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A Ground state wave function from Euclidean path integral

  1. Dec 9, 2016 #1

    ShayanJ

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    From the path integral approach, we know that ## \displaystyle \langle x,t|x_i,0\rangle \propto \int_{\xi(0)=x_i}^{\xi(t_f)=x} D\xi(t) \ e^{iS[\xi]}##. Now, using ## |x,t\rangle=e^{-iHt}|x,0\rangle ##, ## |y\rangle\equiv |y,0\rangle ## and ## \sum_b |\phi_b\rangle\langle \phi_b|=1 ## where ## \{ \phi_b \} ## are the energy eigenstates we have:

    ## \langle x,t|x_i\rangle =\langle x| e^{iHt}|x_i\rangle=\sum_b \langle x|e^{iHt}|\phi_b\rangle \langle \phi_b|x_i\rangle=\sum_b \langle x|\phi_b\rangle e^{iE_bt}\langle \phi_b|x_i\rangle=\sum_b \phi_b(x) \phi_b^*(x_i) e^{iE_bt} ##

    Now by doing a Wick rotation, ## t=it_E ## and taking the limit ## t_E\to \infty ##, we'll have:

    ##\displaystyle \langle x,i\infty|x_i\rangle \propto \phi_0(x) \Rightarrow \phi_0(x) \propto \int_{\xi(0)=x_i}^{\xi(t_E=\infty)=x} D\xi(t) \ e^{-S[\xi]} ##

    Using a similar argument we can find:

    ##\displaystyle \phi_0^*(x) \propto \int_{\xi(t_E=-\infty)=x}^{\xi(0)=x_i} D\xi(t) \ e^{-S[\xi]} ##

    The problem is, everywhere that I see this, the path integral for ## \phi_0(x) ## is from ## -\infty ## to 0 and the path integral for ## \phi_0^*(x) ## is from 0 to ## \infty ##, the opposite of what I got. What am I doing wrong?

    Thanks
     
  2. jcsd
  3. Dec 9, 2016 #2

    stevendaryl

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    I'm not super-competent at path integrals, but it seems to me that starting with your expression:

    [itex]\langle x,t|x_i\rangle = \sum_b \phi_b(x)^* \phi_b(x_i) e^{i E_b t}[/itex]

    Then letting [itex]t = i \beta[/itex], we have:
    [itex]G(x, x_i, \beta) \equiv \sum_b \phi_b(x)^* \phi_b(x_i) e^{- E_b \beta}[/itex]

    (I'm introducing the Green function [itex]G[/itex] just because it's easier to write than the sum or the path integral)

    If [itex]\beta \gg 1[/itex], then [itex]G(x,x_i,\beta) \approx \phi_0^*(x) \phi_0(x_i) e^{-E_0 \beta}[/itex]

    Given this asymptotic expression, we can get [itex]E_0[/itex] as a limit:

    [itex]E_0 = lim_{\beta \rightarrow \infty} \frac{-log(G(x,x_i,\beta))}{\beta}[/itex]

    In terms of [itex]E_0[/itex], we can get another limit:

    [itex]\phi_0^*(x) \phi_0(x_i) = lim_{\beta \rightarrow \infty} G(x,x_i,\beta) e^{+\beta E_0}[/itex]

    So taking the limit as [itex]\beta \rightarrow \infty[/itex] or [itex]t \rightarrow +i \infty[/itex] gives us information about the product [itex]\phi_0^*(x) \phi_0(x_i)[/itex]. I don't see how it gives us either [itex]\phi_0^*(x)[/itex] or [itex]\phi_0(x_i)[/itex] separately.
     
  4. Dec 9, 2016 #3

    stevendaryl

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    You could get [itex]\phi_0(x)[/itex] up to a phase:

    [itex]|\phi_0(x)|^2 = lim_{\beta \rightarrow \infty} G(x,x,\beta) e^{+\beta E_0}[/itex]

    Then you take the square-root to get [itex]\phi_0(x)[/itex] up to an unknown phase.
     
  5. Dec 9, 2016 #4

    ShayanJ

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    ## \phi_0(x_i) ## and ## \phi_0^*(x_i) ## are just the value of some function at some point and so both are constant. And since I the formulas I wrote are proportionalities instead of equalities, I don't think they make any difference.
     
  6. Dec 9, 2016 #5

    stevendaryl

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    Well, in that case, it seems to me that you have both:

    [itex]\phi_0^*(x) \propto lim_{\beta \rightarrow +\infty} G(x,x_i,\beta) e^{+\beta E_0}[/itex] (considering [itex]x_i[/itex] a constant)

    and

    [itex]\phi_0(x_i) \propto lim_{\beta \rightarrow +\infty} G(x,x_i,\beta) e^{+\beta E_0}[/itex] (considering [itex]x[/itex] a constant)

    I don't understand why the limits would be different.
     
  7. Dec 9, 2016 #6

    ShayanJ

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    The Wick rotation changes ## \sum_b \phi_b(x)\phi_b^*(x_i) e^{iE_bt} ## to ## \sum_b \phi_b(x)\phi_b^*(x_i) e^{-E_bt_E} ##. So the phase is now a decaying exponential and its limit at ## \infty ## is zero.

    But if you do the calculation starting with ## \langle x_i|x,t\rangle ##, you'll have:
    ## \langle x_i|x,t\rangle=\langle x_i|e^{-iHt}|x\rangle=\sum_b \langle x_i|e^{-iHt}|\phi_b\rangle \langle \phi_b|x\rangle=\sum_b \phi_b(x_i) \phi_b^*(x) e^{-iE_bt} ##.

    The Wick rotation changes ## \sum_b \phi_b(x_i) \phi_b^*(x) e^{-iE_bt} ## to ## \sum_b \phi_b(x_i) \phi_b^*(x) e^{E_bt_E} ##. Now the exponential function is not decaying and the limit we have to take is ## t_E\to -\infty ##.

    P.S.
    Your approach makes sense but I'm doing this to make sense of the calculations of a paper and to match those calculations, I need one of the boundaries to be at ## t_E=-\infty ##. Although I have some doubts about what you do because our starting point is not symmetric w.r.t. ## x \leftrightarrow x_i ## but at the end you're assuming that the whole process has that symmetry.
     
  8. Dec 9, 2016 #7

    stevendaryl

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    Just a point about definitions: Normally, the definition of the Green function, or propagator, is:

    [itex]\langle x|e^{-i H t}|x_i \rangle[/itex]

    not

    [itex]\langle x|e^{+i H t}|x_i \rangle[/itex]

    Are you sure about the plus sign?
     
  9. Dec 9, 2016 #8

    ShayanJ

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    Actually I didn't start from any definition. The standard derivation of the path integral starts from ## \langle x,t|x_i \rangle ##. Using ## |x,t\rangle=e^{-iHt}|x\rangle \Rightarrow \langle x,t|=\langle x|e^{iHt} ##, we can write it as ## \langle x|e^{iHt}|x_i\rangle ##. Then by breaking down the intervals, they derive the result ##\displaystyle \langle x,t|x_i \rangle\propto \int_{\xi(0)=x_i}^{\xi(t)=x} D\xi(t) \ e^{iS[\xi]} ##. This formula will give us ## \phi_0(x) ## as I explained in my previous posts.
    But because I also needed to derive the same result for ## \phi_0^*(x) ##, I thought maybe I can do the same calculations, this time starting from ## \langle x_i|x,t\rangle ##. But this time we have ## |x,t\rangle=e^{-iHt}|x\rangle ## and so ## \langle x_i|x,t\rangle=\langle x_i|e^{-iHt}|x\rangle ##.
     
  10. Dec 9, 2016 #9

    stevendaryl

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    Okay, but the usual interpretation of the path integral is that it gives the probability amplitude of going from some initial point [itex]x_i[/itex] at time [itex]t_0[/itex] to some final point [itex]x_f[/itex] at time [itex]t_f[/itex], which would be [itex]\langle x_f| e^{-i H (t_f - t_i)} |x_i\rangle[/itex]. I thought that's what you were doing, with [itex]x_f \Rightarrow x[/itex] and [itex]t_i \Rightarrow 0[/itex] and [itex]t_f \Rightarrow t[/itex]. That would give: [itex]\langle x| e^{-i H t} |x_i\rangle[/itex]. With the + sign, it's the complex conjugate of the amplitude to go from [itex]x[/itex] at time 0 to [itex]x_i[/itex] at time [itex]t[/itex]. So it seems backwards to me.

    So letting the usual Green function be [itex]G(x,x_i, t) = \langle x | e^{-iHt} | x_i \rangle[/itex], the expression you wrote is
    [itex]G(x, x_i, -t)[/itex] or [itex]G(x_i, x, t)^*[/itex].

    I guess it doesn't matter how they relate to the usual Green function, but in any case, you can do either of these
    • Start with [itex]\langle x|e^{-iHt}|x_i\rangle[/itex], and take the limit [itex]t \rightarrow -i \infty[/itex]
    • Start with [itex]\langle x|e^{+iHt}|x_i\rangle[/itex], and take the limit [itex]t \rightarrow +i\infty[/itex]
    Those both give exactly the same thing, which is [itex]\phi_0(x) \phi_0^*(x_i)[/itex]

    Or you can switch [itex]x[/itex] and [itex]x_i[/itex] (which is the same as taking the complex conjugate and flipping the time):
    • Start with [itex]\langle x_i|e^{-iHt}|x\rangle[/itex], and take the limit [itex]t \rightarrow -i \infty[/itex]
    • Start with [itex]\langle x_i|e^{+iHt}|x\rangle[/itex], and take the limit [itex]t \rightarrow +i\infty[/itex]
    Those give you the complex conjugate: [itex]\phi_0^*(x) \phi_0(x_i)[/itex]
     
  11. Dec 9, 2016 #10

    vanhees71

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    There seems to be some confusion concerning the time-evolution pictures here. I suppose by ##|x,t \rangle## you mean the position eigenvector in the Heisenberg picture of time evolution. It is defined such that the position operator obeys the equation of motion (setting ##\hbar=1##)
    $$[\hat{x},\hat{H}]=\mathrm{i} \dot{\hat{x}}.$$
    This implies that
    $$\hat{x}(t)=\exp(\mathrm{i} \hat{H} t) \hat{x}_0 \exp(-\mathrm{i} \hat{H}t),$$
    because then
    $$\dot{\hat{x}}=-\mathrm{i} \exp(\mathrm{i} \hat{H} t) [\hat{x}_0,\hat{H}] \exp(-\mathrm{i} \hat{H}t) = -\mathrm{i} [\hat{x},\hat{H}].$$
    The eigenvector is defined such that for all ##t##
    $$\hat{x}(t)|x,t \rangle = x|x,t \rangle \qquad (*)$$
    with ##x \in \mathbb{R}## fixed. Now we have
    $$\hat{x}(t) |x,t \rangle=\exp(\mathrm{i} \hat{H} t) \hat{x}_0 \exp(-\mathrm{i} \hat{H}t)|x,t \rangle,$$
    So if we define (!)
    $$\exp(-\mathrm{i} \hat{H} t) |x,t \rangle=|x,0 \rangle \; \Rightarrow \; |x,t \rangle=\exp(+\mathrm{i} \hat{H} t) |x_0,t \rangle,$$
    This fulfills (*), and that's how one defines (up to a phase) the Heisenberg position eigenvectors.

    Now we can write the wave function (which is independent of the picture of time evolution) as
    $$\psi(t,x)=\langle x,t|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x' \langle x,t|x',0 \rangle \langle x',0|\psi \rangle=\int_{\mathbb{R}} G(t,x,x') \psi_0(x'),$$
    i.e., the Green's function is
    $$G(t,x,x')=\langle x,t|x',0 \rangle=\langle x,0|\exp(-\mathrm{i} \hat{H} t|x',0 \rangle$$
    as written by @stevendaryl in posting #7.

    Obviously it fulfills
    $$\mathrm{i} \partial_t G(t,x,x') = \langle x,t|\hat{H}|x',0 \rangle, \qquad G(0^+,x,x')=\delta(x-x'),$$
    i.e., the Green's function fullfills the Schrödinger Equation with the appropriate boundary condition.
     
  12. Dec 14, 2016 #11

    samalkhaiat

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    Be careful here. If the state [itex]|\psi ;t \rangle[/itex] evolves “forward” in time, i.e., [tex]|\psi ;t \rangle = e^{-iHt}|\psi ; 0\rangle ,[/tex] the position eigen-ket [itex]|x,t\rangle[/itex] must evolve “backward” in time, i.e., [tex]|x,t\rangle = e^{iHt} |x,0\rangle = e^{-iH(-t)}|x,0\rangle .[/tex] This is because the wave-function [itex]\psi(x,t)[/itex] is defined by picture-independent inner product: [tex]
    \begin{align*}
    \psi(x,t) &= \langle x , 0 |\psi ; t \rangle \\
    &= \langle x , 0 |\left( e^{-iHt} |\psi ;0 \rangle \right) \\
    &= \left( \langle x , 0 | e^{-iHt} \right) |\psi ; 0 \rangle \\
    &= \langle x , t |\psi ; 0 \rangle ,
    \end{align*}
    [/tex]
    where [itex]\langle x , 0 |\psi ; t \rangle[/itex] is the inner product of the time-fixed position bra [itex]\langle x | \equiv \langle x , 0 |[/itex] with the “forward”-evolving state [itex]|\psi ; t \rangle[/itex], and [itex]\langle x , t |\psi ; 0 \rangle[/itex] is the inner product of the Heisenberg position bra [itex]\langle x , t |[/itex], which evolves “backward” in time, with the time-fixed Heisenberg state [itex]|\psi ; 0\rangle[/itex].
    Now, at any given time [itex]t’[/itex], the Heisenberg set [itex]\{ |x , t’\rangle \}[/itex] is complete. So, for [itex]t > t’ > 0[/itex], we can write
    [tex]
    \begin{align*}
    \psi(x,t) &= \int dx’ \langle x , t |x’ , t’ \rangle \langle x’ , t’ | \psi ; 0 \rangle \\
    &= \int dx’ \ K(x , t ; x’ , t’) \ \psi(x’,t’) .
    \end{align*}
    [/tex]
    Now, by inserting complete energy eigenstates [itex]\{ |\varphi_{a}\rangle \}[/itex] in the propagator, we get
    [tex]
    \begin{align*}
    K( x , t ; x’ , t’) &= \langle x , t |x’ , t’ \rangle \\
    &= \sum_{a} \langle x , 0| e^{-iHt}|\varphi_{a}\rangle \langle \varphi_{a}|e^{iHt’}|x’, 0\rangle \\
    &= \sum_{a} e^{-iE_{a}(t-t’)} \langle x |\varphi_{a}\rangle \langle \varphi_{a}|x’\rangle .
    \end{align*}
    [/tex]
    Thus [tex]K(x , t ; x’ , 0) = \sum_{a} e^{-iE_{a}t} \ \varphi_{a}(x) \varphi^{*}_{a}(x’) .[/tex] Now, let [itex]t \to \infty[/itex] (or if you like [itex]t \to –i\infty[/itex]). Assuming there is a gap between [itex]E_{0}[/itex] and [itex]E_{1}[/itex], the main contribution to the sum will come from the ground state [itex]\varphi_{0}(x)[/itex]. As [itex]t[/itex] gets large, all terms above the ground state [itex](n > 0)[/itex] will oscillate very fast and will not contribute. Also, in the limit [itex]t \to –i\infty[/itex], any term with [itex]n>0[/itex] will die out exponentially relative to the vacuum state. So, [tex]\lim_{t \to \infty} K(x , t ; x , 0) = \varphi_{0}(x) \varphi^{*}(x) ,[/tex] and, therefore, [tex]\lim_{t \to \infty} K(x,t ; 0,0) \sim \varphi_{0}(x) = \langle x | 0 \rangle .[/tex]
     
  13. Dec 15, 2016 #12

    vanhees71

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    It should be stressed again that of course you have to stay in one picture of time evolution. In the following I conveniently assume that the Schrödinger and Heisenberg picture coincide at ##t=0##, which is arbitrary and convenient. For the Schrödinger picture you have (for a time-independent Hamiltonian)
    $$|\psi,t \rangle_S=\exp(-\mathrm{i} \hat{H} t) |\psi,0 \rangle, \quad |\vec{x},t \rangle_S=|\vec{x},0 \rangle,$$
    and for the Heisenberg picture
    $$|\psi,t \rangle_H=|\psi,0 \rangle, \quad |\vec{x},t \rangle=\exp(+\mathrm{i} \hat{H}t) |\vec{x},0 \rangle.$$
    Of course, the wave function is picture independent, because
    $$\psi(t,\vec{x})=_S\langle \vec{x},t|\psi \rangle_S = \langle \vec{x},0|\exp(-\mathrm{i} \hat{H} t)| \psi,0 \rangle = \langle \exp(+\mathrm{i} \hat{H} t) \vec{x},0|\psi,0 \rangle = _H \langle \vec{x},t |\psi,t \rangle_H.$$
     
  14. Dec 26, 2016 #13

    ShayanJ

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    How should do these calculations if the Hamiltonian is time dependent?
     
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