# A Ground state wave function from Euclidean path integral

1. Dec 9, 2016

### ShayanJ

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. Dec 9, 2016

### stevendaryl

Staff Emeritus
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.

3. Dec 9, 2016

### stevendaryl

Staff Emeritus
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.

4. Dec 9, 2016

### ShayanJ

$\phi_0(x_i)$ 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.

5. Dec 9, 2016

### stevendaryl

Staff Emeritus
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.

6. Dec 9, 2016

### ShayanJ

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.

7. Dec 9, 2016

### stevendaryl

Staff Emeritus
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?

8. Dec 9, 2016

### ShayanJ

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 $\langle x_i|x,t\rangle=\langle x_i|e^{-iHt}|x\rangle$.

9. Dec 9, 2016

### stevendaryl

Staff Emeritus
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: $\phi_0^*(x) \phi_0(x_i)$

10. Dec 9, 2016

### vanhees71

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.

11. Dec 14, 2016

### samalkhaiat

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, $$\lim_{t \to \infty} K(x,t ; 0,0) \sim \varphi_{0}(x) = \langle x | 0 \rangle .$$

12. Dec 15, 2016

### vanhees71

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

13. Dec 26, 2016

### ShayanJ

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