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Hermitian Conjugates and Operators

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the following hermitian conjugates and show if they are hermitian operators:

    i) xp
    ii) [x , p]
    iii) xp + px

    Where x is the position operator and p is the momentum operator.

    2. Relevant equations

    <f|Qg> = <Q[itex]^{t}[/itex]f|g>
    Q = Q[itex]^{t}[/itex] Hermitian operator
    p = -ih(d/dx)

    3. The attempt at a solution

    For the first case I have applied the condition of hermitian operators where I get <f|-ihx(dg(x)/dx)> and then get the form <f|-ih(dxg(x)/dx)>which leads to the integral ∫f*(-ih(d/dx)(xg(x)dx but am not sure how you bring the operator to f*. Do you just do the product rule and bring the product or to f* or do you need to do integration by part?

    For the case of [x , p] you get [x, p] = -xih(d/dx) +ih(dx/dx) and find I apply this to the hermitian conjugate I get <-ihf|g> but was told something was wrong since I got it from <-xih(dg*/dx)+ih(g +x(dg/dx)>. For the third part I'm not getting anything that make sense when I apply -(xih(d/dx) + ih(dx/dx)) and got ∫f*(ih(g + 2x(dg/dx))dx but according to my notes xp + px should be hermitian.

    Thanks to anybody that helps.
     
  2. jcsd
  3. Apr 13, 2013 #2
    Sorry, but I think I'm confused on what you have to even do for the hermitian conjugate. Do you even need to apply the <f|Qg) = <Q*f|g> to show the hermitian conjugate or do you just need to do the complex conjugate of the operator and if it isn't hermtian it will be something different when you apply <f|Qg) = <Q*f|g>?

    For example, in the case of xp if I apply the complex conjugate I will get ihx(d/dx) instead of -ihx(d/dx) meaning it isn't a hermitian operator. When I applied the condition for hermitian conjugates and get <f|(-ihx(dg/dx)>, I then bring the brak-ket notation to the integral form ∫f*(-ihx(dg/dx)dx where -ihx can be brough to f in the form ∫(ihxf)*(dg/dx)dx where integration by parts is the process to bring the derivative to f. The problem I have here is with x since you then need to do the product rule for (ihxf)* where I get the result -∫(ihf - ihx(df/dx))gdx which can be changed into form -<ih(f - x(df/dx))|g> which is obviously not the same as the original form. Is this correct and the answer like that was produced under the brak-ket notation because xp isn't hermitian?
     
  4. Apr 13, 2013 #3

    Dick

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    Yes, xp isn't Hermitian. You use integration by parts to move the derivatives around and the x factor will block that. [x,p] should be easier, it's a constant but it's pure imaginary. For xp+px you can just deal on the abstract level with the operators. AB+BA is always hermitian no matter what the hermitian operators A and B are. Can you tell me why?
     
    Last edited: Apr 13, 2013
  5. Apr 14, 2013 #4
    Alright, I think I understand now. If I apply a function to [x, p] I then find that [x, p]f = ihf so [x, p] = ih. When I apply this to the hermitian condition I get negative sign produced when I bring the i to the other function so [x, p] isn't hermitian.

    For the case of xp +px if I consider it to be in form AB + BA, I can then use (AB)[itex]^{t}[/itex] = B[itex]^{t}[/itex]A[itex]^{t}[/itex] meaning that (AB + BA)[itex]^{t}[/itex] = B[itex]^{t}[/itex]A[itex]^{t}[/itex] + A[itex]^{t}[/itex]B[itex]^{t}[/itex] = BA + AB = BA +AB. This means that (xp + px)[itex]^{t}[/itex] = (xp + px) so xp + px is a hermitian operator.
     
  6. Apr 14, 2013 #5

    Dick

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    Yeah, that's it exactly.
     
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