Position and momentum spaces

Most often in Lagrangian mechanics, the Lagrangian L(q, dq/dt, t) is in configuration space, where q = (q₁, q₂,..., q_n) is an n-tuple of the generalized coordinates. The Euler–Lagrange equations of motion are

$${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {q}}_{i}}}={\frac {\partial L}{\partial q_{i}}}\,,\quad {\dot {q}}_{i}\equiv {\frac {dq_{i}}{dt}}\,.}$$

(One overdot indicates one time derivative). Introducing the definition of canonical momentum for each generalized coordinate

$${\displaystyle p_{i}={\frac {\partial L}{\partial {\dot {q}}_{i}}}\,,}$$

the Euler–Lagrange equations take the form

$${\displaystyle {\dot {p}}_{i}={\frac {\partial L}{\partial q_{i}}}\,.}$$

The Lagrangian can be expressed in momentum space also,[2] L′(p, dp/dt, t), where p = (p₁, p₂, ..., p_n) is an n-tuple of the generalized momenta. A Legendre transformation is performed to change the variables in the total differential of the generalized coordinate space Lagrangian;

$${\displaystyle dL=\sum _{i=1}^{n}\left({\frac {\partial L}{\partial q_{i}}}dq_{i}+{\frac {\partial L}{\partial {\dot {q}}_{i}}}d{\dot {q}}_{i}\right)+{\frac {\partial L}{\partial t}}dt=\sum _{i=1}^{n}({\dot {p}}_{i}dq_{i}+p_{i}d{\dot {q}}_{i})+{\frac {\partial L}{\partial t}}dt\,,}$$

where the definition of generalized momentum and Euler–Lagrange equations have replaced the partial derivatives of L. The product rule for differentials[nb 1] allows the exchange of differentials in the generalized coordinates and velocities for the differentials in generalized momenta and their time derivatives,

$${\displaystyle {\dot {p}}_{i}dq_{i}=d(q_{i}{\dot {p}}_{i})-q_{i}d{\dot {p}}_{i}}$$

$${\displaystyle p_{i}d{\dot {q}}_{i}=d({\dot {q}}_{i}p_{i})-{\dot {q}}_{i}dp_{i}}$$

which after substitution simplifies and rearranges to

$${\displaystyle d\left[L-\sum _{i=1}^{n}(q_{i}{\dot {p}}_{i}+{\dot {q}}_{i}p_{i})\right]=-\sum _{i=1}^{n}({\dot {q}}_{i}dp_{i}+q_{i}d{\dot {p}}_{i})+{\frac {\partial L}{\partial t}}dt\,.}$$

Now, the total differential of the momentum space Lagrangian L′ is

$${\displaystyle dL'=\sum _{i=1}^{n}\left({\frac {\partial L'}{\partial p_{i}}}dp_{i}+{\frac {\partial L'}{\partial {\dot {p}}_{i}}}d{\dot {p}}_{i}\right)+{\frac {\partial L'}{\partial t}}dt}$$

so by comparison of differentials of the Lagrangians, the momenta, and their time derivatives, the momentum space Lagrangian L′ and the generalized coordinates derived from L′ are respectively

$${\displaystyle L'=L-\sum _{i=1}^{n}(q_{i}{\dot {p}}_{i}+{\dot {q}}_{i}p_{i})\,,\quad -{\dot {q}}_{i}={\frac {\partial L'}{\partial p_{i}}}\,,\quad -q_{i}={\frac {\partial L'}{\partial {\dot {p}}_{i}}}\,.}$$

Combining the last two equations gives the momentum space Euler–Lagrange equations

$${\displaystyle {\frac {d}{dt}}{\frac {\partial L'}{\partial {\dot {p}}_{i}}}={\frac {\partial L'}{\partial p_{i}}}\,.}$$

The advantage of the Legendre transformation is that the relation between the new and old functions and their variables are obtained in the process. Both the coordinate and momentum forms of the equation are equivalent and contain the same information about the dynamics of the system. This form may be more useful when momentum or angular momentum enters the Lagrangian.

In Hamiltonian mechanics, unlike Lagrangian mechanics which uses either all the coordinates or the momenta, the Hamiltonian equations of motion place coordinates and momenta on equal footing. For a system with Hamiltonian H(q, p, t), the equations are

$${\displaystyle {\dot {q}}_{i}={\frac {\partial H}{\partial p_{i}}}\,,\quad {\dot {p}}_{i}=-{\frac {\partial H}{\partial q_{i}}}\,.}$$

Suppose we have a three-dimensional wave function in position space ψ(r), then we can write this functions as a weighted sum of orthogonal basis functions ψ_j(r):

$${\displaystyle \psi (\mathbf {r} )=\sum _{j}\phi _{j}\psi _{j}(\mathbf {r} )}$$

or, in the continuous case, as an integral

$${\displaystyle \psi (\mathbf {r} )=\int _{\mathbf {k} {\text{-space}}}\phi (\mathbf {k} )\psi _{\mathbf {k} }(\mathbf {r} )\mathrm {d} ^{3}\mathbf {k} }$$

It is clear that if we specify the set of functions ${\displaystyle \psi _{\mathbf {k} }(\mathbf {r} )}$, say as the set of eigenfunctions of the momentum operator, the function ${\displaystyle \phi (\mathbf {k} )}$ holds all the information necessary to reconstruct ψ(r) and is therefore an alternative description for the state ${\displaystyle \psi }$.

In quantum mechanics, the momentum operator is given by

$${\displaystyle \mathbf {\hat {p}} =-i\hbar {\frac {\partial }{\partial \mathbf {r} }}}$$

(see matrix calculus for the denominator notation) with appropriate domain. The eigenfunctions are

$${\displaystyle \psi _{\mathbf {k} }(\mathbf {r} )={\frac {1}{({\sqrt {2\pi }})^{3}}}e^{i\mathbf {k} \cdot \mathbf {r} }}$$

and eigenvalues ħk. So

$${\displaystyle \psi (\mathbf {r} )={\frac {1}{({\sqrt {2\pi }})^{3}}}\int _{\mathbf {k} {\text{-space}}}\phi (\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }\mathrm {d} ^{3}\mathbf {k} }$$

and we see that the momentum representation is related to the position representation by a Fourier transform.[6]

Conversely, a three-dimensional wave function in momentum space ${\displaystyle \phi (\mathbf {k} )}$ can be expressed as a weighted sum of orthogonal basis functions ${\displaystyle \phi _{j}(\mathbf {k} )}$,

$${\displaystyle \phi (\mathbf {k} )=\sum _{j}\psi _{j}\phi _{j}(\mathbf {k} ),}$$

or as an integral,

$${\displaystyle \phi (\mathbf {k} )=\int _{\mathbf {r} {\text{-space}}}\psi (\mathbf {r} )\phi _{\mathbf {r} }(\mathbf {k} )\mathrm {d} ^{3}\mathbf {r} .}$$

The position operator is given by

$${\displaystyle \mathbf {\hat {r}} =i\hbar {\frac {\partial }{\partial \mathbf {p} }}=i{\frac {\partial }{\partial \mathbf {k} }}}$$

with eigenfunctions

$${\displaystyle \phi _{\mathbf {r} }(\mathbf {k} )={\frac {1}{\left({\sqrt {2\pi }}\right)^{3}}}e^{-i\mathbf {k} \cdot \mathbf {r} }}$$

and eigenvalues r. So a similar decomposition of ${\displaystyle \phi (\mathbf {k} )}$ can be made in terms of the eigenfunctions of this operator, which turns out to be the inverse Fourier transform,[6]

$${\displaystyle \phi (\mathbf {k} )={\frac {1}{({\sqrt {2\pi }})^{3}}}\int _{\mathbf {r} {\text{-space}}}\psi (\mathbf {r} )e^{-i\mathbf {k} \cdot \mathbf {r} }\mathrm {d} ^{3}\mathbf {r} .}$$