Useful references: Sakurai or Shankar for bra-ket notation, which we use heavily here. Chaichian/Demichev v1 for the derivation of path integrals. For a more rigorous approach, Techniques and Applications of Path Integration by Larry Schulman.

First, we need the concept of the time-evolution operator for the (time-dependent) Schroedinger equation. Recall that the Schroedinger equation is

(1)Consider a one-parameter family of operators $\mathcal{U} \left( t \right)$ on quantum states $\psi \left( x \right)$. By "one-parameter family of operators," we mean that for any value of the parameter $t$, $\mathcal{U} \left( t \right)$ is a function taking the state $\psi \left( x \right) \in L^2$ to some other state $\psi \left( x ; t \right) \in L^2$.

As you might guess, the operator $\mathcal{U} \left( t \right)$ will be set up to take the state $\psi \left( x \right)$ as an initial state at time $t = 0$, and give us an evolved state $\mathcal{U} \left(t \right) \cdot \psi \left( x \right) = \psi \left( x ; t \right)$ which satisfies the Schroedinger equation at time $t$. In other words, $\mathcal{U} \left( t \right)$ treats $\psi \left( x \right)$ as an initial snapshot and extrapolates forward in time to give us a state evolving in a manner consistent with the Schroedinger equation:

(2)To get this, we need $\mathcal{U}$ to satisfy the operator differential equation

(3)with initial value $\mathcal{U} \left( 0 \right) = \mathbb{I}$, the identity operator (so that $\mathcal{U} \left( 0 \right) \cdot \psi \left( x \right) = \psi \left( x \right)$).

We can easily show that the operator $\mathcal{U} \left( t \right) = \exp \left[ - \frac{i \hat H t }{\hbar} \right]$ solves 3. First, expand as a Taylor series of operators:

(4)which yields

(5)as desired.

Suppose we want to evolve the state $\lvert \alpha \rangle$ (in bra-ket notation). First write $\lvert \alpha \rangle = \int dx \lvert x \rangle \langle x \rvert \alpha \rangle$ (note that here we are explicitly assuming that the appropriate measure for the $L^2$ space of quantum states is $dx$, but we can and will modify this in the future). Then

(6)or to put this new evolved state into a position representation (still using bra-ket notation)

(7)and we can now see that

(8)is the integral kernel (propagator) which gives us $\langle y \vert \alpha ; t \rangle$ from $\langle x \vert \alpha \rangle$. It is also the transition amplitude for evolving from the localized state $\lvert x \rangle$ (delta function centered at $x$) to the localized state $\lvert y \rangle$ (delta function centered at $y$) between time $0$ and time $t$.

The amplitude for a system to evolve from the localized state $\lvert x_i \rangle$ at time $T_i$ to the localized state $\lvert x_f \rangle$ at time $T_f$ is

(9)For any time $t_1$ between $T_i$ and $T_f$, we can break the transition amplitude $\langle x_f ; T_f \vert x_i ; T_i \rangle$ into two pieces:

(10)What this says is that at time $t_1$, we don't know where the particle was (what its position $x_1$ was), but we can account for all possibilities by integrating over every value in $x$-space.

Another way to see this is to notice that

(11)Writing the identity operator in this way as an integral over projection operators is called a "resolution of the identity." What we have inserted in 10 is just a form of the identity operator, so we have changed nothing.

In 10 we have broken our transition amplitude into two transition amplitudes over shorter time intervals. The way to get the path integral is to continue this process ad infinitum, and then get an expression for the "short-time propagator" (i.e., the transition amplitude between two states over an infinitesimally small time interval). Break the interval $\left[ T_i , T_f \right]$ up as

(12)Then using the same reasoning as for 10, we have

(13)where $x_0 \equiv x_i$ and $x_N \equiv x_f$.

Now we just have to evaluate the short-time propagator:

(14)where here we have used a resolution of identity in momentum $p_j$ at time $t_j$ (you can think of this as an integration over projection operators in momentum coordinates rather than position coordinates). Recall that a momentum eigenstate (momentum eigenket in position basis) is

(15)(You can verify that this is a momentum eigenstate by using the Schroedinger representation and verifying that if we take $\psi \left( x \right) = e^{ipx / \hbar}$, then $-i \hbar \frac{d}{dx} \psi \left( x \right) = p \psi \left( x \right)$.)

So we have

(16)Here $\mathcal{O} \left( \varepsilon^2 \right)$ denotes terms containing two or more factors of $\varepsilon$. We assume that epsilon is small enough that we can neglect these terms. As we will see later, these kinds of considerations are crucial when we begin to consider how the factor ordering problem affects the definition of path integrals. For now, we are glossing over these issues in order to lay out the central idea of the path integral formulation, but we will return to them.

Having Taylor expanded the exponentiated operator above, we evaluate $\langle p_j \rvert \mathbb{I} - i \hat H \left( \hat p , \hat q \right) \varepsilon / \hbar + \mathcal{O} \left( \varepsilon^2 \right) \lvert x_j \rangle$ to be $\left[ 1 - i H \left( p_j , x_j \right) \varepsilon / \hbar + \mathcal{O} \left( \varepsilon \right) \right] \langle p_j \vert x_j \rangle$. Again, here we assume there are no factor-ordering issues. This is fine for the case when $\hat H$ contains no products of $x$ and $p$. For the case when there are products of $x$ and $p$, this expression can be viewed as assuming an ordering where all $\hat p$'s are placed to the left and all $\hat x$'s to the right. This issue is the biggest one we will return to when we stop ignoring factor ordering.

OK, but for now we're ignoring it, so now we write

(17)and

(18)- Exercise: Use the fact that $\langle x \vert p \rangle = e^{ixp / \hbar}$ to show $\langle p \vert x \rangle = e^{-ixp / \hbar}$.

If the momentum term in $H$ is of the form $p^2 /2m$ (no factors of $x$), then the integral over $p_j$ is Gaussian and can be explicitly performed to yield a normalization factor $\sqrt{ \frac{m}{2 \pi i \hbar \epsilon} }$, times $e^{ \frac{i m \varepsilon}{2 \hbar} \left( \frac{x_{j+1} - x_j}{\varepsilon} \right)^2 }$. Putting everything together, we get

(19)where $V \left( x \right)$ is the potential term of the Hamiltonian. Now what we really want to to is let $\varepsilon$ go to 0 (equivalently $N \to \infty$ - i.e. we let the short time-steps of our mini-propagators go to 0, so that

(20)and notice that the exponentiated expression inside the integrals is actually the Riemann sum for

(21)where $L$ is the Lagrangian for our system.

This is the path integral expression for the transition amplitude. What people usually do is think of the limiting measure

(22)as defining a measure $\mathcal{D} \left[ x \left( \tau \right) \right]$ on the space of paths $x \left( \tau \right)$ between the points $\left( x_i , T_i \right)$ and $\left( x_f , T_f \right)$, since in the limiting case, we are allowing for any value $x$ at each time $\tau$. Before we take the limit, we have a piecewise version which allows for any position at the subdivided times

(23)Note that we glossed over things to do with operator ordering and also things to do with whether all of this is well-defined. In particular, it is very problematic that we have an $i$ in the exponent, which will make for an oscillatory integrand. For a more rigorous definition, we can instead use so-called "imaginary time," which is a transformation of variables to $s = i t$. You can think of it as like a u-substitution.

Even with these modifications, making the limit of the measures rigorous is a huge challenge. The path integral approach is quite beautiful because it tells you that in some sense, the particle takes every possible route between $\left( x_i , T_i \right)$ and $\left( x_f , T_f \right)$, but with a probability weighting determined by the action ($e^{\frac{i}{\hbar} \int L}$). When we do the imaginary-time version, it will be very clear that the classical path is the most likely trajectory, and paths farther and farther from this are less likely. The price we pay for this beautiful idea, though, is that we still don't know exactly how to make it precise.