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Sakurai 1.27

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that [tex]f\left(A\right)[/tex] is a function of a Hemitian operator A with the property [tex]A|a'\rangle=a'|a'\rangle[/tex]. Evaluate [tex]\langle b''|f\left(A\right)|b'\rangle[/tex] when the transformation matrix from the a' basis to the b' basis is known.

    [tex]\langle\beta|A|\alpha\rangle=\int dx' \int dx'' \langle\beta|x'\rangle\langle x'|A|x''\rangle \langle x''|\alpha\rangle=\int dx' \int dx'' \psi_\beta^*\left(x'\right)\langle x'|A|x''\rangle\psi_\alpha\left(x''\right)[/tex]


    3. The attempt at a solution

    Transformation from [tex]a'[/tex] to [tex]b'[/tex] basis:

    [tex]U|a'\rangle = |b'\rangle[/tex]

    Multiply both sides by [tex]a''[/tex]:

    [tex]\langle a''|U|a'\rangle = \langle a''|b'\rangle[/tex]

    This is the transformation matrix.

    Now insert [tex]|a''\rangle\langle a''|[/tex] into [tex]\langle b''|f\left(A\right)|b'\rangle[/tex]:

    [tex]\langle b''|f\left(A\right)|a''\rangle\langle a''|b'\rangle[/tex]

    Now insert [tex]|a'\rangle\langle a'|[/tex] into the above:

    [tex]\langle b''|a'\rangle\langle a'|f\left(A\right)|a''\rangle\langle a''|b'\rangle[/tex]

    This is a transformation matrix: [tex]\hat{T}=\langle a''|b'\rangle[/tex]

    And so is this: [tex]\hat{T}^*=\langle b''|a'\rangle[/tex].

    So then:

    [tex]\langle b''|f\left(A\right)|b'\rangle = \hat{T}^* \langle a'|f\left(A\right)|a''\rangle \hat{T}[/tex]
     
    Last edited: Oct 26, 2009
  2. jcsd
  3. Oct 26, 2009 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    No it isn't. In fact, it isn't even a matrix; just an inner product between two states.

    What makes you think you are allowed to do this?


    Again, not matrices. Also, [itex]\langle a''|b'\rangle^{\dagger}=\langle b'|a''\rangle\neq\langle b''|a'\rangle[/itex]
     
  4. Oct 26, 2009 #3
    If you have [tex]\langle b''|f\left(A\right)|b'\rangle[/tex], why not just insert [tex]\sum_{a'}|a'\rangle\langle a'|[/tex] in there so you can get the eigenvalue of [tex]f(A)[/tex]:

    [tex]
    \sum_{a'}\langle b''|f(A)|a'\rangle\langle a'|b'\rangle=\sum_{a'}f(a')\langle b''|a'\rangle\langle a'|b'\rangle
    [/tex]
     
    Last edited: Oct 26, 2009
  5. Oct 26, 2009 #4
    Ahhh...I looked at Sakurai a little closer. You're right.

    Section 1.5:

    "The matrix elements of the U operator are built up of the inner products of old base bras and new base kets."


    If Chuck Norris can divide by zero, then I can do that.
     
  6. Oct 26, 2009 #5
    If

    [tex]A|a'\rangle=a'|a'\rangle[/tex]

    Then

    [tex]f\left(A\right)|a'\rangle=f\left(a'\right)|a'\rangle[/tex] ??

    I was wondering how that eigenvalue relationship came into play in this problem.

    Thanks.
     
  7. Oct 26, 2009 #6
    Yeah, you can prove this by Taylor expanding some function and then applying the ket, for example the exponential function:

    [tex]
    \begin{array}{ll}\exp[mA]|a'\rangle&=\left(1+mA+\frac{m^2A^2}{2!}+\cdots\right)|a'\rangle
    \\ \,&=|a'\rangle+mA|a'\rangle+\frac{m^2}{2!}A^2|a'\rangle+\cdots
    \\ \,&=|a'\rangle+ma'|a'\rangle+\frac{m^2}{2!}(a')^2|a'\rangle+\cdots
    \\ \,&=\left(1+ma'+\frac{(ma')^2}{2!}+\cdots\right)|a'\rangle
    \\ \,&=\exp[ma']|a'\rangle
    [/tex]
     
  8. Oct 26, 2009 #7
    So here's what I came up with. Please let me know how it looks:

    [tex]\langle b^{(m)}|f\left(A\right)|b^{(n)}\rangle[/tex]

    Insert the following:

    [tex]\sum_k |a^{(k)}\rangle \langle a^{(k)}|[/tex]
    [tex]\sum_l |a^{(l)}\rangle \langle a^{(l)}|[/tex]

    [tex]\sum_k \sum_l \langle b^{(m)}|a^{(k)}\rangle \langle a^{(k)}|f\left(A\right)|a^{(l)}\rangle \langle a^{(l)}|b^{(n)}\rangle [/tex]

    [tex]=\sum_k \sum_l f\left(a^{(l)}\right) \langle b^{(m)}|a^{(k)}\rangle \langle a^{(k)}|a^{(l)}\rangle \langle a^{(l)}|b^{(n)}\rangle [/tex]

    [tex]=\sum_k \sum_l f\left(a^{(l)}\right) \langle b^{(m)}|a^{(k)}\rangle \delta_{kl} \langle a^{(l)}|b^{(n)}\rangle [/tex]

    [tex]=\sum_k f\left(a^{(k)}\right) \langle b^{(m)}|a^{(k)}\rangle \langle a^{(k)}|b^{(n)}\rangle [/tex]

    [tex]=\sum_k f\left(a^{(k)}\right) \langle a^{(m)}|U^{\dagger}|a^{(k)}\rangle \langle a^{(k)}|U|a^{(n)}\rangle [/tex]

    [tex]=f\left(a^{(k)}\right) \langle a^{(m)}|U^{\dagger}U|a^{(n)}\rangle [/tex]

    [tex]=f\left(a^{(k)}\right) \langle a^{(m)}|a^{(n)}\rangle [/tex]

    [tex]=f\left(a^{(k)}\right) \delta_{nm} [/tex]
     
  9. Oct 26, 2009 #8
    It looks like you don't actually need to add [tex]|a^{(k)}\rangle\langle a^{(k)}|[/tex] in the above, but otherwise seems fine to me.
     
  10. Oct 26, 2009 #9
    Any ideas on how to do part (b)?

    I'm approaching it the same way, with some differences:

    [tex]\langle \textbf{p}''|F\left(r\right)|\textbf{p}'\rangle[/tex]

    [tex]= \sum_{\textbf{x}'} \sum_{\textbf{x}''}\langle \textbf{p}''|\textbf{x}''\rangle \langle \textbf{x}'' |F\left(r\right)|\textbf{x}'\rangle \langle \textbf{x}'|\textbf{p}'\rangle[/tex]

    [tex]\langle \textbf{x}'|\textbf{p}' \rangle = \frac{1}{\left(2\pi\hbar\right)^{\frac{3}{2}}}e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}[/tex]

    [tex]= \sum_{\textbf{x}'} \sum_{\textbf{x}''}\frac{1}{\left(2\pi\hbar\right)^{\frac{3}{2}}}e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \langle \textbf{x}'' |F\left(r\right)|\textbf{x}'\rangle \frac{1}{\left(2\pi\hbar\right)^{\frac{3}{2}}}e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}[/tex]

    [tex]= \frac{1}{\left(2\pi\hbar\right)^{3}} \sum_{\textbf{x}'} \sum_{\textbf{x}''}e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \langle \textbf{x}'' |F\left(r\right)|\textbf{x}'\rangle e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}[/tex]

    Now I'm stuck.
     
  11. Oct 26, 2009 #10
    I guess those sums should be integrals.
     
  12. Oct 26, 2009 #11
    [tex]\frac{1}{\left(2\pi\hbar\right)^{3}} \int \int e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \langle \textbf{x}'' |F\left(r\right)|\textbf{x}'\rangle e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}d^3x' d^3x''[/tex]
     
  13. Oct 26, 2009 #12
    Does this work here?

    [tex]F(r)|\textbf{x}'\rangle = F(x')|\textbf{x}'\rangle[/tex]
     
  14. Oct 26, 2009 #13
    If it does, then:

    [tex]\frac{1}{\left(2\pi\hbar\right)^{3}} \int \int F\left(x'\right)e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \langle \textbf{x}'' |\textbf{x}'\rangle e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}d^3x' d^3x''[/tex]

    [tex]=\frac{1}{\left(2\pi\hbar\right)^{3}} \int \int F\left(x'\right)e^{-i \frac{\textbf{p}''\cdot\textbf{x}''}{\hbar}} \delta^3\left(\textbf{x}''-\textbf{x}'\right) e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}d^3x' d^3x''[/tex]

    [tex]=\frac{1}{\left(2\pi\hbar\right)^{3}} \int F\left(x'\right)e^{-i \frac{\textbf{p}''\cdot\textbf{x}'}{\hbar}} e^{i \frac{\textbf{p}'\cdot\textbf{x}'}{\hbar}}d^3x'[/tex]

    [tex]=\frac{1}{\left(2\pi\hbar\right)^{3}} \int F\left(x'\right)e^{-i \frac{\textbf{x}'}{\hbar}\cdot\left(\textbf{p}''-\textbf{p}'\right)}d^3x'[/tex]
     
  15. Oct 26, 2009 #14
    while [tex]r=\sqrt{x^2+y^2+z^2}[/tex], you can't make this claim. You'l have to solve this integral as a function of [tex]F(\mathbf{r})[/tex]. Also, an integral is also a sum so

    [tex]
    \sum_{a'}|a'\rangle\langle a'|=\int|a'\rangle\langle a'|\,da'=1
    [/tex]

    Basically your last post has it mostly right, though you can't use [tex]F(x')[/tex]. Expanding the integral for spherical coordinates:

    [tex]
    \langle \mathbf{p}''|F(\mathbf{r})|\mathbf{p}'\rangle=\frac{1}{\left(2\pi\hbar\right)^3}\int_0^{2\pi}d\phi'\int_{-1}^1d(\cos\theta)\int_0^\infty (r')^2F(r')\exp\left[\frac{i|\mathbf{p}'-\mathbf{p}''|\cdot\mathbf{r}}{\hbar}\right]\,dr'
    [/tex]
     
  16. Oct 26, 2009 #15
    As before, I don't think you need the double sum. You should just have

    [tex]
    \begin{array}{ll}\langle \textbf{p}''|F\left(\mathbf{r}\right)|\textbf{p}'\rangle&= \sum_{\textbf{r}'} \langle \textbf{p}''|F\left(\mathbf{r}\right)|\textbf{r}'\rangle \langle \textbf{r}'|\textbf{p}'\rangle
    \\ &=\sum_{r'}F(\mathbf{r}')\langle\mathbf{p}''|\mathbf{r}'\rangle\langle\mathbf{r}'|\mathbf{p}'\rangle\end{array}
    [/tex]

    note the use of [tex]\mathbf{r}[/tex] instead of [tex]\mathbf{x}[/tex] because [tex]F[/tex] is a function of [tex]r[/tex], not [tex]x[/tex].
     
    Last edited: Oct 26, 2009
  17. Oct 26, 2009 #16
    Actually, this last line should read

    [tex]
    \langle \mathbf{p}''|F(\mathbf{r})|\mathbf{p}'\rangle=\frac{1}{\left(2\pi\hbar\right)^3}\int_0^{2\pi}d\phi\int_{-1}^1d(\cos\theta)\int_0^\infty (r')^2F(r')\exp\left[\frac{i|\mathbf{p}'-\mathbf{p}''|r'\cos\theta}{\hbar}\right]\,dr'
    [/tex]

    I just forgot to remove the vector off of [tex]\mathbf{r}'[/tex], but is otherwise correct as is
     
  18. Oct 26, 2009 #17
    Yeah...I made a mistake on this post and since we can't delete them, I'm just erasing it all. You should be fine with what I had written in posts 15 and 16.
     
    Last edited: Oct 26, 2009
  19. Oct 27, 2009 #18
    When I say [tex]\textbf{x}[/tex], I mean [tex]\textbf{x}=x\hat{\textbf{x}}+y\hat{\textbf{y}}+z\hat{\textbf{z}}[/tex]

    And [tex]\textbf{x}'=x'\hat{\textbf{x}}+y'\hat{\textbf{y}}+z'\hat{\textbf{z}}[/tex]
     
  20. Oct 27, 2009 #19
    So evaluate this:

    [tex]\langle\mathbf{p}''|F(\mathbf{r})|\mathbf{p}'\rangle=\frac{1}{\left(2\pi\hbar\right)^3}\int_0^{2\pi}d\phi\int_{-1}^1d(\cos\theta)\int_0^\infty (r')^2F(r')\exp\left[\frac{i|\mathbf{p}'-\mathbf{p}'|r'\cos\theta}{\hbar}\right]\,dr'[/tex]

    And get this:

    [tex]\langle\mathbf{p}''|F(\mathbf{r})|\mathbf{p}'\rangle=\frac{4\pi}{\left(2\pi\hbar\right)^3} \int_0^\infty (r')^2F(r')\exp\left[\frac{i|\mathbf{p}'-\mathbf{p}'|r'\cos\theta}{\hbar}\right]\,dr'[/tex]

    What next? integrate by parts?
     
  21. Oct 27, 2009 #20
    Looks like integrating by parts will get rid of the exponential leaving me something in terms of [tex]F(r')[/tex].
     
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