The Feynman path integral, The Trotter Product Formula

The Transition Amplitude as a Path Integral

In quantum mechanics, the probability amplitude for a particle to propagate from an initial position $\displaystyle x_i$ at time $\displaystyle t_i$ to a final position $\displaystyle x_f$ at time $\displaystyle t_f$ is given by the matrix element of the time-evolution operator:

$\displaystyle K(x_f, t_f; x_i, t_i) = \langle x_f | \hat{U}(t_f - t_i) | x_i \rangle = \langle x_f | e^{-i \hat{H}(t_f - t_i)/\hbar} | x_i \rangle$

where $\displaystyle \hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})$ is the system’s Hamiltonian. The path integral is derived by evaluating this matrix element through a process of “time-slicing.”

1.1 Time-Slicing the Evolution Operator

We first divide the total time interval $\displaystyle T = t_f - t_i$ into $\displaystyle N$ infinitesimally small time slices, each of duration $\displaystyle \epsilon = T/N$ . The total evolution operator can then be expressed as a product of $\displaystyle N$ identical operators, each governing the evolution for a single slice:

$\displaystyle e^{-i\hat{H}T/\hbar} = (e^{-i\hat{H}\epsilon/\hbar})^N$

To break down the propagator calculation, we insert a complete set of position states between each of these operators. The completeness relation for position eigenstates $\displaystyle |x\rangle$ is $\displaystyle \int dx, |x\rangle\langle x| = \hat{1}$ . Inserting $\displaystyle N-1$ such relations, one for each intermediate point in time, we get:

$\displaystyle K = \langle x_f | (\int dx_{N-1} |x_{N-1}\rangle\langle x_{N-1}|) e^{-i\hat{H}\epsilon/\hbar} \ldots e^{-i\hat{H}\epsilon/\hbar} (\int dx_1 |x_1\rangle\langle x_1|) e^{-i\hat{H}\epsilon/\hbar} |x_i \rangle$

By defining $\displaystyle x_0 = x_i$ and $\displaystyle x_N = x_f$ , we can rearrange this into a chain of integrals over all intermediate positions:

$\displaystyle K = \int dx_1 \ldots dx_{N-1} \prod_{j=0}^{N-1} \langle x_{j+1} | e^{-i\hat{H}\epsilon/\hbar} | x_j \rangle$

This expression is still exact. We have transformed the calculation of a single, complicated matrix element into the calculation of $\displaystyle N$ simpler “infinitesimal propagators,” $\displaystyle \langle x_{j+1} | e^{-i\hat{H}\epsilon/\hbar} | x_j \rangle$ , integrated over all possible intermediate steps.

1.2 The Infinitesimal Propagator:

The central challenge is evaluating the infinitesimal propagator. The kinetic energy operator $\displaystyle \hat{T} = \frac{\hat{p}^2}{2m}$ and the potential energy operator $\displaystyle \hat{V} = V(\hat{x})$ do not commute, meaning $\displaystyle [\hat{T}, \hat{V}] \ne 0$ . This prevents us from simply writing $\displaystyle e^{-i(\hat{T} + \hat{V})\epsilon/\hbar} = e^{-i\hat{T}\epsilon/\hbar} e^{-i\hat{V}\epsilon/\hbar}$ .

However, for a very small time step $\displaystyle \epsilon$ , we can use an approximation based on the Trotter product formula (discussed in Section 3). This formula justifies splitting the exponential, with an error that vanishes as $\displaystyle \epsilon \to 0$ :

$\displaystyle e^{-i\hat{H}\epsilon/\hbar} = e^{-i(\hat{T} + \hat{V})\epsilon/\hbar} \approx e^{-i\hat{V}\epsilon/\hbar} e^{-i\hat{T}\epsilon/\hbar} + O(\epsilon^2)$

We can now evaluate the matrix element of this approximate operator. To handle the momentum operator, we insert a completeness relation for momentum states, $\displaystyle \int \frac{dp}{2\pi\hbar} |p\rangle\langle p| = \hat{1}$ :

$\displaystyle \langle x_{j+1} | e^{-i\hat{H}\epsilon/\hbar} | x_j \rangle \approx \int \frac{dp_j}{2\pi\hbar} \langle x_{j+1} | e^{-i\hat{V}\epsilon/\hbar} | p_j \rangle \langle p_j | e^{-i\hat{T}\epsilon/\hbar} | x_j \rangle$

Position and momentum states are eigenstates of their respective operators: $\displaystyle \hat{V}(\hat{x})|x\rangle = V(x)|x\rangle$ and $\displaystyle \hat{T}(\hat{p})|p\rangle = \frac{p^2}{2m}|p\rangle$ .

The potential operator acts on the position eigenstate to its left:

$\displaystyle \langle x_{j+1} | e^{-iV(\hat{x})\epsilon/\hbar} = e^{-iV(x_{j+1})\epsilon/\hbar} \langle x_{j+1} |$

The kinetic operator acts on the momentum eigenstate to its right:

$\displaystyle e^{-i\frac{\hat{p}^2}{2m}\epsilon/\hbar} |p_j\rangle = e^{-i\frac{p_j^2}{2m}\epsilon/\hbar} |p_j\rangle$

Substituting these back into the integral gives:

$\displaystyle \int \frac{dp_j}{2\pi\hbar} e^{-iV(x_{j+1})\epsilon/\hbar} \langle x_{j+1} | p_j \rangle e^{-i\frac{p_j^2}{2m}\epsilon/\hbar} \langle p_j | x_j \rangle$

Next, we use the explicit form of the inner product between position and momentum states, $\displaystyle \langle x | p \rangle = e^{ipx/\hbar}$ :

$\displaystyle \int \frac{dp_j}{2\pi\hbar} e^{-iV(x_{j+1})\epsilon/\hbar} e^{ip_j x_{j+1}/\hbar} e^{-i\frac{p_j^2}{2m}\epsilon/\hbar} e^{-ip_j x_j/\hbar}$

Combining the terms in the exponential:

$\displaystyle e^{-iV(x_{j+1})\epsilon/\hbar} \int \frac{dp_j}{2\pi\hbar} \exp\left[ -\frac{i\epsilon}{\hbar} \frac{p_j^2}{2m} + \frac{i}{\hbar} p_j (x_{j+1} - x_j) \right]$

This is a standard (complex) Gaussian integral. Completing the square in the exponent with respect to $\displaystyle p_j$ :

$\displaystyle -\frac{i\epsilon}{\hbar} \left[ p_j^2 - \frac{2m}{\epsilon}(x_{j+1} - x_j)p_j \right] = -\frac{i\epsilon}{\hbar} \left[ \left(p_j - \frac{m}{\epsilon}(x_{j+1} - x_j)\right)^2 - \left(\frac{m}{\epsilon}(x_{j+1} - x_j)\right)^2 \right]$

So the integral becomes:

$\displaystyle \int \frac{dp_j}{2\pi\hbar} \exp\left( -\frac{i\epsilon}{\hbar} \cdot \frac{1}{2m} \left(p_j - \frac{m}{\epsilon} \Delta x_j \right)^2 \right) \exp\left( \frac{im}{2\hbar\epsilon} (\Delta x_j)^2 \right)$

where $\displaystyle \Delta x_j = x_{j+1} - x_j$ .

The first part is a shifted Gaussian integral $\displaystyle \int du, e^{-au^2} = \sqrt{\pi/a}$ with $\displaystyle a = \frac{i\epsilon}{\hbar} \cdot \frac{1}{2m}$ . Its value is:

$\displaystyle \sqrt{\frac{2\pi i\hbar\epsilon}{m}}$

The second exponential is:

$\displaystyle \exp\left( \frac{i m}{2\hbar\epsilon} (x_{j+1} - x_j)^2 \right)$

Therefore, the infinitesimal propagator is:

$\displaystyle \langle x_{j+1} | e^{-i\hat{H}\epsilon/\hbar} | x_j \rangle \approx \sqrt{\frac{m}{2\pi i\hbar\epsilon}} \exp\left[ \frac{i\epsilon}{\hbar} \left( \frac{m}{2} \left( \frac{x_{j+1} - x_j}{\epsilon} \right)^2 - V(x_{j+1}) \right) \right]$

There is a choice of squareroot that has to be chosen carefully here.

1.3 The Limit to the Path Integral

Now we reconstruct the full propagator $\displaystyle K$ by substituting this result back into the product over all $\displaystyle j$ :

$\displaystyle K = \lim_{N\to\infty} \int dx_1 \ldots dx_{N-1} \left(\frac{m}{2\pi i\hbar\epsilon}\right)^{N/2} \prod_{j=0}^{N-1} \exp\left[\frac{i\epsilon}{\hbar} L(x_j, \dot{x}_j)\right]$

We can pull the constant factors out and combine the exponentials:

$\displaystyle K = \lim_{N\to\infty} \left(\frac{m}{2\pi i\hbar\epsilon}\right)^{N/2} \int dx_1 \ldots dx_{N-1} \exp\left[\frac{i}{\hbar} \sum_{j=0}^{N-1} \epsilon L(x_j, \dot{x}_j)\right]$

In the limit $\displaystyle N\to\infty$ (and thus $\displaystyle \epsilon\to 0$ ):

The sum becomes a Riemann integral over time. This is the classical action $\displaystyle S$ :

$\displaystyle \sum_{j=0}^{N-1} \epsilon L(x_j, \dot{x}j) \rightarrow \int{t_i}^{t_f} L(x(t), \dot{x}(t)), dt \equiv S[x(t)]$

The chain of integrals over all intermediate positions, along with the normalization constant, is symbolically written as the path integral measure $\displaystyle \int\mathcal{D}x(t)$ :

$\displaystyle \lim_{N\to\infty} \left(\frac{m}{2\pi i\hbar\epsilon}\right)^{N/2} \int dx_1 \ldots dx_{N-1} \rightarrow \int \mathcal{D}x(t)$

This measure represents an integral over the infinite-dimensional space of all possible paths $\displaystyle x(t)$ that start at $\displaystyle x_i$ and end at $\displaystyle x_f$ .

This leads to the final, elegant expression for the Feynman Path Integral:

$\displaystyle K(x_f, t_f; x_i, t_i) = \int \mathcal{D}x(t), e^{\frac{i}{\hbar} S[x(t)]}$

This remarkable formula states that the quantum amplitude is found by summing the contributions of every possible path. Each path contributes a phase factor $\displaystyle e^{iS/\hbar}$ , determined by its classical action.

2. Time-Ordered Amplitudes

A key strength of the path integral formalism is its natural handling of time-ordered products of operators. Suppose we wish to calculate the expectation value of an operator $\displaystyle \hat{O}$ acting at time $\displaystyle t_k$ , where $\displaystyle t_i < t_k < t_f$ . In the discretized derivation, this corresponds to inserting the operator at the k-th time slice.

$\displaystyle \langle x_f | \ldots e^{-iH\epsilon/\hbar} \hat{O}(\hat{x}) e^{-iH\epsilon/\hbar} \ldots | x_i \rangle$

When this insertion is made between slices k and k+1, the relevant term in the product becomes:

$\displaystyle \langle x_{k+1} | e^{-iH\epsilon/\hbar} \hat{O}(\hat{x}) | x_k \rangle$

Since $\displaystyle |x_k\rangle$ is an eigenstate of $\displaystyle \hat{x}$ , the action is simple: $\displaystyle \hat{O}(\hat{x})|x_k\rangle = O(x_k)|x_k\rangle$ . The operator $\displaystyle \hat{O}(\hat{x})$ becomes its classical value $\displaystyle O(x_k)$ evaluated at that point in the path.

The net effect is that the classical value $\displaystyle O(x(t_k))$ is inserted into the integrand of the path integral. The chronological construction of the integral automatically enforces the correct time ordering. For multiple operators, the general result is:

$\displaystyle \langle x_f | T{ \hat{O}_1(t_1)\ldots \hat{O}_n(t_n) } | x_i \rangle = \int \mathcal{D}x(t), O_1(x(t_1))\ldots O_n(x(t_n)) e^{\frac{i}{\hbar} S[x]}$

3. Rigorous Justification: The Trotter Product Formula

The entire derivation hinges on the approximation $\displaystyle e^{\epsilon(\hat{A} + \hat{B})} \approx e^{\epsilon \hat{A}} e^{\epsilon \hat{B}}$ for small $\displaystyle \epsilon$ . The rigorous justification is the Trotter product formula.

Theorem (Trotter): For self-adjoint operators $\displaystyle \hat{A}$ and $\displaystyle \hat{B}$ such that $\displaystyle \hat{A} + \hat{B}$ is also self-adjoint, the following limit holds:

$\displaystyle e^{-it(\hat{A} + \hat{B})/\hbar} = \lim_{N\to\infty} \left(e^{-it \hat{A}/(N\hbar)} e^{-it \hat{B}/(N\hbar)}\right)^N$

To see where the error comes from in a single step, let $\displaystyle \epsilon = t/N$ . We compare the Taylor series expansions:

$\displaystyle e^{\epsilon(\hat{A} + \hat{B})} = \hat{1} + \epsilon(\hat{A} + \hat{B}) + \frac{\epsilon^2}{2} (\hat{A} + \hat{B})^2 + O(\epsilon^3) = \hat{1} + \epsilon(\hat{A} + \hat{B}) + \frac{\epsilon^2}{2}(\hat{A}^2 + \hat{A}\hat{B} + \hat{B}\hat{A} + \hat{B}^2) + O(\epsilon^3)$

And the product of exponentials:

$\displaystyle e^{\epsilon \hat{A}} e^{\epsilon \hat{B}} = (\hat{1} + \epsilon \hat{A} + \frac{\epsilon^2}{2} \hat{A}^2)(\hat{1} + \epsilon \hat{B} + \frac{\epsilon^2}{2} \hat{B}^2) + O(\epsilon^3) = \hat{1} + \epsilon(\hat{A} + \hat{B}) + \frac{\epsilon^2}{2}(\hat{A}^2 + 2\hat{A}\hat{B} + \hat{B}^2) + O(\epsilon^3)$

The difference between the two expressions at order $\displaystyle \epsilon^2$ is:

$\displaystyle \text{Error} = \frac{\epsilon^2}{2}(\hat{B}\hat{A} - \hat{A}\hat{B}) = -\frac{\epsilon^2}{2}[\hat{A}, \hat{B}]$

The error in a single step is proportional to $\displaystyle \epsilon^2$ and the commutator of the operators. When we concatenate $\displaystyle N$ steps, the total error is of order $\displaystyle N\times\epsilon^2 = N\times(T/N)^2 = T^2/N$ . As $\displaystyle N\to\infty$ , this total error vanishes, making the formula exact in the limit.

4. A Deeper Issue: Unbounded Operators

The simple Taylor series proof above is only strictly valid for bounded operators—those whose action cannot “blow up” the norm of a state vector. However, the fundamental operators of quantum mechanics, position ( $\displaystyle \hat{x}$ ) and momentum ( $\displaystyle \hat{p}$ ), are unbounded.

An operator $\displaystyle \hat{O}$ is unbounded if there is no finite constant $\displaystyle C$ such that $\displaystyle | \hat{O} \psi | \leq C |\psi|$ for all states $\displaystyle \psi$ .

For the position operator $\displaystyle \hat{x}$ , we can imagine a state $\displaystyle \psi(x)$ that is increasingly localized at a very large position $\displaystyle x_0 \to \infty$ . The expectation value $\displaystyle \langle \hat{x} \rangle$ for this state can be made arbitrarily large.

Similarly, for the momentum operator $\displaystyle \hat{p}$ , a state can be chosen to represent a particle with arbitrarily high momentum, making $\displaystyle \langle \hat{p} \rangle \to \infty$ .

Because $\displaystyle \hat{x}$ and $\displaystyle \hat{p}$ (and thus $\displaystyle \hat{T}$ and $\displaystyle \hat{V}$ ) are unbounded, the simple proof of the Trotter formula does not apply. This is a deep mathematical problem that requires the machinery of functional analysis. The rigorous justification is provided by the Trotter-Kato Theorem. This theorem extends the formula to unbounded operators under certain conditions, most critically that the total Hamiltonian $\displaystyle \hat{H} = \hat{T} + \hat{V}$ is a well-behaved self-adjoint operator.

The self-adjointness of the Hamiltonian is a non-negotiable physical requirement, as it guarantees that time evolution is unitary, meaning total probability is conserved over time. The theorem ensures that the physically intuitive path integral rests on solid mathematical ground.

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