# Ground state wave function from Euclidean path integral

• A
• ShayanJ
In summary: This time, the breaking down will give us ## \phi_0^*(x) ##. So far so good. But if you compare this with the first result we got, you'll see that the ## e^{iE_bt} ## in the first one is changed to ## e^{-iE_bt} ## in the second one. This is because of the definition of the Green function. Because of this change, the limits of the path integrals are changed too. And it seems that this change is making a problem for me. I hope you understood my problem.In summary, the path integral approach is a useful tool for understanding quantum mechanics. By using the Green function, the path integral can be written as a
ShayanJ
Gold Member
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

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

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

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

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

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

Given this asymptotic expression, we can get $E_0$ as a limit:

$E_0 = lim_{\beta \rightarrow \infty} \frac{-log(G(x,x_i,\beta))}{\beta}$

In terms of $E_0$, we can get another limit:

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

So taking the limit as $\beta \rightarrow \infty$ or $t \rightarrow +i \infty$ gives us information about the product $\phi_0^*(x) \phi_0(x_i)$. I don't see how it gives us either $\phi_0^*(x)$ or $\phi_0(x_i)$ separately.

stevendaryl said:
In terms of $E_0$, we can get another limit:

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

So taking the limit as $\beta \rightarrow \infty$ or $t \rightarrow +i \infty$ gives us information about the product $\phi_0^*(x) \phi_0(x_i)$. I don't see how it gives us either $\phi_0^*(x)$ or $\phi_0(x_i)$ separately.

You could get $\phi_0(x)$ up to a phase:

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

Then you take the square-root to get $\phi_0(x)$ up to an unknown phase.

## \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.

ShayanJ said:
## \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.

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

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

and

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

I don't understand why the limits would be different.

stevendaryl said:
I don't understand why the limits would be different.
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.

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

$\langle x|e^{-i H t}|x_i \rangle$

not

$\langle x|e^{+i H t}|x_i \rangle$

Are you sure about the plus sign?

Demystifier and vanhees71
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 ##.

ShayanJ said:
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]} ##.

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

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

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 $\langle x|e^{-iHt}|x_i\rangle$, and take the limit $t \rightarrow -i \infty$
• Start with $\langle x|e^{+iHt}|x_i\rangle$, and take the limit $t \rightarrow +i\infty$
Those both give exactly the same thing, which is $\phi_0(x) \phi_0^*(x_i)$

Or you can switch $x$ and $x_i$ (which is the same as taking the complex conjugate and flipping the time):
• Start with $\langle x_i|e^{-iHt}|x\rangle$, and take the limit $t \rightarrow -i \infty$
• Start with $\langle x_i|e^{+iHt}|x\rangle$, and take the limit $t \rightarrow +i\infty$
Those give you the complex conjugate: $\phi_0^*(x) \phi_0(x_i)$

ShayanJ
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.

ShayanJ
ShayanJ said:
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

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

vanhees71 and ShayanJ
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
$$\psi(t,\vec{x})=_S\langle \vec{x},t|\psi \rangle_S = \langle \vec{x},0|\exp(-\mathrm{i} \hat{H} t)| \psi,0 \rangle = \langle \exp(+\mathrm{i} \hat{H} t) \vec{x},0|\psi,0 \rangle = _H \langle \vec{x},t |\psi,t \rangle_H.$$

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

## 1. What is the ground state wave function?

The ground state wave function is a mathematical description of the lowest energy state of a quantum mechanical system. It represents the probability amplitude of finding a particle in a specific location and with a specific energy in the ground state of the system.

## 2. What is the Euclidean path integral?

The Euclidean path integral is a mathematical tool used in quantum field theory and statistical mechanics to calculate the probability amplitudes of different quantum states. It involves summing over all possible paths that a particle could take between two points in space and time.

## 3. How is the ground state wave function related to the Euclidean path integral?

The ground state wave function can be obtained from the Euclidean path integral by taking the limit of the path integral as the time interval between the two points approaches zero. This yields a stationary phase approximation, which gives the most probable path that the particle would take in the ground state.

## 4. What does the ground state wave function tell us about a system?

The ground state wave function provides information about the spatial distribution and energy of a system in its lowest energy state. It can also reveal the probability of finding a particle at a specific location and energy, and how that probability changes over time.

## 5. How is the ground state wave function useful in science?

The ground state wave function is fundamental to understanding the behavior of quantum mechanical systems. It allows scientists to calculate and predict the properties and behavior of particles at the smallest scales, which has numerous applications in fields such as particle physics, chemistry, and material science.

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