My question is about the two versions of the path integral, Hamiltonian and Lagrangian, that show up in most derivation of path integral quantum mechanics, but specifically in this case the derivation presented in Altland and Simons pg. 98-101.

They consider the propagator $\langle x_f, t_f |\ U\left(t_f; t_i\right) | x_i, t_i \rangle$, discretise the path and the time evolution operator $U$ into $N$ steps and insert a lot of identities of the form $\int dx |x \rangle\langle x|$ and $\int dp |p \rangle\langle p|$ to arrive at an expression for the propagator given by:

$$\langle x_f, t_f |\ U\left(t_f; t_i\right) | x_i, t_i \rangle = \int \mathcal{Dx} \exp \left[ \frac{i}{\hbar} \int_0^t dt' ( p \dot q - H(p,q) ) \right] \ \ \text{with} H = T+V \tag{1}$$

where the $\mathcal{Dx}$ stands for a path integral over BOTH $p$ and $q$, i.e. $\mathcal{Dx} = \lim_{N \to \infty} dq_1...dq_N dp_1...dp_N$. They call this the Hamiltonian/Phase Space Path Integral and this seems to be the standard way of deriving it looking at different literature.

They then go on to draw a connection to the classical Hamiltonian and Lagrangian where $L = p\dot{q} - H(p,q)$ and inspired by this they set out to find a Lagrangian path integral. Restricting themselves to Hamiltonians quadratic in $p$ they perform the $p$ integrals of $\mathcal{Dx}$ and find a path integral with an expression in the exponent that can be identified as the classical Lagrangian:

$$\langle x_f, t_f |\ U\left(t_f; t_i\right) | x_i, t_i \rangle = \int \mathcal{Dq} \exp \left[ \frac{i}{\hbar} \int_0^t dt' L(q,\dot{q}) \right] \ \ \text{with} \ \ L = T - V \tag{2}$$

which they appropriately call the Lagrangian/Configuration Space Path Integral, which also seems to be the standard way of deriving it.

My question is:

Given that during the derivation we replaced all the operators $\hat p$ and $\hat q$ with "actual values" along a path in phase space, why can we not just directly identify $L = p\dot{q} - H(p,q)$ and apply what we have learned from Lagrangian mechanics to immediately write $L = T-V$ in equ. (1) without having to do the cumbersome $p$ integrals? Obviously this would give us something different from equ. (2):

$$\langle x_f, t_f |\ U\left(t_f; t_i\right) | x_i, t_i \rangle = \int \mathcal{Dx} \exp \left[ \frac{i}{\hbar} \int_0^t dt' ( L(q,\dot q ) \right] \tag{3}$$

  • 2
    $\begingroup$ Lagrangian $=L(q,\dot q)\neq$ hamiltonian Lagrangian $=L_H(q,p):=p\dot q-H(q,p)$. $\endgroup$ Commented Feb 1, 2021 at 20:15
  • $\begingroup$ Minor comment to the post (v2): The LHS of eqs. (1)-(3) in Altland & Simons is $\langle q_f, t_f |\ U\left(t_f; t_i\right) | q_i, t_i \rangle$ not $\langle x_f, t_f |\ U\left(t_f; t_i\right) | x_i, t_i \rangle$. $\endgroup$
    – Qmechanic
    Commented Feb 2, 2021 at 1:28
  • $\begingroup$ Performing the dp integration in the path integral only gives you 1/2 m xdot^2 if H = p^2/2m + V(x) (which is a Gaussian integral). Surprisingly there is a difference for general H. $\endgroup$
    – lalala
    Commented Oct 19, 2021 at 18:34
  • $\begingroup$ related: physics.stackexchange.com/q/637531/226902 and physics.stackexchange.com/q/134714/226902 $\endgroup$
    – Quillo
    Commented Sep 19, 2022 at 7:59

2 Answers 2


Well, there are a ton of reasons. Let's just mention a few:

  1. In the Feynman path integral one should sum over all histories/off-shell configurations, not just insert the stationary values.

  2. If we start from the Lagrangian path integral, at the physics level of rigor, we would have to insert by hand the ad-hoc Feynman fudge factor, cf. e.g this Phys.SE post. On the other hand, this fudge factor is neatly explained from the Hamiltonian path integral via the Gaussian $p$-integrations.

  3. The Lagrangian kinetic energy is more subtle than it Hamiltonian counterpart due to the time ordering prescription, cf. e.g. this Phys.SE answer.


I think that the key issue is what you already pinpointed in your last paragraph. The integrand, that is the $e^{iS}$ part, isn't the only result of a derivation of a path integral formula. Another equally important ingredient that has to be derived is the path measure ${\cal D}q$ or ${\cal D}x$.

It is perfectly possible to go from one formulation to the other, but you have to put in a little more effort than blunt insertion into the integrand. When you write a statement like $L = p\dot{q} - H(p,q)$, some care has to be taken. You cannot consider $\dot{q}$ to be an independent variable in this relation. You must use the Legendre transform to get from the variable pair $(q,\dot{q})$ to the pair $(q,p)$, like $p(q,\dot{q}) = \frac{\partial L}{\partial \dot{q}}(q,\dot{q})$ or its inverse.

With that, going from the path of a lifted curve $(q(t),\dot{q}(t))$ to a phase space path $(q(t),p(t))$, you change coordinates, which results in a change of integration measure ${\cal D}x$ to ${\cal D}q$. It is a bit involved to describe this transformation of integral measures by the measures alone, so it is usually easier to just do the derivation twice for two different sets of variables.

Mathematically, this is much like the fact that you have to include the Jacobian determinant when changing coordinates in a normal $n$-dimensional integral $$\int_{\Phi(\Omega)}\textrm{d}x^n\, f(x) = \int_{\Omega}\textrm{d}y^n\, |\det D\Phi(y)|f(\Phi(y)).$$


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