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Expectation value of 1/(1+x)

  1. Apr 4, 2015 #1

    blue_leaf77

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    I have an inner product ## \langle \alpha|f| \beta \rangle## where ##f## is an operator that is a function of position ##x## operator (1D). According to the book I read (and I'm sure in any other book as well), that inner product can be written in position representation as ## \int u^*_{\alpha}(x) f(x) u_{\beta}(x) dx ##. This book doesn't discuss how to get this integral expression but I can guess how it's done, namely sandwiching the operator ##f## by two completeness relations of position eigenkets. In doing so we get ##f## sandwiched by a ket and a bra of ##x## and since ##f## is a function of ##x## operator, it can be expanded into power series of ##x##. This way we will have a dirac delta ##\delta (x'-x") ## which will simplify to the integral above. But my question is what if ##f## is, for example, ##1/(1+x) ## whose power series only converges between x=-1 and 1? What will happen with the integral of all space in the expectation value of ##f(x)##? Or is my argument expanding f into power series wrong?
     
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  3. Apr 4, 2015 #2

    mfb

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    Why do you want to expand f in a power series of x? That expansion happens in terms of different values of x, not in terms of functions xn.

    (I'm not sure if f(x)=1/(1+x) would give a well-defined expression here but there are other functions where the power series does not converge to the function).
     
  4. Apr 4, 2015 #3

    blue_leaf77

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    So what I want to do is to get ## \int u^*_{\alpha}(x) f(x) u_{\beta}(x) dx ## from ## \langle \alpha |\hat{f}(\hat{x})| \beta \rangle ## with hat represents operator. In my mind, the way I get the integral expression is that by first placing two ## \int dx |x\rangle \langle x| ## to the left and to the right of ##\hat{f}##. This way I will get the matrix element ## \langle x' |\hat{f} | x" \rangle## in my expression. Now I can simplify further if I expand f into powers of x operator then let the operators ##x^n## act on ##|x"\rangle##, and I will end up with the integral expression in the end. Is my way correct? So in short I want to derive the expression for an expectation value in term of wavefunctions from that in term of ket and bra.
    Why in terms of different values of x? If I expand a function in its Taylor terms I will get terms of powers of x. Did you misunderstand it as an expansion into bases of ##|x\rangle##? If so that's not what I meant, f is an operator not a state.
     
    Last edited: Apr 4, 2015
  5. Apr 4, 2015 #4

    mfb

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    ##\langle x' |\hat{f} | x" \rangle## looks like ##\delta(x'-x'') f(x')##. Otherwise I don't understand what your operator is doing.
    Sure, but the |x> are.
     
  6. Apr 5, 2015 #5

    DrDu

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    ##(1+x)^{-1}## is the inverse of 1+x, no?
     
  7. Apr 5, 2015 #6

    blue_leaf77

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    That's my point, basically what I was trying to understand is how I can get the latter expression from the former. Are there any intermediate steps in between, or are they simply defined to be equal, namely ## \langle x'| \hat{f} | x" \rangle = f(x) \delta(x'-x") ##?
     
  8. Apr 5, 2015 #7

    vanhees71

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    Indeed, you have, for a true pure state, i.e., for a normalizable ket with ##\langle \psi|\psi \rangle=1##,
    $$\langle f(x) \rangle=\langle \psi|f(\hat{x}) \psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle \psi|f(\hat{x}) x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \langle \psi| x \rangle \langle x|\psi \rangle f(x)=\int_{\mathbb{R}} \mathrm{d} x |\psi^2(x)|f(x).$$
    The same you can achieve by using the matrix elements of your operator wrt. the position eigenbasis
    $$f(x',x)=\langle x' |f(\hat{x}) x \rangle=f(x) \langle x'|x \rangle=f(x) \delta(x-x').$$
    Now you have
    $$\langle f(x) \rangle=\langle \psi|f(\hat{x}) \psi \rangle=\int_{\mathbb{R}}\mathrm{d} x' \int_{\mathbb{R}}\mathrm{d} x \langle \psi|x' \rangle \langle x'|f(\hat{x}) x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}}\mathrm{d} x' \int_{\mathbb{R}}\mathrm{d} x \psi^*(x') f(x',x) \psi(x)=\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2 f(x).$$
    All this shows from the general form of Born's rule that for the position representation ##|\psi(x)^2## is the probability density for the position of the particle.
     
  9. Apr 5, 2015 #8

    blue_leaf77

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    In your second equation line you wrote ## f(\hat{x}) |x \rangle = f(x) |x \rangle ## where the x in the variable of f is changed from an operator (denoted by the hat) to a mere number. I know that the ket the operator ##f(\hat{x})## acts on is of position eigenket, but what is the mathematical proof which guarantees that we can just change the x in f from an operator to a number while we don't need to know the particular functional form of ##f(\hat{x})##? I can accept it if, for example, ##f(\hat{x}) = \hat{x}^2## since in that case it's clear that ##\hat{x}## will act twice on its own eigenket giving a number ##x^2##. But what if, as in my original problem, ##f(\hat{x}) = 1/(1+\hat{x})##. What shall I do to know the manner how it would act on ##|x \rangle##?
     
  10. Apr 5, 2015 #9

    DrDu

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    This is in deed a problem for generak functions. I tried to give you a hint, how to proceed. From your argument, it is clear that 1+x is diagonal in position representation. Show that this will also hold true for the inverse operator.
     
  11. Apr 5, 2015 #10

    vanhees71

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    I'd think, that it's the definition of what's meant by a function of an operator. Of course, what I've done in my posting above is not mathematically rigorous. The position eigenvectors are highly non-trivial. Particularly they are of course no Hilbert-space vectors. For a more rigorous treatment of all this, see, e.g., the textbook by Galindo and Pacual.
     
  12. Apr 5, 2015 #11

    blue_leaf77

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    Thanks for your hint DrDU.
    A product between two diagonal matrices proceeds like element-wise product, that is the element of the resulting matrix is the product between the corresponding elements of the composite matrices. The inverse matrix of A is defined to be A-1A=AA-1=I, so if A is diagonal, so is its inverse. Ok up to this point I can see why the operator 1/(1+x) also has the same eigenkets as the operator x.
    I just want to confirm this, is it not always correct expanding an operator ##f(\hat{x})## into its power terms as we do for exponential operator (e.g. rotation operator and translation operator)?
     
  13. Apr 5, 2015 #12

    blue_leaf77

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    I presume it can be a general problem that can happen to finite and denumerable bases too, for example spin eigenstates. The most common example is spin rotation operator ##\exp(iS_z \phi/\hbar)##, to know how that operator would act on spin up or down state people expand it in terms of power series of Sz so the problem will become trivial. So again my question is can we always expand any function of operator into power series?
     
  14. Apr 5, 2015 #13

    vanhees71

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    For finite-dimensional vector spaces, this is not such a complicated issue. The matrices build algebraically a ring, and you can define norms that are compatible with this algebraic structure, e.g., the induced norm from the scalar product
    $$\|\hat{A} \|=\sup \{\|A |u \rangle \|; \|u\| \}=1$$
    or a pratically more simple norm
    $$\|\hat{A} \|^2=\mathrm{Tr} (\hat{A} \hat{A}^{\dagger})=\sum_{j,k=1}^n |A_{jk}|^2.$$
    Then you can define functions as in analysis with numbers via polynomials, which are defined algebraically and power series, etc.

    Particularly the matrix exponential function is well defined for all matrices, since the defining series is absolutely convergent (in the sense of the matrix norm).
     
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