex]p=-i\hbar*d/dx[/tex] be the usual momentum operator . State whether each of the following operators is self-adjoint, anti-self-adjoint([tex]A^{\dagger}
    =A[/tex] , unitary([tex] A^{\dagger}A=1[/tex], or, if none of the above , what the adjoint is:

    xxp, xpx, xpp+ppx, d^2/dx^2 , d^3/dx^3,e^p
    2. Relevant equations



    3. The attempt at a solution

    I will perform the first operator xxp. [tex]p=-i\hbar*d/dx[/tex], therefore xxp=x^2*[tex]p=-i\hbar*d/dx[/tex]=2*x*[tex]-\hbar*i[/tex]? So would this operator be a self-adjointed one ?
     
  2. jcsd
  3. Sep 14, 2009 #2
    That depends on whether it equals the adjoint. Compute the operator and its adjoint and see if they are equal or not. Remember that [itex](AB)^\dagger = B^\dagger A^\dagger[/itex].
    This is chaotic and probably false. Why would you state that [itex]xxp = -i \hbar \frac{\textrm{d}}{\textrm{d}x} [/itex], when p is [itex]-i \hbar \frac{\textrm{d}}{\textrm{d}x}[/itex]?
     
  4. Sep 14, 2009 #3

    gabbagabbahey

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    First, this is incredibly sloppy notation. Usually one denotes operators with uppercase letters, and an operator's expansion into a certain basis (the x-basis in this case[/itex] is denoted with the symbol [itex]\to[/itex] instead of an equal sign. So, [itex]P\to -i\hbar \frac{d}{dx}[/itex] and [itex]X\to x[/itex] in the x-basis.

    Second, the order of multiplication between operators is important (sound familiar?).

    [tex]XXP\to x^2\left(-i\hbar \frac{d}{dx}\right)\neq \left(-i\hbar \frac{d}{dx}\right)x^2[/tex]

    To determine the nature of its adjoint, you might consider actually computing the adjoint!:wink:...In the x-basis, you have:

    [tex](XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}[/tex]
     
  5. Sep 14, 2009 #4
    would I need to prove(or disprove) that [tex] (xxp)^{\dagger}=(xxp),xxp*(xxp)^{\dagger}=1, , and (xxp)^{\dagger}=-(xxp)[/tex]
     
  6. Sep 14, 2009 #5

    gabbagabbahey

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    Well, it turns out that [itex]XXP[/itex] is anti-self-adjoint....to show that, just calculate [tex]\left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}[/tex] and compare it to the x-basis representation of [itex]XXP[/itex]...
     
  7. Sep 14, 2009 #6

    [tex]xxp^{\dagger}=(-i\hbar*d/dx)^{\dagger}*x^{\dagger}x^{\dagger}=xxp, since x=x^{/dagger}, p=p^{/dagger} [/tex]; If PF doesn't display my Latex code properly, Here is the code I was orginally trying to translate through Latex xxp^{\dagger}=(-i\hbar*d/dx)^{\dagger}*x^{\dagger}x^{\dagger}=xxp, since x=x^{/dagger}, p=p^{/dagger}
     
  8. Sep 14, 2009 #7

    gabbagabbahey

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    [itex]P\neq P^{\dagger}[/itex]...why would you think that they were equal?

    And again, order is important (drill that into your head):

    [tex](XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}=x^{\dagger}x^{\dagger}\left(-i\hbar \frac{d}{dx}\right)^{\dagger}\neq\left(-i\hbar \frac{d}{dx}\right)^{\dagger}x^{\dagger}x^{\dagger}[/tex]

    The [itex]\dagger[/itex] symbol tells you to take the complex conjugate transpose of each quantity...so do just that! The quantities [itex]x[/itex] and [itex]-i\hbar\frac{d}{dx}[/itex] are scalars, so taking their transpose is trivial....but what about their complex conjugates?
     
  9. Sep 14, 2009 #8
    But according to my textbook[tex](AB)^{\dagger} =(B)^{\dagger}A^{\dagger}[/tex] so therefore [tex](xxp)^{\dagger}=p^{\dagger}x^{\dagger}x^{\dagger}[/tex] so I don't understand why my expression would be in correct and don't understand why you think your expression is correct
     
  10. Sep 14, 2009 #9

    gabbagabbahey

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    Yes, sorry , my mistake.

    [tex](XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}=\left(-i\hbar \frac{d}{dx}\right)^{\dagger}x^{\dagger}x^{\dagger}[/tex]

    but my other two comments still stand.
     
  11. Sep 14, 2009 #10
    I am kinda of confused because my book(Peebles) gives two definitions for a self-adjoint operator. [tex]Q^{\dagger}=Q[/tex] and the other definition is [tex]c^{\dagger}=c*[/tex] where c* represent the adjoint of ca complex conjugate. For the latter case, the x matrices will not change since they are real numbers and contain no imaginary terms right?
     
  12. Sep 14, 2009 #11
    An operator adjoint to A in a Hilbert space with the product [itex]\langle \ldots;\ldots \rangle[/itex] is an operator B such that for all vectors [itex]\varphi,\psi[/itex]:

    [itex]\langle \varphi; A \psi \rangle = \langle B\varphi; \psi \rangle[/itex]

    (You can try writing this out in x-representation.)
     
  13. Sep 14, 2009 #12
    [tex](XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}=\left(-i\hbar \frac{d}{dx}\right)^{\dagger}x^{\dagger}x^{\dagger }=\left(i\hbar\frac{d}{dx}\right)x*x=2*i\hbar*x[/tex] since by definition, [tex]c=c*[/tex]
     
  14. Sep 14, 2009 #13

    gabbagabbahey

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    [itex]Q^{\dagger}=Q[/itex] is the definition of a self-adjoint operator.

    [itex]c^{\dagger}=c^{*}[/itex] just tells you how to take the adjoint of a complex scalar [itex]c[/itex]. (The fact that [itex]c[/itex] is a scalar tells you [itex]c^{T}=c[/itex], so you don't need to worry about the transpose, just the complex conjugation). [itex]x[/itex] is such a scalar,and is also real valued, so [itex]x^{\dagger}=x^{*}=x[/itex]

    Right, [itex](XXP)^{\dagger}=2i\hbar x[/itex]...does that equal [itex]XXP[/itex]? does it equal [itex]-XXP[/itex]? Does the product [itex](XXP)(XXP)^{\dagger}[/itex] equal the identity operator?
     
  15. Sep 14, 2009 #14
    ?
    no.


    [tex]XXP(XXP)^{\dagger}=X^2*(-i\hbar*d/dx)2i\hbar x\neq\leftI[/tex] , i.e. not equal to the identity.
     
    Last edited: Sep 14, 2009
  16. Sep 14, 2009 #15

    gabbagabbahey

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    Right, so [itex]XXP[/itex] is neither self-adjoint, anti-self-adjoint or unitary.

    Now move on to the rest of the operators in your question...
     
  17. Sep 15, 2009 #16
    [tex](xpp+ppx)^{\dagger}=(xpp)^{\dagger}+{ppx}^{\dagger}=p^{\dagger}p^{\dagger}x^{\dagger}+x^{\dagger}p^{\dagger}p^{\dagger}[/tex]. [tex]p^{\dagger}=p*,x^{\dagger}=x. (i*\hbar*d/dx)(-i*\hbar)+x(\hbar)^2*d^2/dx^2=0+0[/tex] Therefore, there it is neither

    [tex](xpx)^{\dagger}=x^{\dagger}p^{\dagger}x^{\dagger). xpx=x(-i*\hbar*d/dx)x=-xi\hbar.(xpx)^{\dagger}=x(i*\hbar*d/dx)x=xi\hbar[/tex]. Therefore, operator is anti-self-adjoint.

    [tex]e^p=e^(-i*\hbar*d/dx). (e^p)^{\dagger)=e^(i*\hbar*d/dx). (e^p)(e^p)^{\dagger}=e^0=1[/tex]. Therefore, [tex]e^p[/tex] is unitary. not sure how to prove how [tex] d^2/dx^2[\tex] and [tex] d^3/dx^3 [/tex] are adjoint, anti-self-adjoint or unitary.
     
  18. Sep 15, 2009 #17

    gabbagabbahey

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    [itex](xpp+ppx)^{\dagger}=(xpp)^{\dagger}+(ppx)^{\dagger}=p^{\dagger}p^{\dagger}x^{\dagger}+x^{\dagger}p^{\dagger}p^{\dagger}[/itex] is correct, but I'm not sure what you're doing after that. (Your [itex]\LaTeX[/itex] is very messy!)

    Just use [itex]x^{\dagger}=x[/itex] (since [itex]x[/itex] is a real valued scalar) and [itex]p^{\dagger}=-p[/itex] (Since [itex]\left[-i\hbar\frac{d}{dx}\right]^{\dagger}=i\hbar\frac{d}{dx}[/itex] )

    That gives you,

    [tex](xpp+ppx)^{\dagger}=p^{\dagger}p^{\dagger}x^{\dagger}+x^{\dagger}p^{\dagger}p^{\dagger}=(-p)(-p)x+x(-p)(-p)=ppx+xpp[/tex]
     
  19. Sep 15, 2009 #18
    [tex]
    e^p=e^(p). (e^p)^{\dagger)=1? since p^{dagger}=-p, then (e^p)(e^-p)^{\dagger}=e^0=1
    [/tex] Therefore , [tex]e^p[/tex] is unitary
    Don't know how o show whether or not [tex] d^2/dx^2[/tex] or [tex] d^3/dx^3[/tex] is adjoint, self-adjoint ,unitary , or neither
     
  20. Sep 15, 2009 #19

    gabbagabbahey

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    Let's see,

    [tex](e^p)^{\dagger}=\left(\sum_{n=0}^{\infty}\frac{p^n}{n!}\right)^{\dagger}=\sum_{n=0}^{\infty}\frac{(p^n)^{\dagger}}{n!}=\sum_{n=0}^{\infty}\frac{(p^{\dagger})^n}{n!}=\sum_{n=0}^{\infty}\frac{(-p)^n}{n!}=e^{-p}[/tex]

    So yes, I would say [itex]e^{p}[/itex] is unitary.

    I'd start by defining the operator [itex]D\leftrightarrow\frac{d}{dx}[/itex], then [itex]\frac{d^2}{dx^2}\leftrightarrow D^2=DD[/itex], then just apply the same reasoning as before.
     
    Last edited: Sep 15, 2009
  21. Sep 15, 2009 #20
    i.e. show that [tex]D^{\dagger}=D[/tex] and [tex](D^2)^{\dagger}=D^{\dagger}D^{\dagger}[/tex]?
     
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