# A Question about equivalence of Path Integral and Schrodinger

#### stevendaryl

Staff Emeritus
I've seen a proof that the path integral formulation of quantum mechanics is equivalent to solving Schrodinger's equation. However, it appears to me that the proof actually depended on the Hamiltonian having a particular form. I'm wondering how general is the equivalence.

Let me sketch a handwavy derivation of path integrals from Schrodinger's equation to illustrate where I think the problem might lie.

Suppose we want to find a green function $G(x,y,t)$ with the property that if $\psi(x,t)$ is a solution to Schrodinger's equation, then $\psi(x,t) =\int G(x,y,(t-t_0)) \psi(y,t_0) dy$.

Formally, we can write down the answer as follows:

$G(x,y,t) = e^{-iHt/\hbar} \delta(x-y)$

where $H$ is the hamiltonian. To see that this solves the equation:

$\int G(x,y,(t-t_0)) \psi(y,t_0) dy = \int e^{-iH(t-t_0)/\hbar} \delta(x-y) \psi(y,t_0) dy = e^{-iH(t-t_0)/\hbar} \int \delta(x-y) \psi(y,t_0) dy = e^{-iH(t-t_0)/\hbar} \psi(x,t_0)$

Now, the question is to evaluate $e^{-iHt/\hbar} \delta(x-y)$. We can easily prove the following fact about Green functions:

$G(x,y,t) = \int dx_1 dx_2 ... dx_{N-1} G(x, x_{N-1}, \frac{t}{N}) G(x_{N-1}, x_{N-2}, \frac{t}{N}) ... G(x_1,y, \frac{t}{N})$

So we can pick $N$ large enough that $\delta t \equiv \frac{t}{N}$ is small. So the question really becomes expressing $G(x,y,\delta t)$ in the limit as $\delta t$ is small. If we know what $G$ is like for small times, then we can use the above integral to compute it for larger times. Going back to our original expression for $G$, we have:

$G(x,y,\delta t) = e^{-iH \delta t/\hbar} \delta(x-y)$

We assume that $H$ is an expression $H(p,x)$ depending on the operators $p$ and $x$. Here's the first approximation (so far, everything has been exact):

We assume that if you have a function $f(x,t)$ which is strongly localized near $x=y$ at time $t=0$, and you let it evolve under Schrodinger's equation, then after a small amount of time $\delta t$, the function will still be strongly localized near $x=y$. In that case, the spatial dependence of the hamiltonian is likely to be irrelevant. If $H(p,x) = \frac{p^2}{2m} + V(x)$, then as long as the wave function is strongly localized near $x=y$, then there should be a negligible error if you replace $V(x)$ by $V(y)$. This can probably be made more rigorous, but I'm not too concerned about it. So our first approximation is this:

$G(x,y,\delta t) = e^{-iH(p,x) \delta t/\hbar} \delta(x-y) \approx e^{-iH(p,y) \delta t/\hbar} \delta(x-y)$

Since the delta function is so strongly localized, it only "feels" the potential near $x=y$, so we make a negligible error by replacing $H(p,x)$ by $H(p,y)$. So we're replacing the potential by a constant potential.

At this point, $H(p,y)$ no longer involves $x$, which suggests that we should evaluate it when applied to momentum eigenstates. We use the usual representation of $\delta(x-y)$:

$\delta(x-y) = \frac{1}{2\pi \hbar} \int dp e^{i p (x-y)/\hbar}$

Then we can write:

$e^{-iH(p,y)\delta t/\hbar} \delta(x-y) = \frac{1}{2\pi \hbar} \int dp e^{-iH(p,y)\delta t/\hbar} e^{i p (x-y)/\hbar}$

I was a little sloppy in distinguishing between $\hat{p}$, the operator, and $p$ the variable of integration. However, since

$\hat{p} e^{i p(x-y)/\hbar} = p e^{i p(x-y)/\hbar}$,

bringing the operator $e^{-iH(p,y)\delta t/\hbar}$ inside the integral allows us to replace $\hat{p}$ by $p$, its eigenvalue.

So now we have an expression for $G(x,y,\delta t)$, valid for small $\delta t$, which no longer involves any operators at all:

$G(x,y,\delta t) \approx \frac{1}{2\pi \hbar} \int dp e^{-iH(p,y) \delta t/\hbar + i p (x-y)/\hbar}$

Now, to get further, we have to do another approximation to reduce that integral to something doable. We use the method of stationary phase. Write:

$p\ (x-y) - H(p,y) \delta t \approx p_0 (x-y) - H(p_0, y) \delta t + \delta p ((x-y) - \frac{\partial H}{\partial p}|_{p=p_0} \delta t) - \frac{\delta p^2}{2} \frac{\partial^2 H}{\partial p^2}|_{p=p_0} \delta t + Q(\delta p, x, y, \delta t)$

I'm writing the expression as a Taylor series about the point $p=p_0$, and $Q$ represents third-order terms and higher. I'm defining $\delta p = p - p_0$.

Here's where an unexpected connection with Lagrangian mechanics comes in. If $H(p,y)$ is treated as a classical Hamiltonian, then the corresponding velocity is $v_c(p) \equiv \frac{\partial H}{\partial p}$. The velocity is implicitly a function of momentum. If we choose $p_0$ so that $v_c(p_0) = \frac{x-y}{\delta t}$, then the expression becomes

$p\ (x-y) - H(p,y) \delta t \approx (p_0 v_c(p_0) - H(p_0, y)) \delta t - \frac{\delta p^2}{2} \frac{\partial^2 H}{\partial p^2}|_{p=p_0} \delta t + Q(\delta p, x, y, \delta t)$

The linear term vanishes. Then amazingly, the first expression is just the classical lagrangian:

$L(y,v) = p v - H$

So we have our expression for $G$:

$G(x,y,\delta t) \approx e^{i L(y, v) \delta t/\hbar} \frac{1}{2\pi \hbar} \int dp e^{-i\delta t/\hbar \frac{\delta p^2}{2} \frac{\partial^2 H}{\partial p^2}|_{p=p_0} - i/\hbar Q(\delta p, x,y, \delta t)}$

We've been completely general so far. This expression simplifies enormously if we assume that $H$ is quadratic in $p$ and that $\frac{\partial^2 H}{\partial p^2}$ is a constant (for classical mechanics, it is just $\frac{1}{m}$). In that case, the integral is some function of $\delta t$ alone, and we can write:

$G(x,y,\delta t) \approx e^{i L(y,v) \delta t/\hbar} f(\delta t)$

for some unknown function $f(\delta t)$. Under the assumption that $H = \frac{p^2}{2m} + V(y)$, the integral becomes a well-known gaussian integral, except for the peculiarity that the exponent is imaginary. If you assume that it can be analytically continued from the real case, you get:

$f(t) = \sqrt{\frac{m}{2\pi i \hbar \delta t}}$

However, if $H$ is not simply quadratic in $p$, then the integral becomes some complex function:

$G(x,y,\delta t) \approx e^{i L(y, v) \delta t/\hbar} f(x,y,\delta t)$

So for more complex hamiltonians, the path integral cannot simply be written in terms of the Lagrangian.

Putting the expression for $G$ back to get the value for larger times, we have:

$G(x,y,t) \approx \int dx_1 dx_2 ... dx_{N-1} e^{i/\hbar \sum_j L(x_j, v_j) \delta t} \Pi_j f(x_{j}, x_{j-1}, \delta t)$

Which can be written as:

$G(x,y,t) \approx \int dx_1 dx_2 ... dx_{N-1} e^{i/\hbar \int L(x(t), v(t)) dt} \Pi_j f(x_{j}, x_{j-1}, \delta t)$

So that's almost the path integral representation, except for the pesky functions $f$. I don't see how to evaluate that product, in general, so I don't see how this actually becomes the path integral expression for the Green function (unless the $f$ does not depend on position).

Of course, we can always fold the unknown functions into the definition of the path integral: Define:

$\int \mathcal{D}x = dx_1 dx_2 ... dx_{N-1} \Pi_j f(x_{j}, x_{j-1}, \delta t)$

then we have the expression

$G(x,y,t) \approx \int \mathcal{D}x e^{i/\hbar \int L(x(t), v(t)) dt}$

but that buries the problem into the expression $\mathcal{D}x$

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#### king vitamin

Gold Member
If you begin with a completely general Hamiltonian of the form $H = H(\hat{p},\hat{x})$, it's simply true that sometimes you end up with $x$-dependent normalization factors. For example, in Zinn-Justin's textbook on path integrals he shows that even with a Hamiltonian quadratic in momenta (but where the coefficient of the $p^2$ term may be multiplied by the coordinates), the resulting path integral has a finite quantum correction of exactly the form you've written, which needs to be considered in addition to the classical action. These sorts of Hamiltonians occur if you try to write down the path integral for the quantum O(N) rigid rotator for example.

You should always consider the continuum path integral to be a mnemonic for the discretized version before your last step. In particular, you should also be careful in keeping track of the exact approximations you've made, because they can blow up in certain situations (I believe this can always be traced to some ambiguity in in operator ordering/quantization, but I'm not familiar enough with your approach to see how this comes in). Zinn-Justin discusses how you can sometimes get wrong results if you don't think carefully about this.

#### ftr

this book
https://www.amazon.com/dp/0201360799/?tag=pfamazon01-20

does QFT in all the three usual pictures but says some picture is more suitable for certain problems. They can be solved in other pictures but gets very complicated/messy.

#### Demystifier

2018 Award
I have never seen a completely satisfying systematic derivation of path integrals for fermionic fields, without any ad hoc axioms about "Grassmann integrals". Any literature suggestions in that respect?

#### A. Neumaier

I have never seen a completely satisfying systematic derivation of path integrals for fermionic fields, without any ad hoc axioms about "Grassmann integrals". Any literature suggestions in that respect?
Grassmann integrals are (in finite dimensions) mathematically as well-defined as Lebesgue integrals. So what is it that you ask for?

#### Demystifier

2018 Award
Grassmann integrals are (in finite dimensions) mathematically as well-defined as Lebesgue integrals. So what is it that you ask for?
What? I have never seen a definition of Grassmann integral as a Lebesgue integral. But if that exists (a reference please!), it could be exactly what I look for.

But actually, I want the following. Is there a general non-perturbative proof (at least for systems with finite number of degrees of freedom) that canonical and path-integral quantization of fermions are equivalent? By non-perturbative, I mean an approach that doesn't use the idea that both approaches eventually give the same Feynman rules for Feynman diagrams.

#### A. Neumaier

What? I have never seen a definition of Grassmann integral as a Lebesgue integral. But if that exists (a reference please!), it could be exactly what I look for.
I didn't claim that they are defined in terms of Lebesgue integrals, only that the Grassmann integral is as well-defined as the latter, but of course with a different definition.
But actually, I want the following. Is there a general non-perturbative proof (at least for systems with finite number of degrees of freedom) that canonical and path-integral quantization of fermions are equivalent? By non-perturbative, I mean an approach that doesn't use the idea that both approaches eventually give the same Feynman rules for Feynman diagrams.
For imaginary time, I think there is, but I don't have a reference at hand. Will try to find find one.

Note that even in the bosonic case, the rigorous nonperturbative approach (Feynman-Kac formula) only works for the Euclidean case - i.e., a parabolic differential equation in imaginary time in place of the Schroedinger equation. The real time ''derivation'', given, e.g., in Wikipedia, is not rigorous. A fermionic analogue of such a heuristic derivation in real time can be found, e.g., in notes by Bastianelli.

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#### stevendaryl

Staff Emeritus
The real time ''derivation'', given, e.g., in Wikipedia, is not rigorous.
That's basically the same as my original post. The key fact that I glossed over was this:

$e^{A+B} \psi = lim_{N \rightarrow \infty} (e^{A/N} e^{B/N})^N$

That justifies the approximation, for small $\delta t$, that:

$e^{-i H(\hat{x},\hat{p}) \delta t/\hbar} \delta(x-y) \approx e^{-i H(y, \hat{p}) \delta t/\hbar} \delta(x-y)$

at least for hamiltonians of the form $H(x,p) = \frac{p^2}{2m} + V(x)$

But the Wikipedia derivation also doesn't address the issue of more complicated Hamiltonians (ones that are not quadratic in $p$, or whose second derivative with respect to $p$ is non-constant).

#### stevendaryl

Staff Emeritus
What? I have never seen a definition of Grassmann integral as a Lebesgue integral.
The treatments I've seen (in Hatfield's quantum field theory book, for example), just proceed via stating the properties that we want an integral to have, and then showing that there is only one possibility for Grassman variables. So it's just axiomatic, rather than being derived from anything else.

2018 Award

#### samalkhaiat

$G(x,y,t) = e^{-iHt/\hbar} \delta(x-y)$

where $H$ is the hamiltonian.
This expression is totally ambiguous: $H$ cannot be the Hamiltonian operator because the Green function is a c-number “function”. Also, it cannot be the (classical) Hamiltonian function $H(x,p)$ because the Green function $G(x,t ; y,0)$ does not depend on the momentum $p$.
What you need to consider is the path-integral in phase space also known as the Hamiltonian path-integral.
Recall that the Green function associated with the Schrödinger equation (in the coordinate representation) is given by the following matrix element of the evolution operator $$G(x,t ; y,0) = \langle x | e^{-i t \hat{H}} | y \rangle . \ \ \ \ \ \ \ (1)$$ Now, path-integral arises from the identity $$e^{\hat{A}} = \left( e^{\hat{A}/N} \right)^{N} , \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ in the limit $N \to \infty$. Next, we insert $N-1$ identity operator: $$\int dx_{j} \ |x_{j}\rangle \langle x_{j}| = 1 \ , \ \ j = 1 , \cdots , N-1,$$ between the factors in (2) to obtain
$$G(x,t ; y,0) = \int_{\mathbb{R}^{N-1}} \prod_{j=1}^{N-1}dx_{j} \ \prod_{k = 1}^{N} \langle x_{k}| e^{-i \tau_{N} \hat{H}}| x_{k-1} \rangle ,$$ where $x_{0} = y$, $x_{N} = x$ and $\tau_{N} = t/N$. In this, we also insert the integral $\int_{\mathbb{R}} dx_{N} \delta (x_{N} - x ) = 1$ so that the range of indices on the product signs becomes identical:
$$G(x,t ; y,0) = \int_{\mathbb{R}^{N}} \prod_{j=1}^{N}dx_{j} \ \delta (x_{N} - x ) \prod_{k = 1}^{N} \langle x_{k}| e^{-i \tau_{N} \hat{H}}| x_{k-1} \rangle . \ \ \ \ \ \ (3)$$ From this point on, I will not bother myself with convergence and other mathematical problems because everything can be made rigorous. For large $N$, we write
$$\langle x_{k}| e^{-i \tau_{N} \hat{H}}| x_{k-1} \rangle = \langle x_{k}| (1 - i \tau_{N} \hat{H}) | x_{k-1}\rangle = \delta (x_{k} - x_{k-1}) - i \tau_{N} \langle x_{k}|\hat{H}|x_{k-1}\rangle . \ \ \ (4)$$ The delta function can be written as
$$\delta (x_{k} - x_{k-1}) = \frac{1}{2\pi} \int_{\mathbb{R}} dp_{k} \ e^{ip_{k}(x_{k} - x_{k-1})} . \ \ \ \ \ (5)$$ So, all we need is an expression for the integral kernel (i.e., the matrix element) of the Hamiltonian operator. Since $\langle x_{k}| \hat{H}| x_{k-1}\rangle$ is a c-number, we may regard it as “function” of two variables, and denote it by $h (x_{k}, x_{k-1})$. Now, every function of two variables can be identical represented as a function of the difference between, and average of, those variables
$$\langle x_{k}| \hat{H}| x_{k-1}\rangle = h(x_{k}, x_{k-1}) = h \left( (x_{k} - x_{k-1}) , \frac{1}{2}(x_{k} + x_{k-1})\right) .$$ Next, we assume that the dependence of this function on $(x_{k} - x_{k-1})$ can be represented by Fourier integral of the phase-space function $H\left( p_{k} , \frac{1}{2}(x_{k} + x_{k-1} \right)$:
$$\langle x_{k}| \hat{H}| x_{k-1}\rangle = \frac{1}{2\pi} \int_{\mathbb{R}} dp_{k} \ e^{ip_{k}(x_{k} - x_{k-1})} \ H\left( p_{k} , \frac{x_{k} + x_{k-1}}{2} \right) . \ \ \ \ \ \ (6)$$ This is the famous Weyl quantization formula of 1925. Substituting (5) and (6) in (4) and writing $(1 - i \tau_{N} H ) = e^{-i \tau_{N}H}$, we obtain
$$\langle x_{k}| e^{-i \tau_{N}\hat{H}}| x_{k-1}\rangle = \int_{\mathbb{R}} \frac{dp_{k}}{2\pi} \ e^{ip_{k}(x_{k}-x_{k-1}) - i\tau_{N}H\left(p_{k} , \frac{1}{2}(x_{k}+x_{k-1})\right) } .$$ Thus
$$\prod_{k = 1}^{N}\langle x_{k}| e^{-i \tau_{N}\hat{H}}| x_{k-1}\rangle = \int_{\mathbb{R}^{N}} \prod_{j = 1}^{N} \left( \frac{dp_{j}}{2 \pi}\right) \exp \left[ i \sum_{k = 1}^{N}\tau_{N} \{ p_{k} \cdot \frac{x_{k} - x_{k-1}}{\tau_{N}} - H ( p_{k} , \frac{ x_{k} + x_{k-1}}{2}) \} \right]$$
Substituting this back in (3), taking the formal limits $N \to \infty , \tau_{N} \to 0$, introducing the phase-space trajectory $\left(p_{k},x_{k}\right) \to \left(p(\tau) , q(\tau) \right)$ and writing $\sum_{k}^{N}\tau_{N} \to \int_{0}^{t} d\tau$ we obtain a path-integral in phase-space with fixed initial $q(0) = y$ and final $q(t) = x$ points
$$G(x , t ; y ,0) = \int_{M} \mathcal{D}q(\tau) \mathcal{D}p(\tau) \ e^{i S[p(\tau) , q(\tau)]}, \ \ \ \ \ \ \ \ \ (7)$$ where
$$\mathcal{D}q(\tau) \mathcal{D}p(\tau) = \prod_{\tau} \left( \frac{dp(\tau) dq(\tau)}{2\pi} \right) ,$$ is the measure, and
$$S = \int_{0}^{t} d\tau \ \big[p(\tau)\dot{q}(\tau) - H\left(p(\tau) , q(\tau) \right) \big],$$ is the action integral written in terms of the classical Hamiltonian function $H(p,q)$.
If the Hamiltonian is of the form $H = \frac{p^{2}}{2m} + V(q)$, then (7) reduces to the ordinary path-integral of Feynman. To see this, change the variables according $$p \to p + m\dot{q}, \ \ q \to q,$$ so that $$p\dot{q} - H \to L(q , \dot{q}) - \frac{p^{2}}{2m},$$ where $$L = \frac{m}{2}\dot{q}^{2} - V(q).$$ So now, the integral over $p$ can be done because it is just the usual Gaussian integral.

It is remarkable we have derived the path-integral quantization of Feynman (1948) from the Weyl quantization formula (6) of (1925). If the physics community listened, understood or tried to understand Weyl, the path-integral formalism would have been discovered 23 years earlier.

#### Demystifier

2018 Award
The treatments I've seen (in Hatfield's quantum field theory book, for example), just proceed via stating the properties that we want an integral to have, and then showing that there is only one possibility for Grassman variables. So it's just axiomatic, rather than being derived from anything else.
Yes, that's the standard approach, which I don't like. In axiomatic approach we assume that a path-integral formulation exists (which allows us to make natural axioms about those integrals), but a priori it is not at all obvious that this assumption is true.

#### stevendaryl

Staff Emeritus
This expression is totally ambiguous: $H$ cannot be the Hamiltonian operator because the Green function is a c-number “function”. Also, it cannot be the (classical) Hamiltonian function $H(x,p)$ because the Green function $G(x,t ; y,0)$ does not depend on the momentum $p$.
I don't understand that objection. You have a function $\psi(x,t)$ with the property that $e^{-iHt/\hbar} \psi(x,0) = \psi(x,t)$. $H$ is an operator, an expression involving both $\hat{x}$ and $\hat{p}$. But of course $\psi(x,0)$ is a c-number function. There is no ambiguity.

Of course, the expression $e^{-iHt/\hbar} \delta(x-y)$ is not rigorous, because $\delta(x-y)$ is not an actual function, but it's not problematic from the point of view of ambiguity between operators and c-numbers. $H$ is an operator, and $\delta(x-y)$ is a c-number function.

#### samalkhaiat

I don't understand that objection. You have a function $\psi(x,t)$ with the property that $e^{-iHt/\hbar} \psi(x,0) = \psi(x,t)$. $H$ is an operator,
I understand this as follows
$$e^{-iHt}|\psi (0) \rangle = | \psi (t)\rangle.$$ Or, in terms of wave functions as
$$\psi (x,t) = \int dy \ \langle x | e^{-iHt}| y \rangle \psi (y , 0)$$ You wrote the Green function as
$$G(x,t;y,0) \equiv \langle x | e^{-iHt}|y\rangle \overset{?} {=} e^{-iHt} \langle x|y\rangle .$$ So, what is the justification of the last step? And if you can justify it, why should the functional dependence of $H(x , -i\partial_{x})$ on its arguments stays the same after you have pulled it out?

#### DarMM

Gold Member
This can be a complicated subject, I'll probably repeat stuff you know. Expanding on what @A. Neumaier said only the Euclidean path integral has a rigorous non-perturbative definition, the Minkowski one can be proven not to converge.

Technically in the bosonic case we have the Wightman-Gårding axioms which are conditions that the correlation functions have to satisfy. Then there are Wightman axioms which are conditions on the Hilbert space and field operators (position operator in 0+1 dimensions, i.e. Non-Relativistic QM). You can then prove these are fully equivalent.

Then you can show that the analytic continuation of these correlation functions gives another set of correlation functions (the Schwinger functions) on Euclidean space. A natural set of axioms on distributions in Euclidean space (the Osterwalder-Schrader axioms) allow you to prove the other direction, i.e. given a set of Euclidean correlation functions obeying the Osterwalder-Schrader axioms, you can prove that they are the analytic continuation of the Wightman correlation functions of a quantum field theory on Minkowski space. However this isn't an equivalence, not all quantum field theories have a Euclidean continuation.

Finally one can then show that there is a Markov process which can generate these Schwinger functions. The Markov process must obey a set of conditions in order to correctly give rise to the Schwinger functions (Nelson's axioms, also called Fröhlich's axioms). This Markov process can be presented in differential form (generalized stochastic equation) or integral form, which in standard terminology is the Euclidean path integral.

However the implication once again only goes one way. There are sets of Schwinger functions (themselves analytic continuations of QFT correlators) that don't originate in a path integral. This is because Markov processes can only have second order differential operators as their generators, e.g. a $p^4$ term won't result in a rigorously defined path integral. See Barry Simon's "The $\mathcal{P}\left(\phi\right)_2$ (Euclidean) Quantum Field Theory", p.127 for an example and discussion. So it is essentially related to the issue you raised.

As for fermions, there is a rigorous demonstration that the Grassman Euclidean path integral gives the (analytic continuation of the) correlation functions of a Fermionic Field theory, see Osterwalder, K. and Schrader, R. "Euclidean Fermi Fields and a Feynman-Kac Formula for Boson-Fermion Models", Helvitica Physica Acta, 46, p.277

However there are many complications, for example for bosons the restriction to a fixed time $\phi(\underline{x},0)$ makes sense, not so for fermions. The Grassman algebra has to be introduced to prevent a contradiction between Euclidean covariance and the required form of the two-point function, ultimately encoding Fermi statistics.

#### stevendaryl

Staff Emeritus
I understand this as follows
$$e^{-iHt}|\psi (0) \rangle = | \psi (t)\rangle.$$ Or, in terms of wave functions as
$$\psi (x,t) = \int dy \ \langle x | e^{-iHt}| y \rangle \psi (y , 0)$$ You wrote the Green function as
$$G(x,t;y,0) \equiv \langle x | e^{-iHt}|y\rangle \overset{?} {=} e^{-iHt} \langle x|y\rangle .$$ So, what is the justification of the last step? And if you can justify it, why should the functional dependence of $H(x , -i\partial_{x})$ on its arguments stays the same after you have pulled it out?
I don't understand what your objection is.

If $\psi(x,t)$ is a solution to Schrodinger's equation, then $\psi(x,t) = e^{-iH(x,p) t/\hbar} \psi(x,0)$. If you want to write that as $\langle x| e^{-iHt/\hbar} |\psi\rangle$, fine, but the only meaning to that expression to me is to understand it as $e^{-iH(x,p) t/\hbar} \psi(x,0)$.

There is no need to introduce bras and kets except possibly to make the expressions look neater. There is certainly no mathematical necessity.

#### Demystifier

2018 Award
Yes, corrected in post #7.
Thanks, this is exactly what I was looking for. I have no further objections on fermion path integrals.

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#### A. Neumaier

In axiomatic approach we assume that a path-integral formulation exists (which allows us to make natural axioms about those integrals), but a priori it is not at all obvious that this assumption is true.
The Lebesgue integral can be defined axiomatically, as a translation invariant monotone linear functional normalized such that the characteristic function of [0,1] integrates to 1. From this one can derive the usual rules, and then prove that the integral exists and is unique.

For the Grassmann integral, one does something completely analogous, one gives the axioms and then proves that the integral exists and is unique.

These constructions are completely independent of the functional integral, which is a different thing both in the bosonic and the fermionic case. It is (in its usual presentation) ill-defined in Minkowski space (real time), but well-defined in Euclidean space (imaginary time).

#### Demystifier

2018 Award
These constructions are completely independent of the functional integral
From the lectures by Bastianelli you linked now I understand it. The Grassmann integrals appear only from the insertions of the unit operator represented as a formal Grassmann integral over fermion coherent states. The content of those lectures should be a standard textbook stuff, but unfortunately it isn't.

"Question about equivalence of Path Integral and Schrodinger"

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