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Prove that any Hermitian operator is linear

  1. Oct 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Simply--Prove that any Hermitian operator is linear


    2. Relevant equations
    Hermitian operator defined by: int(f(x)*A*g(x)dx)=int(g(x)*A*f(x)dx)

    Linear operator defined by: A[f(x)+g(x)]=Af(x)+Ag(x)

    Where A is an operator


    3. The attempt at a solution

    I am at a complete loss of how to even start this. I know I need to get from int(f(x)*A*g(x)dx)=int(g(x)*A*f(x)dx) to A[f(x)+g(x)]=Af(x)+Ag(x) where A is the Hermitian operator, but I dont know how/what to manipulate to get here....
     
    Last edited: Oct 24, 2008
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  3. Oct 24, 2008 #2

    morphism

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    For future reference, it's helpful if you actually post the definitions you're given. Usually the word "operator" means a linear operator, so a Hermitian operator would be linear by definition. From what you posted, I take it when you say A is Hermitian, then you mean A is defined on some space of functions, and [itex]\int f \overline{(Ag)} = \int (Af) \overline{g}[/itex], where the bar denotes complex conjugation and the integral presumably will always make sense.

    Anyway, define <f,g> to be [itex]\int f\overline{g}[/itex]. So we have that <Af,g>=<f,Ag>. Consider

    [tex]\langle A(f+\lambda g) - (Af + \lambda Ag), A(f+\lambda g) - (Af + \lambda Ag) \rangle.[/tex]
     
  4. Oct 24, 2008 #3
    Thank you for the response and sorry I did not explicity post the definitions.

    I understand that you defined A to be hermitian as demonstrated by your first set where <Af,g>=<f,Ag>, but I have no idea what the second part means. What is lambda? Why is the same thing on both sides and why did you choose to subtract (Af + \lambda Ag)?
     
  5. Oct 24, 2008 #4

    morphism

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    Lambda is just some scalar. We want to prove that A(f + cg) = Af+cAg, or written differently, A(f+cg) - (Af+cAg) = 0, for all functions f and g and all scalars c.

    If you're familiar with inner/dot products you'll know why I did what I did: if <v,v>=0 then v=0.
     
  6. Oct 24, 2008 #5
    My class isnt using dot products or anything like that to do anything. We have just been mired in calculus with little to no instruction on how to use it.

    Why do you want to involve a scalar? Is it possible to show this without using dot products/other 3D tools?
     
  7. Oct 24, 2008 #6

    Fredrik

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    I didn't read the whole thing, but it looks like lambda is just an arbitrary complex number. When you see someone put the same thing on both sides of the scalar product, it's usually because they want to use that [itex]\langle f|f\rangle \geq 0[/itex] for arbtrary non-zero f.

    Edit: Much too slow :frown:
     
  8. Oct 24, 2008 #7

    morphism

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    We want to involve a scalar because we want to prove that A is linear - remember, this means that A(cg) = cAg for all scalars c, as well as A(f+g)=Af+Ag.

    And you don't really need any 3D tools - I was just using them to guide my intuition.

    For now let's ignore the scalar and focus on proving that A(f+g)-(Af+Ag)=0.

    If we can prove that

    [tex]\int (A(f+g)-(Af+Ag)) \overline{(A(f+g)-(Af+Ag))} = \int |A(f+g)-(Af+Ag)|^2 = 0[/itex]

    then this will do the job (why?). To accomplish this, expand the left-hand side, and use the fact that A is Hermitian.
     
  9. Oct 24, 2008 #8
    I guess what I mean is that seems really arbitrary to pick what you did and I do not understand why/how you would come to pick that since my class has solely been working with integrals etc and hasnt even touched vectors/dot products.

    Edit--I am sorry but my level of understanding is really low. How did you synthesize that integral?

    And I also apologize but with my level of math I am unsure with what to do to the left side. I know it looks like I am begging for an answer, but I literally am lost in this class since it assumes we know a level of math that wasnt required to take it. How does the complex conjugate effect how it expands it and how can that be simplified any further?
     
    Last edited: Oct 24, 2008
  10. Oct 24, 2008 #9

    morphism

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    Yes, I guess it does look like I'm pulling a rabbit out of a hat if you don't know what dot products are. It's strange that your class is discussing Hermitian operators without mentioning anything about dot products.
     
  11. Oct 24, 2008 #10

    Fredrik

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    This is wrong, or at least incomplete. It should be A[af+bg]=aAf+bAg, where a and b are arbitrary complex numbers and f and g are arbitrary functions. (I'm in a nitpicking mood, so I'll also point out that it's wrong to write f(x) instead of f here, since f(x) is a number and f is a function).

    I would write it as <f,Ag>=<A*f,g>, where <f,g> is defined as [itex]\int f(x)^*g(x)dx[/itex], but you can keep expressing it as an integral if you want to.
     
  12. Oct 24, 2008 #11
    Like I said, we havent addressed how this applies to space, only in paper/calculus form so I dont know any better. And I am pretty far removed from the realm of proper math expression (over 3 yrs since I took any math class) so forgive me for these technical errors.
     
  13. Oct 24, 2008 #12
    So does that mean that [tex]\int (A(f+g)-(Af+Ag)) \overline{(A(f+g)-(Af+Ag))} = \int |A(f+g)-(Af+Ag)|^2 = 0[/itex] becomes [tex]\int (A(f+g)-(Af+Ag)) \overline{(A(f+g)-(A(f+g)} = \int |A(f+g)-(Af+Ag)|^2 = 0[/itex]
    [tex]\int (A(f+g)-(Af+Ag)) \overline{(0))} = \int |A(f+g)-(Af+Ag)|^2 = 0[/itex]

    Since if A is linear it follows that Af+Ag=A(f+g)? Isnt this approaching the question backwards though, because now I assuming it is linear and showing it is Hermitian instead of the other way around.
     
  14. Oct 24, 2008 #13

    Fredrik

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    You're not supposed to use the linearity. You're supposed to show that the left hand side (or the right-hand side, since it's the same thing) is =0 without using the linearity. This would prove that A(f+g)=Af+Ag.

    The calculation is a bit awkward when we have to manipulate the integrals instead of abstract scalar products. I'm going to show you how I would prove the linearity using the abstract notation and the properties of a scalar product. Mostly because I'm really bored right now. This isn't a complete solution to your homework problem, since if you do it this way, you would have to prove that the "scalar product" defined using the integrals really is a scalar product, and do a few more things to have a complete proof.

    [tex]\langle f,A(ag+bh)\rangle =\langle A^*f,ag+bh\rangle =a\langle A^*f,g\rangle+b\langle A^*f,h\rangle[/tex]

    [tex]=a\langle f,Ag\rangle+b\langle f,Ah\rangle =\langle f,aAg+bAh\rangle[/tex]

    This proves the linearity if the following is true: If <x,y>=<x,z> for all x, then y=z. You can prove that as an excercise if you want. If you don't feel like doing it this way, don't worry about it. The method that morphism suggested is fine too.
     
  15. Oct 25, 2008 #14
    Hmm without using that linearity equation then I would look for something to expand or factor to see if I can get some sort of cancellation, but since we are working with operators that strategy doesnt work out. I think there might be some way to split the left side integral apart based on the Hermiticty of A, but I dont see how/where since A is always near both f and g and they cant seem to separate.
     
  16. Oct 25, 2008 #15

    Fredrik

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    The idea is to use the Hermiticity of the operator to rewrite the expression so that you can use the linearity of the integral (or scalar product). Example:

    [tex]\int\overline{A(f+g)}A(f+g) =\int\overline{A^*A(f+g)}(f+g) =\int\overline{A^*A(f+g)}f+\int\overline{A^*A(f+g)}g[/tex]

    [tex]=\int\overline{A(f+g)}Af+\int\overline{A(f+g)}Ag =\int\overline{A(f+g)}(Af+Ag)[/tex]

    Now rewrite this as

    [tex]=\overline{\int\overline{(Af+Ag)}A(f+g)}[/tex]

    and repeat the same procedure to get

    [tex]\cdots=\overline{\int\overline{(Af+Ag)}(Af+Ag)} =\int\overline{(Af+Ag)}(Af+Ag)[/tex]

    The calculation is pretty long. One of the reasons is that this method proves that <A(f+g),A(f+g)>=0 instead of separately proving that <h,A(f+g)>=<h,Af+Ag> for all h and that this implies that A(f+g)=Af+Ag. Morphism came up with a trick to do it all at once, but it makes the calculations longer. So you might want to stick with the method I used in #13 and just figure out how to express it using integrals.
     
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