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Path integral propagator as Schr's eq Green's function

  1. Nov 3, 2015 #1

    ShayanJ

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    Its usually said that the propagator ## K(\mathbf x'',t;\mathbf x',t_0) ## that appears as an integral kernel in integrals in the path integral formulation of QM, is actually the Green's function for the Schrodinger equation and satisfies the equation below:
    ## \left[ -\frac{\hbar^2}{2m} \nabla^{''2} +V(\mathbf x'')-i\hbar \frac{\partial}{\partial t} \right] K(\mathbf x'',t;\mathbf x',t_0)=-i\hbar \delta^3(\mathbf x''-\mathbf x')\delta(t-t_0)##.

    My problem with this statement, is that the method of Green's functions is for solving inhomogeneous differential equations, but the Schrodinger equation is a homogeneous equation and I don't understand how using a Green's function can help us solve it. So I'm beginning to doubt that the above kernel is a green's function. But this is something that I read in lots of references. What's going on?
    Thanks
     
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  3. Nov 3, 2015 #2

    stevendaryl

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    The way that I've seen Green's functions used in QM is for perturbation theory.

    Suppose you know the solutions to the Schrodinger equation for a simple hamiltonian [itex]H_0[/itex]:

    [itex](H_0 - i \hbar \dfrac{\partial}{\partial t}) \psi_0 = 0[/itex]

    But you want to know how [itex]\psi_0[/itex] is changed in the presence of a perturbing potential [itex]V[/itex]. The full wave function [itex]\psi[/itex] satisfies:

    [itex](H_0 - i \hbar \dfrac{\partial}{\partial t} + V) \psi = 0[/itex]

    We can write that as an inhomogeneous equation as follows:

    [itex](H_0 - i \hbar \dfrac{\partial}{\partial t}) \psi = - V \psi[/itex]

    That doesn't actually work as an inhomogeneous equation, because the right-hand side involves the same unknown function [itex]\psi[/itex], whereas in a real inhomogeneous equation, the right-hand side is a known function. However, in the can-do spirit of physics, where we pretend that series always converge, we can come up with a sequence of approximations to [itex]\psi[/itex] using our known solution [itex]\psi_0[/itex]:

    1. [itex](H_0 - i \hbar \dfrac{\partial}{\partial t}) \psi_1 = - V \psi_0[/itex]
    2. [itex](H_0 - i \hbar \dfrac{\partial}{\partial t}) \psi_2 = - V \psi_1[/itex]
    3. [itex](H_0 - i \hbar \dfrac{\partial}{\partial t}) \psi_3 = - V \psi_2[/itex]
    4. etc.
    Each of these equations is solvable as inhomogeneous equations. Then [itex]\psi[/itex] is in some sense the limit: [itex]lim_{n \rightarrow \infty} \psi_n[/itex]

    Then, the question becomes: How do you use [itex]K_0[/itex] to solve the inhomogeneous equation?
    where [itex]K_0[/itex] satisfies the equation

    [itex](H_0 - i \hbar \dfrac{\partial}{\partial t}) K_0(x,t, x', t') = -i \hbar \delta(x-x') \delta(t-t')[/itex]

    I think the idea is this: If you have an inhomogeneous equation of the form:

    [itex](H_0 - i \hbar \dfrac{\partial}{\partial t}) \psi(x,t)= S(x,t)[/itex]

    where [itex]S(x,t)[/itex] is some known function, then you can try the guess:

    [itex]\psi(x,t) = C \int dx' dt' K(x,t, x', t') S(x',t')[/itex]

    Then if you operate on both sides by [itex](H_0 - i \hbar \dfrac{\partial}{\partial t})[/itex] you get:
    [itex](H_0 - i \hbar \dfrac{\partial}{\partial t}) \psi = (H_0 - i \hbar \dfrac{\partial}{\partial t}) C \int dx' dt' K(x,t, x', t') S(x',t')[/itex]
    which using the definition of [itex]K[/itex] gives us:

    [itex](H_0 - i \hbar \dfrac{\partial}{\partial t}) \psi = -i\hbar C \int dx' dt' \delta(x-x') \delta(t- t') S(x',t')[/itex]
    [itex]= -i\hbar C S(x,t)[/itex]

    So that's our solution, with [itex]C = \frac{1}{-i \hbar}[/itex]:

    [itex]\psi(x,t) = \frac{1}{-i \hbar} \int dx' dt' K(x,t, x', t') S(x',t')[/itex]
     
  4. Nov 3, 2015 #3

    vanhees71

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    That's only one use of the (free) Green's function. You can also use it to describe the intial-value problem for the full Schrödinger equation. For simlicity let's assume the Hamiltonian is not explicitly time dependent, e.g., as in the OP:
    $$\hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+V(\hat{\mathbf{x}}).$$
    Then let's work in the Schrödinger picture. The state ket then obeys
    $$\mathrm{i} \partial_t |\psi(t) \rangle=\hat{H} |\psi(t) \rangle.$$
    The solution of the initial-value problem then obviously is
    $$|\psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi(0)\rangle.$$
    For the wave function, inserting a unity operator in terms of the completeness relation for the position eigenvectors, you then find
    $$\psi(t,\mathbf{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \mathbf{x}' \langle \mathbf{x}|\exp(-\mathrm{i} \hat{H} t)|\mathbf{x}' \rangle \langle \mathbf{x}'|\psi(0) \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \mathbf{x}' K(t,\mathbf{x},0,\mathbf{x}') \psi(0,\mathbf{x}).$$
    The time-evolution kernel thus fulfills the time-dependent Schrödinger equation in position representation with the initial condition
    $$K(0,\mathbf{x};0,\mathbf{x}')=\delta^{(3)}(\mathbf{x}-\mathbf{x}').$$
     
  5. Nov 3, 2015 #4

    ShayanJ

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    @stevendaryl : If one follows your perturbative description, the final result would be:
    ## \psi(x,t)=\frac{1}{i\hbar}\int dx' dt' K(x,t,x',t') V(x') \psi(x',t') ##.
    So the kernel wouldn't be the Green's function, but the potential multiplied by the Green's function. So it wouldn't lead to path integrals.

    @vanhees71 : I know that part, it just seems to me that the kernel that appears in the integral, shouldn't be the Green's function for the Schrodinger's equation. But I always read in textbooks that it is. I'm confused.
    My point is, the Schrodinger equation is a homogeneous equation. But Green's functions are used for inhomogeneous equations. So how that kernel can be the Green's function while the equation we're solving is a homogeneous one?
     
  6. Nov 3, 2015 #5

    samalkhaiat

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    Not at all. As pointed out by vanhees71, in general, the solution to a given field equation will have contributions from the initial conditions, from boundary conditions and from the sources (on the RHS of the field equation). Green’s function method allows you to express all these contributions in terms of the same Green’s function.
    The free Shrodinger equation, the heat conduction equation, the diffusion equation and many other are homogeneous differential equations and they all have well defined solutions in terms of Green’s functions.
    For example, consider the equation [tex](\partial_{t} - a \nabla^{2}) \Psi(\mathbf{x},t) = \rho(\mathbf{x},t) ,[/tex] with the following initial and boundary conditions [tex]\Psi(\mathbf{x},t) = \psi (\mathbf{x}) , \ \ \mbox{at} \ \ t = 0,[/tex] [tex]\frac{\partial \Psi}{\partial n} - b \Psi = \Phi , \ \ \mbox{is given on the boundary} .[/tex]
    Using Green’s function satisfying [tex]\frac{\partial G}{\partial n} - b G = 0 ,[/tex] the problem is easily solved by
    [tex]\Psi (\mathbf{x},t) = \int d^{3}\bar{\mathbf{x}} \ G(\mathbf{x}- \bar{\mathbf{x}} , t) \psi(\bar{\mathbf{x}}) + a B + C ,[/tex] where [tex]B = \int_{0}^{t} d\bar{t} \oint d\bar{S} \ G(\mathbf{x} - \bar{\mathbf{x}}, t - \bar{t}) \ \Phi(\bar{\mathbf{x}}, \bar{t}) ,[/tex] gives the contribution from the boundary, and [tex]C = \int_{0}^{t} d\bar{t} \int d^{3}\bar{\mathbf{x}} \ G(\mathbf{x} - \bar{\mathbf{x}}, t - \bar{t}) \ \rho(\bar{\mathbf{x}}, \bar{t}) ,[/tex] represents the contribution from the source. Clearly, setting [itex]\rho = 0[/itex], i.e., dealing with homogeneous equation, will only lead to [itex]C = 0[/itex] and the Green’s function method is still applicable.
     
  7. Nov 3, 2015 #6

    ShayanJ

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    Thanks Sam.
    Do you know any math book(preferably written for physicists), that treats Green's functions in the sense you described them for general PDEs?
     
  8. Nov 3, 2015 #7

    samalkhaiat

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    Any text on initial/boundary problems, potential theory and of course on differential equations should describe the Green's function method. I learnt about Green's function from nice old text by P. M. Morse and H. Feshback, "Methods of Theoretical Physics" , McGraw-Hill 1953.
     
  9. Nov 3, 2015 #8

    stevendaryl

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    There are two different functions that are sometimes referred to as "Green's functions".

    One is [itex]G(x,t, x',t')[/itex], which can be defined heuristically in this way:

    [itex]G(x,t',x',t') = e^{-i H (t - t')/\hbar} \delta(x-x')[/itex]

    The other is [itex]K(x,t',x,t')[/itex], which can be defined via a differential equation:

    [itex](H - i \frac{\partial}{\partial t}) K(x,t,x',t') = -i \hbar \delta(x-x') \delta(t-t')[/itex]

    Vanhee's post I think was about [itex]G[/itex], rather than [itex]K[/itex]. I don't think that [itex]G[/itex] satisfies

    [itex](H - i \frac{\partial}{\partial t}) G(x,t,x',t') = -i \hbar \delta(x-x') \delta(t-t')[/itex]

    In the simplest case of a free, non-relativistic spinless particle, we can actually come up with explicit expressions for [itex]G[/itex] and [itex]K[/itex]:

    [itex]G(x,t, x', t') = \frac{1}{2\pi} \int dk e^{i k (x-x') -i \frac{\hbar k^2}{2m} (t-t')}[/itex]

    [itex]K(x,t, x', t') = (\frac{1}{2\pi})^2 \int dk d\omega (\frac{-i\hbar}{\frac{\hbar^2 k^2}{2m} - \hbar \omega}) e^{i k (x-x') -i \omega (t-t')}[/itex]
     
  10. Nov 4, 2015 #9

    ShayanJ

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    Sakurai and Ballentine beg to differ!
     
  11. Nov 4, 2015 #10

    stevendaryl

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    About which point?

    Point 1: Let [itex]G(x,t,x',t') = \frac{1}{2\pi} \int dk e^{-i \frac{\hbar k^2}{2m}(t-t') + i k (x-x')}[/itex]. If [itex]\psi(x,t)[/itex] satisfies Schrodinger's equation [itex]- \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi(x,t) = i \hbar \frac{\partial}{\partial t} \psi(x,t)[/itex], then

    [itex]\psi(x,t) = \int dx' G(x,t, x',t') \psi(x',t')[/itex]
    Point 2: Let [itex]K(x,t,x',t') = \frac{-i \hbar}{(2 \pi)^2} \int dk d\omega \dfrac{e^{-i \omega (t-t') + i k (x-x')}}{(\frac{\hbar^2 k^2}{2m} - \hbar \omega)}[/itex]. Then

    [itex](- \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} - i \hbar \frac{\partial}{\partial t}) K(x,t, x',t') = -i \hbar \delta(x-x') \delta(t-t')[/itex]

    Point 3: [itex]G(x,t,x',t') \neq K(x,t,x',t')[/itex]

    Point 4: Both [itex]G(x,t,x',t')[/itex] and [itex]K(x,t,x',t')[/itex] are sometimes called "Green's functions".

    Which of those 4 do Sakurai and Ballentine say are wrong? Or are you saying that they disagree with each other about it?
     
  12. Nov 4, 2015 #11

    ShayanJ

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    I really should practice being more clear about what I mean. Sorry!

    Both Sakurai and Ballentine use the Green's function for Schrodinger's equation(Green's function in the mathematical sense, which is your K), interchangeably with your G. They actually never define something like G.

    Where definition 4.40 is ## G(x,t;x',t_0)=\langle x| U(t,t_0)|x'\rangle ##.
    And the K he talks about, appears in the equation below:
    ##\psi(\mathbf x'',t)=\int d^3x'K(\mathbf x'',t,\mathbf x',t_0)\psi(\mathbf x',t_0)##
     
  13. Nov 4, 2015 #12

    stevendaryl

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    According to Wikipedia: https://en.wikipedia.org/wiki/Propagator#Non-relativistic_propagators, I have the symbols switched: They use [itex]G[/itex] for the definition I gave for [itex]K[/itex], and vice-versa. They also give a simple relationship between the two:

    [itex]G(x,t,x',t') = \frac{1}{i \hbar} K(x,t,x',t') \theta(t-t')[/itex]

    The [itex]\theta[/itex] function is necessary to give a [itex]\delta(t-t')[/itex]
     
  14. Nov 4, 2015 #13

    stevendaryl

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    . But they are equal when [itex]t > t'[/itex], which is the usual case we are interested in. But from the definitions, it's not clear that they are equal.
     
  15. Nov 4, 2015 #14

    ShayanJ

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    But if we use this G as the kernel in the formula below
    ##\psi(\mathbf x'',t)=\int d^3x' G(\mathbf x'',t,\mathbf x',t_0)\psi(\mathbf x',t_0)##
    we can only integrate forward in time. But because the path integral formulation is equivalent to wave mechanics and we surely can integrate backwards in time using the Schrodinger's equation, we should be able to integrate backwards in time using the above formula too. But it seems to me that the Heaviside function is preventing that and I think that's the reason Sakurai and Ballentine consider G and K to be equivalent.
     
  16. Nov 4, 2015 #15

    vanhees71

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    Green's functions are not uniquely defined just by the equation
    $$\left (\mathrm{i}\partial_t-\hat{H} \right) G(t,\vec{x};t',\vec{x}')=\delta(t-t') \delta^{(3)}(\vec{x}-\vec{x}'),$$
    but you need an additional condition, which tells you which Green's function you want, because obviously it's only defined up to an arbitrary solution of the homogeneous equation of the differential operator on the left-hand side. In non-relativistic QT you usually want the retarded Green's function, which is given by stevendaryl. In relativistic physics, for the usual vacuum QFT you need the time-ordered propagators, and in the most general case of non-equilibrium many-body theory you need the propagator on the Schwinger-Keldysh real-time contour, which can be split into a 2x2-matrix which contains the time-ordered and the anti-time ordered propagators as well as the fixed-ordered Wightman functions (which can be transformed into the retarded, advanced, and the Keldysh two-point function).

    If ##\hat{H}## does not explicitly depend on time, one has
    $$G(t,\vec{x};t,\vec{x}')=G(t-t';\vec{x},\vec{x}'),$$
    and you can use the Fourier representation with respect to time
    $$G(t-t';\vec{x},\vec{x}')=\int_{\mathbb{R}} \frac{\mathrm{d} \omega}{(2 \pi)} \tilde{G}(\omega;\vec{x},\vec{x}') \exp(-\mathrm{i} \omega t).$$
    Plugging this into the defining equation above, leads to
    $$(\omega-\hat{H}) \tilde{G}(\omega;\vec{x},\vec{x}')=\delta^{(3)}(\vec{x}-\vec{x}').$$
    That cries for expressing ##\tilde{G}## in terms of the energy eigenfunctions, ##u_{\alpha}(\vec{x})##, where ##\alpha## is a label, which can be continuous or discrete. For free particles, you can use ##\alpha=\vec{p}## or any other complete set of conserved quantities. If the Hamiltonian is non-degenerate it's sufficient to use the energy eigenvalues for ##\alpha##. No matter what, we can make the ansatz
    $$\tilde{G}(\omega;\vec{x},\vec{x}')=\sum_{\alpha} \tilde{g}(\omega,\alpha; \vec{x}') u_{\alpha}(\vec{x}).$$
    If there are continuous parameters ##\alpha##, the sum becomes an integral, but it doesn't change the conceptual idea. Using the above equation for ##\tilde{G}## and using the completeness relation of the energy-eigenfunctions,
    $$\sum_{\alpha} u_{\alpha}^*(\vec{x}') u_{\alpha}(\vec{x})=\delta^{(3)}(\vec{x}-\vec{x}'),$$
    leads to
    $$[\omega-E(\alpha)] \tilde{g}(\omega,\alpha;\vec{x}')=u_{\alpha}^*(\vec{x}').$$
    Now it's clear that for the Fourier integral you have to make sense of the denominator, and depending on how you deform the integration path along the real ##\omega## axis you get either the retarded or advanced Green's function. Alternatively you can shift the poles into the upper or lower complex ##\omega## plane by an infinitesimal amount. Then you get
    $$\tilde{g}^{(\pm)}(\omega,\alpha;\vec{x}') = \frac{u_{\alpha}^*(\vec{x}')}{\omega-E_{\alpha} \pm \mathrm{i} 0^+}.$$
    To evaluate the Fourier integral, you like to use the theorem of residues, by closing the contour with a large semicircle in the upper or lower half-plane, but you have to close it in the upper (lower) plane if ##t-t'<0## (##t-t'>0##), from which you get
    $$g^{(\pm)}(t-t',\alpha;\vec{x}) = \mp \mathrm{i} \Theta[\pm (t-t')] u_{\alpha}^{*}(\vec{x}') \exp(-\mathrm{i} E_{\alpha} t).$$
    Since both ##g^{(pm}## solve the time-dependent Schrödinger equation for ##t \neq t'##, for these values both Green's functions are identical with the time-evolution kernel you can use both to solve the initial-value problem for the time-dependent Schrödinger equation. Of course you have to use the retarded (+) Green's function for ##t>t'## and the advanced (-) Green's function for ##t<t'##, because otherwise you get the trivial solution ##\psi=0##.

    The original Green's functions now read
    $$G^{(\pm)}(t-t';\vec{x},\vec{x}')=\mp \mathrm{i} \Theta[\pm(t-t')] \sum_{\alpha} u_\alpha^*(\vec{x}') u_{\alpha}(\vec{x}) \exp[-\mathrm{i} E_{\alpha}(t-t')],$$
    and it's very easy to verify that both solve the Green's-function equation by using
    $$\partial_t \Theta [\pm(t-t')]=\pm \delta(t-t'), \quad \hat{H} u_{\alpha}(\vec{x})=E_{\alpha} u_{\alpha}(\vec{x}).$$
     
  17. Nov 4, 2015 #16

    ShayanJ

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    Good clarification vanhees, thanks.
    You said the inhomogeneous equation with dirac deltas in the RHS doesn't uniquely determine the Green's function. But it seems to me that to uniquely determine the Green's function, we should choose a solution of the homogeneous equation and add it to the Green's function we found. But you're not determining that additive function, but a mutiplier. I can still say that I can add a solution of the homogeneous equation to your Green's functions and don't change anything. So you didn't determine it uniquely.
     
  18. Nov 4, 2015 #17

    stevendaryl

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    Hmm. Getting back to the two defining equations for a Green's function:

    1. [itex](H - i \frac{\partial}{\partial t}) G(x,t,x',t') = -i \hbar \delta(x-x') \delta(t-t')[/itex]
    2. [itex]\psi(x,t) = \int dx' G(x,t,x',t') \psi(x',t')[/itex] (for every [itex]\psi[/itex] satisfying the schrodinger equation)
    Clearly, equation 1 is arbitrary to within an additive solution of the homogeneous equation. But it seems to me that equation 2 will not hold with an arbitrary additive function. If [itex]\phi(x,t,x',t')[/itex] satsifies [itex](H - i \frac{\partial}{\partial t}) \phi(x,t, x',t') = 0[/itex], then replacing [itex]G(x,t,x',t')[/itex] by [itex]G(x,t,x',t') + \phi(x,t,x',t')[/itex] will change equation 2 into:

    [itex]\psi(x,t) \ [/itex]?[itex]= (\int dx' G(x,t,x',t') \psi(x',t')) + (\int dx' \phi(x,t,x',t') \psi(x',t')) [/itex]
    [itex] = \psi(x,t) + (\int dx' \phi(x,t,x',t') \psi(x',t')) [/itex]

    Since this equation has to hold for every possible solution [itex]\psi(x,t)[/itex], I think that this implies:
    [itex]\phi(x,t,x',t') = 0[/itex] almost everywhere.

    So the defining integral for [itex]G[/itex] determines [itex]G[/itex] uniquely, it seems to me.
     
  19. Nov 4, 2015 #18

    samalkhaiat

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    What additive solution is that? The equation [tex](H - i \partial_{t}) \ G(x,t ; x_{0} , t_{0}) = 0 ,[/tex] is the Shrodinger equation that you are trying to solve. In fact, for [itex](x,t) \neq (x_{0},t_{0})[/itex], [itex]\psi(x,t) = G(x,t ; x_{0},t_{0})[/itex]. To see that, suppose that the system is located at [itex]x_{0}[/itex] at time [itex]t_{0}[/itex]. So, [itex]| \psi(t_{0}) \rangle = |x_{0}\rangle[/itex] and, therefore [tex]\psi(\bar{x}, t_{0}) = \langle \bar{x}| \psi(t_{0}) \rangle = \langle \bar{x} | x_{0} \rangle = \delta(\bar{x} - x_{0}) .[/tex] Inserting this in the solution
    [tex]\psi(x,t) = \int d \bar{x} \ G(x,t;\bar{x},t_{0}) \ \psi(\bar{x},t_{0}) ,[/tex] we get [tex]\psi(x,t) = G(x,t ; x_{0},t_{0}) .[/tex] This tells us that the wave function at [itex]t[/itex] is the quantum amplitude for a system to go from [itex](x_{0},t_{0})[/itex] to [itex](x,t)[/itex].

    I will clarify the mess in this thread later.
     
  20. Nov 4, 2015 #19

    stevendaryl

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    Thanks for the warning.
     
  21. Nov 5, 2015 #20

    vanhees71

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    I thought in my treatment above, it became very clear that the solution of the initial-value problem
    $$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}), \quad \psi(t=0^+,\vec{x})=\psi_{0}(\vec{x}),$$
    with an arbitrary ##\mathrm{L}^2(\mathbb{R}^3)## function is UNIQUELY given by the retarded propagator, via
    $$\psi(t,\vec{x})=\int_{\mathbb{R}^3} G_{\text{ret}}(t;\vec{x},\vec{x}') \psi_0(\vec{x}'),$$
    where ##G_{\text{ret}} \equiv G^{(+})## in the notation of my previous posting.

    Note that this is the exact propagator, not the free-particle propagator used in the Born series of collision theory (where in fact one must use in general another one, which has the correct asymptotic behavior at inifinite, if you have long-range forces like the Coulomb force, but that's another story).
     
  22. Nov 6, 2015 #21

    samalkhaiat

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    No warning was intended :wink:
     
  23. Nov 6, 2015 #22

    samalkhaiat

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    I will give neat presentation of the subject staying away from any particular representation space as much as I can. At the end, I will translate the operator equations into coordinate space simply by sandwiching them between complete set of position eigenstates. Let us start from the representation-free Schrodinger equation
    [tex]
    (H - i \partial_{t}) | \psi (t)\rangle = 0 . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
    [/tex]
    In operator notation, the Green’s function [itex]G[/itex] for (1) satisfies the equation
    [tex]
    (H - i \partial_{t}) G(t,t_{0}) = - i \mathbb{I} \delta(t-t_{0}) . \ \ \ \ \ \ \ \ \ \ \ (2)
    [/tex]
    So, finding [itex]G[/itex] provides a solution to the time-dependent Schrodinger equation, (1), in the sense that if [itex]|\psi(t_{0}) \rangle[/itex] is the state of the system at [itex]t_{0}[/itex], the state at a later time [itex]t > t_{0}[/itex] is given by
    [tex]
    |\psi(t)\rangle = G(t,t_{0}) \ |\psi(t_{0}) \rangle . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)
    [/tex]
    The restriction [itex]t > t_{0}[/itex] in (3) can be replaced by the step function [itex]\theta(t-t_{0})[/itex] as follow
    [tex]
    \theta(t-t_{0}) \ |\psi(t)\rangle = G(t,t_{0}) \ |\psi(t_{0}) \rangle . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)
    [/tex]
    It is now simple exercise to show that (4) leads to the Schrodinger Green’s function equation (2): just apply the operator [itex](H - i \partial_{t})[/itex] to both sides of equation (4) and use the Schrodinger equation (1).
    Note that [itex]G(t,t_{0}) = 0[/itex] for [itex]t < t_{0}[/itex], so that (in coordinate space) [itex]G[/itex] is the retarded Green’s function.
    For time independent [itex]H[/itex] an operator solution of (2) can be written down
    [tex]
    G(t,t_{0}) = \theta(t-t_{0}) \ e^{- i H(t-t_{0})} \equiv \theta(t-t_{0}) U(t,t_{0}) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)
    [/tex]
    So, for [itex]t > t_{0}[/itex], [itex]G[/itex] coincide with unitary time evolution operator [itex]U[/itex]. In coordinate space, this will translate into a similar relationship between the Green’s function [itex]G(x,t|x_{0},t_{0})[/itex] and the propagator [itex]K(x,t|x_{0},t_{0})[/itex]. Notice that [itex]U \ \mbox{and}\ K[/itex] satisfy the Schrödinger equations, while [itex]G(t,t_{0}) \ \mbox{and} \ G(x,t|x_{0},t_{0})[/itex] satisfy the Schrödinger Green’s function equations.
    Of interest also is the Fourier transform of (5), namely (setting [itex]t_{0}=0[/itex])
    [tex]
    \begin{align*}
    \tilde{G}(\omega) & = \lim_{\epsilon \to 0} \int_{-\infty}^{+\infty} dt \ G(t) e^{ i (\omega + i \epsilon) t} \\
    & = \lim_{\epsilon \to 0} \int_{0}^{+\infty} dt \ e^{ i (\omega - H + i \epsilon) t}
    \end{align*}
    [/tex]
    For [itex]\epsilon > 0[/itex] we have for the energy Green’s function
    [tex]
    \tilde{G}(\omega) = \lim_{\epsilon \to 0} \frac{i}{\omega - H + i \epsilon} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)
    [/tex]
    This equation is the starting point for perturbation theory. Spliting the total Hamiltonian [itex]H[/itex] into unperturbed part [itex]H_{0}[/itex] plus small perturbation [itex]\lambda V[/itex], and using the operator identity
    [tex]
    \frac{i}{A-B} = \frac{i}{A} + \frac{i}{A} (- i B) \frac{i}{A-B} , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7)
    [/tex]
    (to verify this multiply by [itex](A-B)[/itex] from the right) with
    [tex]
    A = \omega - H_{0} + i \epsilon , \ \ \ \ B = \lambda V , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)
    [/tex] we obtain
    [tex]
    \tilde{G}(\omega) = \tilde{G}^{(0)}(\omega) + \tilde{G}^{(0)}(\omega) (- i \lambda V) \tilde{G}(\omega) , \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)
    [/tex]
    where
    [tex]
    \tilde{G}^{(0)}(\omega) = \frac{i}{\omega - H_{0} + i \epsilon} , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)
    [/tex]
    is the energy Green’s function of the unperturbed system. Perturbation series is obtained by iterating (9)
    [tex]
    \tilde{G}(\omega) = \tilde{G}^{(0)} + \tilde{G}^{(0)} (- i \lambda V) \tilde{G}^{(0)} + \tilde{G}^{0} (- i \lambda V) \tilde{G}^{(0)}(- i \lambda V) \tilde{G}^{(0)} + \cdots . \ \ (11)
    [/tex]
    Okay, let us now go back to the beginning and translate our equations to coordinate space. Inserting the unit operator
    [tex]\int dy \ |y \rangle \langle y | = \mathbb{I} ,[/tex] and using [tex]\langle x | H(\hat{q},\hat{p}) | y \rangle = H(x,\partial_{x}) \ \delta(x-y) ,[/tex] equations (1) and (2) become
    [tex](H(x,\partial_{x} ) - i \partial_{t}) \psi(x,t) = 0 , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1a)[/tex]
    [tex]\left(H(x,\partial_{x} ) - i \partial_{t}\right) G(x,t|x_{0},t_{0}) = - i \delta(x-x_{0}) \delta(t-t_{0}) , \ \ \ \ \ \ \ \ \ \ \ \ (2a)[/tex]
    where [itex]G(x,t|x_{0},t_{0}) \equiv \langle x | G(t,t_{0}) | x_{0}\rangle[/itex].

    Sandwiching (5) between [itex]\langle x|[/itex] and [itex]|x_{0}\rangle[/itex] we obtain
    [tex]
    G(x,t|x_{0},t_{0}) = \theta(t-t_{0}) \ K(x,t|x_{0},t_{0}) = \theta(t-t_{0}) \langle x | e^{- i H (t-t_{0})} | x_{0}\rangle , \ \ \ \ \ \ \ \ \ \ \ \ \ (5a)
    [/tex]
    where the propagator [itex]K[/itex] is defined by the matrix element
    [tex]K(x,t|x_{0},t_{0}) = \langle x | U(t-t_{0}) | x_{0}\rangle . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5ab)[/tex]
    Multiplying (4) from the left by [itex]\langle x|[/itex] and inserting the unit [itex]\int dx_{0} \ |x_{0} \rangle \langle x_{0} | = \mathbb{I}[/itex], we get
    [tex]
    \theta(t-t_{0}) \ \psi(x,t) = \int dx_{0} \ G(x,t|x_{0},t_{0}) \ \psi(x_{0},t_{0}) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4a)
    [/tex]
    So, for [itex]t > t_{0}[/itex] we have
    [tex]
    \begin{align*}
    \psi(x,t) & = \int dx_{0} \ G(x,t|x_{0},t_{0}) \ \psi(x_{0},t_{0}) \\
    & = \int dx_{0} \ K(x,t|x_{0},t_{0}) \ \psi(x_{0},t_{0})
    \end{align*} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4ab)
    [/tex]
    You can easily show that the definition of the propagator (5ab) or (4ab) implies a group-like property. That is, for [itex]t_{2}>t_{1}>t_{0}[/itex], the two successive processes
    [tex](x_{0},t_{0}) \to (x_{1},t_{1}) \to (x_{2},t_{2})[/tex] is equivalent to the single process [tex](x_{0},t_{0}) \to (x_{2},t_{2}) .[/tex]
    Thus, the following is satisfied by the propagator [itex]K[/itex] as well as the Green’s function [itex]G[/itex]
    [tex]
    K(x_{2},t_{2}|x_{0},t_{0}) = \int dx_{1} \ K(x_{2},t_{2}|x_{1},t_{1}) \ K(x_{1},t_{1}|x_{0},t_{0}) , \ \ \ \ \ \ \ \ \ (4abc)
    [/tex]
    This simple equation is an important consistency requirement underlying the whole path-integral theory. When the path-integral formalism is applied to Brownian motion it is known as the Chapman-Kolmogorov equation. In fact, any Markovian process is defined by a pair of equations similar to (4ab) and (4abc).
    Suppose we have a complete set of energy eigen-states [itex]H|n\rangle = E_{n}|n\rangle[/itex] and the corresponding wavefunctions [itex]u_{n}(x) = \langle n | x \rangle[/itex], [itex]n=1,2, \cdots [/itex]. Then, for [itex]t > 0[/itex] we have
    [tex]
    G(x, t| x_{0},0) = \sum_{n,m} \langle x | n \rangle \langle n| e^{- i Ht}| m \rangle \langle m | x_{0} \rangle ,
    [/tex]
    or
    [tex]
    G(x, t| x_{0},0) = \sum_{n} u_{n}(x) u_{n}^{*}(x_{0}) \ e^{- i E_{n}t} , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (12)
    [/tex]
    Taking the trace of [itex]G[/itex], i.e., setting [itex]x_{0} = x[/itex] in (12) and integrating over all [itex]x[/itex], and using the normalization of the energy eigen-functions, we find
    [tex]
    \int dx \ G(x,t|x,0) = \sum_{n} e^{ - i E_{n} t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (13)
    [/tex]
    Let’s consider this equation for Euclidean time [itex]\tau = it[/itex]. For large positive [itex]\tau[/itex], the main contribution to the sum in (13) will come from the lowest energy level (the ground state energy):
    [tex]
    \int dx \ G(x, -i \tau |x,0) \approx e^{-E_{0}\tau} , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (14)
    [/tex]
    or
    [tex]
    \lim_{\tau \to \infty} e^{E_{0}\tau}\int dx \ G(x, -i \tau |x,0) = 1 . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (15)
    [/tex]
    If there is no bound state, (15) picks up the bottom of the continuum. The useful Feynman-Kac relation, which relate the asymptotic behaviour of the Green’s function to the ground state energy, can now be obtained by taking the logarithms of (14)
    [tex]
    E_{0} = \lim_{\tau \to \infty} \frac{-1}{\tau} \ln \int dx G(x,-i\tau | x,0) = \lim_{\tau \to \infty} \frac{-1}{\tau} \ln \mbox{Tr}\left(e^{- H \tau}\right) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (16)
    [/tex]
    Finally, we work out the coordinate representation for the energy Green’s function (6) as follow
    [tex]
    \begin{align*}
    \tilde{G}(x,x_{0};\omega) & = \langle x \left| \frac{i}{\omega - H + i \epsilon}\right| x_{0} \rangle \\
    & = \sum_{m,n}\langle x | n \rangle \langle n \left| \frac{i}{\omega - H + i \epsilon}\right| m \rangle \langle m | x_{0}\rangle \\
    & = i \sum_{n} \frac{u_{n}(x) u^{*}_{n}(x_{0})}{\omega - E_{n} + i \epsilon}
    \end{align*}
    [/tex]
    This equation can also be obtained by taking the fourier transform of (12). It allows us to identify the poles in with the location of bound states and the residues with bound states wavefunctions. For continuous spectrum, [itex]\tilde{G}[/itex] has a cut. All this, of course is well known in the usual QM.
    I also wanted to show you a nice and short derivation of the free particle Green’s function but I run out of time, so maybe I do that some other time.
     
    Last edited: Nov 6, 2015
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