Path integral in momentum representation

[jostpuur]

Is it possible to derive the Shrodinger's equation

[tex]
i\hbar\partial_t \Psi(t,p) = \frac{|p|^2}{2m}\Psi(t,p)
[/tex]

in momentum representation directly from a path integral?

If I first fix two points [itex]x_1[/itex] and [itex]x_2[/itex] in spatial space, solve the action for a particle to propagate between these point in time [itex]\Delta t[/itex], and then take the limit [itex]\Delta t\to 0[/itex], the action approaches infinity.

If I instead fix two points [itex]p_1[/itex] and [itex]p_2[/itex] in the momentum space, solve the action for a particle to propagate between these points in time [itex]\Delta t[/itex], and then take the limit [itex]\Delta t\to 0[/itex], the action does not approach infinity, but instead zero. So it looks like stuff goes somehow differently here.

 

[quetzalcoatl9]

first of all, do you agree that

[tex]\Psi(x_f, t_f) = e^{\frac{iSt}{\hbar}}\Psi(x_i, t_i)[/tex]

is equivalent to the TDSE? all you need to do is take the time derivative and know from classical mechanics that [tex]E_f = \frac{\partial S}{\partial t_f}[/tex]

 

[jostpuur]

first of all, do you agree that

[tex]\Psi(x_f, t_f) = e^{\frac{iSt}{\hbar}}\Psi(x_i, t_i)[/tex]

is equivalent to the TDSE? all you need to do is take the time derivative and know from classical mechanics that [tex]E_f = \frac{\partial S}{\partial t_f}[/tex]

Not really, unless there is a misunderstanding with the notation. Wouldn't

[tex]
\Psi(x_f,t_f) = N\int dx_i\; e^{iS/\hbar} \Psi(x_i,t_i) + O((t_f-t_i)^2)
[/tex]

be better?

 

[quetzalcoatl9]

Not really, unless there is a misunderstanding with the notation. Wouldn't

[tex]
\Psi(x_f,t_f) = N\int dx_i\; e^{iS/\hbar} \Psi(x_i,t_i) + O((t_f-t_i)^2)
[/tex]

be better?

sorry, its not [tex]e^{iSt/\hbar}[/tex] it should be [tex]e^{iS/\hbar}[/tex]

 

[quetzalcoatl9]

Not really, unless there is a misunderstanding with the notation. Wouldn't

[tex]
\Psi(x_f,t_f) = N\int dx_i\; e^{iS/\hbar} \Psi(x_i,t_i) + O((t_f-t_i)^2)
[/tex]

be better?

why would you do that? what i originally had (minus the typo) was correct

also, btw, you could do the other way around via

[tex]
Q = e^{-\beta \hat{H}}
[/tex]

impose the trotter thm:

[tex]
e^{A + B} = \lim_{P \rightarrow \infty} (e^{A/P} e^{B/P})^P
[/tex]

to separate the kinetic and potential parts. then use the resolution of the identity to insert the complete position and momenta states. when you're done you will arrive at the path integral.

the P has the significance of being the Trotter ("bead") number, which we tune when doing numerical simulation via path integrals.

 

[jostpuur]

first of all, do you agree that

[tex]\Psi(x_f, t_f) = e^{\frac{iSt}{\hbar}}\Psi(x_i, t_i)[/tex]

is equivalent to the TDSE? all you need to do is take the time derivative and know from classical mechanics that [tex]E_f = \frac{\partial S}{\partial t_f}[/tex]

What is [itex]x_i[/itex] on the right side?

 

[jostpuur]

quetzalcoatl9, the way you insist that the integral is not needed remainds me of an equation that was something like this (it could be I remember some details wrong, I'm not checking now)

[tex]
\int dx_2\; N^2 e^{iS(x_3,x_2)/\hbar} e^{iS(x_2,x_1)/\hbar} = N e^{iS(x_3,x_1)/\hbar}
[/tex]

where [itex]S(y,x)[/itex] means action to move from x to y. Could it be that you are confusing this equation with my OP? Otherwise, I cannot understand what's the idea behind your response.

 

[genneth]

The classical Hamilton function S is defined over two points in position space -- because the Lagrangian is defined over position and the first derivative of position, and not over momentum. How did you calculate the action given momentum of the end points?

 

[jostpuur]

The classical Hamilton function S is defined over two points in position space -- because the Lagrangian is defined over position and the first derivative of position, and not over momentum. How did you calculate the action given momentum of the end points?

I'm interested in the non-relativistic case now, so momentum and velocity are more or less the same thing now.

In position representation we can assume, that the particle travels from point [itex]x_1[/itex] to point [itex]x_2[/itex] linearly, that means like

[tex]
x(t) = \frac{x_2(t-t_0) + x_1(t_0+\Delta t - t)}{\Delta t}.
[/tex]

Now [itex]x(t_0) = x_1[/itex] and [itex]x(t_0+\Delta t)=x_2[/itex]. The Lagrange's function is now defined along this path, in fact it is constant, and the action is integral of the L in time.

Analogously, I thought that in momentum space we should consider linear paths between points [itex]p_1[/itex] and [itex]p_2[/itex]. So the velocity of the particle along this path is

[tex]
v(t) = \frac{v_2(t-t_0) + v_1(t_0+\Delta t - t)}{\Delta t}.
[/tex]

So the action is

[tex]
\int\limits_{t_0}^{t_0+\Delta t} L(t)dt = \int\limits_{t_0}^{t_0+\Delta t} \frac{m}{2}\Big| \frac{v_2(t-t_0) + v_1(t_0+\Delta t-t)}{\Delta t}\Big|^2 dt.
[/tex]

If I did it correctly, then it is

[tex]
S=\frac{m\;\Delta t}{2}\Big(\frac{1}{3}|v_2-v_1|^2 - v_2\cdot(v_2-v_1) + |v_2|^2\Big).
[/tex]

(Written so that [itex]v_1[/itex] dependece is in form of [itex]v_2-v_1[/itex] dependece.)

 

[genneth]

I presume you mean to define L to be p^2/2m, and consider it a function of p and [tex]\dot{p}[/tex]. Unfortunately, as you can see for yourself, applying Euler-Lagrange doesn't give the desired result, giving only that p=0.

Although it is true that in the canonical, i.e. Hamiltonian, approach one can interchange position and momenta freely, it is not true in the Lagrangian aka action approach. In the latter you must define what is position and what is momentum. You can try the following: the interchange of position and momenta in the Hamiltonian will require a generating function, which will be added to the Hamiltonian. You can then try converting that back into a Lagrangian, which will be different to what the naive way will be, and see if you can derive an action that way.

I'm sorry to sound pessimistic -- I'm actually very much intrigue by this.

Good luck!

 

[jostpuur]

I presume you mean to define L to be p^2/2m, and consider it a function of p and [tex]\dot{p}[/tex]. Unfortunately, as you can see for yourself, applying Euler-Lagrange doesn't give the desired result, giving only that p=0.

Good point. I never noticed this before.

Although it is true that in the canonical, i.e. Hamiltonian, approach one can interchange position and momenta freely, it is not true in the Lagrangian aka action approach. In the latter you must define what is position and what is momentum. You can try the following: the interchange of position and momenta in the Hamiltonian will require a generating function, which will be added to the Hamiltonian. You can then try converting that back into a Lagrangian, which will be different to what the naive way will be, and see if you can derive an action that way.

What is this generating function you are talking about?

I'm sorry to sound pessimistic -- I'm actually very much intrigue by this.

Good luck!

I noticed the calculation wasn't working. That's why I posted this! But I was hoping that it could be made to work, and I'm in fact still hoping so. Something should give the Shrodinger's equation in momentum space out of the Lagrange's function. It's just not the same action principle. It should be replaced with something else.

 

[genneth]

The canonical approach is all about transforming things into a more useful coordinate. In fact, the Hamilton function S is exactly that: it specifies a transform where all the position and momenta are constant! That is why we say that it encodes all the dynamics -- it's reduced them to the simplest possible form. I strongly encourage you to learn this stuff: the standard text is Goldstein, Classical Mechanics.

 

[olgranpappy]

Is it possible to derive the Shrodinger's equation

[tex]
i\hbar\partial_t \Psi(t,p) = \frac{|p|^2}{2m}\Psi(t,p)
[/tex]

in momentum representation directly from a path integral?

Yeah it is. Learn from the master himself:

R. P. Feynman (and Hibbs), "Quantum Mechanics and Path Integrals".

 

[jostpuur]

the Hamilton function S is exactly that: it specifies a transform where all the position and momenta are constant!

I thought I knew this stuff when I knew the Hamilton's function H and the action S.

Yeah it is. Learn from the master himself:

R. P. Feynman (and Hibbs), "Quantum Mechanics and Path Integrals".

You wouldn't bother giving some hints on what he does? I'm probably not buying more books very soon.

 

[quetzalcoatl9]

quetzalcoatl9, the way you insist that the integral is not needed remainds me of an equation that was something like this (it could be I remember some details wrong, I'm not checking now)

[tex]
\int dx_2\; N^2 e^{iS(x_3,x_2)/\hbar} e^{iS(x_2,x_1)/\hbar} = N e^{iS(x_3,x_1)/\hbar}
[/tex]

where [itex]S(y,x)[/itex] means action to move from x to y. Could it be that you are confusing this equation with my OP? Otherwise, I cannot understand what's the idea behind your response.

what i wrote was the propagator acting on the initial wavefunction, it's a standard result from QM

 

[quetzalcoatl9]

You wouldn't bother giving some hints on what he does? I'm probably not buying more books very soon.

it's just as well, you can't get it anymore for less than $300 (out of print long ago, plus it's a fantastic book that covers things still, for some reason, neglected in other texts)

 

[Hans de Vries]

You wouldn't bother giving some hints on what he does? I'm probably not buying more books very soon.

It's in Feynman's paper: "Space time approach to non-relativistic quantum mechanics"

The paper is included in many books. probably the most useful of these books is:

"Selected papers on Quantum Electrodynamics" edited by Julian Schwinger, which is a
"must have" collection of historical papers and it goes for as cheap as $6 on Amazon.com :

https://www.amazon.com/dp/0486604446/?tag=pfamazon01-20The paper is also included in "Feynman's Thesis" a book recently brought to the market
by Laurie Brown, (who wrote papers together with Feynman). Feynman's thesis is on
the principle of least action in Quantum Mechanics:

https://www.amazon.com/dp/9812563806/?tag=pfamazon01-20Regards, Hans

 

[jostpuur]

what i wrote was the propagator acting on the initial wavefunction, it's a standard result from QM

For two fixed points [itex]x_i[/itex] and [itex]x_f[/itex], [itex]\exp(\frac{i}{\hbar}S(\Delta t,x_f,x_i))[/itex] is a number. It is not an operator you can put on the left side of the initial wave function alone. You must integrate over the [itex]x_i[/itex] to get the amplitude [itex]\psi(t+\Delta t,x_f)[/itex].

 

[quetzalcoatl9]

For two fixed points [itex]x_i[/itex] and [itex]x_f[/itex], [itex]\exp(\frac{i}{\hbar}S(\Delta t,x_f,x_i))[/itex] is a number. It is not an operator you can put on the left side of the initial wave function alone. You must integrate over the [itex]x_i[/itex] to get the amplitude [itex]\psi(t+\Delta t,x_f)[/itex].

S is the action associated with the path from i to f

[tex]\Psi_f = e^{\frac{i}{\hbar}S} \Psi_i[/tex]

[tex]\frac{\partial \Psi_f}{\partial t_f} = \frac{i}{\hbar} \frac{\partial S}{\partial t_f}e^{\frac{i}{\hbar}S} \Psi_i[/tex]

[tex]\frac{\partial \Psi_f}{\partial t_f} = \frac{i}{\hbar} \frac{\partial S}{\partial t_f} \Psi_f[/tex]

[tex]\frac{\partial S}{\partial t_f} = -E_f[/tex]

[tex]\frac{\partial \Psi_f}{\partial t_f} = -\frac{i}{\hbar} E_f \Psi_f[/tex]

[tex]i \hbar \frac{\partial \Psi_f}{\partial t_f} = E_f \Psi_f[/tex]

 

[jostpuur]

S is the action associated with the path from i to f

[tex]\Psi_f = e^{\frac{i}{\hbar}S} \Psi_i[/tex]

I was talking about action to go from location [itex]x_i[/itex] to location [itex]x_f[/itex], but you have an action to move from quantum mechanical state [itex]\Psi_i[/itex] to state [itex]\Psi_f[/itex]. How is this action defined?

If S is still a number, then [itex]\exp(iS/\hbar)[/itex] is also number, and I don't think that in general [itex]\Psi_i[/itex] and [itex]\Psi_f[/itex] are linearly dependent so that we can get the final state by multiplying the initial state with a number.

 

[jostpuur]

solution?

Now when genneth made the remark that action principle is not even supposed to work the same way in the momentum representation, as it does in the position representation, I think that the answer to the question "how do you solve SE in momentum representation?" is nothing else but "move onto position representation, solve SE, and move back to momentum representation!"

The result is this

[tex]
\psi(t+\Delta t,p) = N\int\frac{d^3x\; d^3x'\; d^3p'}{(2\pi\hbar)^3} e^{iS(x,x')/\hbar} e^{i(p'\cdot x' - p\cdot x)/\hbar} \psi(t,p') + O(\Delta t^2).
[/tex]

Edit: hmhm.. I made it sound funnier than it really was. That's not yet "solving SE". Anyway, something like that... "write down the time evolution in position representation".

 

[olgranpappy]

If S is still a number, then [itex]\exp(iS/\hbar)[/itex] is also number, and I don't think that in general [itex]\Psi_i[/itex] and [itex]\Psi_f[/itex] are linearly dependent so that we can get the final state by multiplying the initial state with a number.

What our Aztec friend wrote is not correct, but neither is what you wrote; certainly, I may obtain the time-evolved state of an eigenstate of the Hamiltonian by simple "multiplication... with a number". I.e., by [tex]e^{-iE(t_2-t_1)}[/tex].

In general the Schrodinger equation is exactly equivalent to an integral equation:

[tex]
\Psi(\vec x;t_f)=\int d^3 y K(\vec x,\vec y;t_f-t_i)\Psi(\vec y;t_i)
[/tex]

where the integration is only over space and thus, as far as time is concerned, this is also in a sense just "multiplication... with a number"... of course, it is much more than that really, and is not at all simply the same as multiplying by e^{iS}. The other important thing to note is that the kernal (K) is homogeneous in time.

Anyways, now the whole problem of quantum mechanics is to calculate K... so what is K, exactly... it's something like:

[tex]
K(x,y;t)=\int_x^y Dq e^{-iS[q]}
[/tex]

where the paths q which I am integrating over have endpoints fixed at q(0)=x and q(t)=y.

Oh, another name for K is a "Green's function".

Cheers.

 

[jostpuur]

W but neither is what you wrote; certainly, I may obtain the time-evolved state of an eigenstate of the Hamiltonian by simple "multiplication... with a number". I.e., by [tex]e^{-iE(t_2-t_1)}[/tex].

To be fully precise, I said that initial and final states are not linearly dependent in general. They may be linearly dependent in some special cases, though, which is not a contradiction.