In quantum mechanics, there is an uncertainty principle between conjugate variables, giving rise to complementary descriptions of a quantum system. But the variables are conjugates in two different mathematical senses.

One sense in which they are conjugates are with respect to the Legendre transform of a Lagrangian into a Hamiltonian (where generalized momentum coordinates are introduced).

And in another sense they are conjugates with respect to a Fourier transform. It seems obvious why being conjugates in this second sense would result in an uncertainty principle, and give rise to two dual descriptions of the system. The same thing happens with any kind of waves. In digital signal processing these are referred to as the frequency domain and the time domain. In solid state physics, k-space is referred to as the "reciprocal lattice" or the "dual lattice", as it's a dual description of position space using the Fourier conjugate wave-number k as a basis instead. Indeed, Fourier transforms are just a special case of Pontryagin duality.

What's not obvious to me is why or how these two different senses of conjugacy are connected. Is there an actual mathematical connection? Or is it just an ad hoc assumption of quantum mechanics that when you see Legendre conjugates you should automatically make them Fourier conjugates by imposing canonical commutation relations? Is there any other way, besides simply accepting this as a postulate, to understand this connection? If not, couldn't there be a consistent version of quantum mechanics where other kinds of pairs of variables were made Fourier conjugates in the Hilbert space, instead of using Legendre conjugates?

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    $\begingroup$ I'll never get tired of saying this: The uncertainty relations of quantum mechanics are far more general than those between Fourier conjugate variables. You can understand why Fourier uncertainty looks like the quantum mechanical uncertainty if you observe that the Stone-von Neumann theorem essentially says that operators with the canonical commutation relations are uniquely represented by Fourier conjugate variables. $\endgroup$
    – ACuriousMind
    Aug 13, 2015 at 18:21
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    $\begingroup$ Why do you think they are connected? Legendre duality is indeed a thing, but the Legendre dual of momentum is velocity, not position. $\endgroup$ Aug 13, 2015 at 18:22

2 Answers 2


I) Well, the Legendre transformation can be e.g. seen as the leading classical tree-level formula of a formal semiclassical Fourier transformation.

This fact is e.g. used in QFT when relating the quantum action $S[\varphi]$, the partition function $Z[J]$, generating functional $W_c[J]$ for connected diagrams, and the effective action $\Gamma[\Phi]$.

II) To see the correspondence in detail, let $x$ and $p$ be the two dual/conjugate variables (in both senses!). Let

$$\tag{1} f(x;\hbar)~\equiv~\sum_{n=0}^{\infty}\hbar^n f_n(x)\quad\text{and}\quad g(p;\hbar)~\equiv~\sum_{n=0}^{\infty}\hbar^n g_n(p)$$

be two formal power series in $\hbar$ with function coefficients. Consider their semiclassical exponentials

$$\tag{2} F(x;\hbar)~:=~e^{if(x;\hbar)/\hbar}\quad\text{and}\quad G(p;\hbar)~:=~e^{ig(p;\hbar)/\hbar}.$$

Now assume that

$$\tag{3} G(p;\hbar)~=~\int \! dx~ e^{-ipx/\hbar} F(x;\hbar)$$

is the Fourier transform of $F(x;\hbar)$. We can use the WKB/stationary phase approximation to deduce that the classical parts $f_0(x)$ and $-g_0(p)$ are then Legendre duals of each other, i.e.

$$\tag{4} g_0(p) ~=~ -px+f_0(x)\quad\text{where}\quad p~=~f_0^{\prime}(x),$$

for a sufficiently nice function $f_0(x)$.

  • $\begingroup$ Thanks, that's very interesting. But I'm still left wondering whether the initial assumption that x and p are Fourier conjugates (ie, that they have the usual canonical commutation relations between them) is an axiom that must be accepted on purely empirical grounds or if it follows somehow from the fact that they are Legendre conjugates. You've started with an assumed connection between x and p and then shown that any functions of them which are Fourier conjugates must be Legendre conjugates at tree level. Can we say anything similar about x and p themselves without being circular? $\endgroup$ Aug 23, 2015 at 17:16

I think I've figured out the answer to my own question.

It seems like the connection between the two originates from Feynman's path integral formulation of quantum mechanics.

To see how, let's pretend for a moment that we've never heard of the Schrödinger Equation and we don't know what the commutation relations are between x and p or that states in those bases are related by Fourier transforms. Instead, all we know is that there is some Lagrangian quadratic in $\dot{x}$ and that the way to evolve an initial state at time t_i to a final state at time t_f is to integrate $\exp(iS)$ over all paths from $x(t_i)$ to $x(t_f)$. Where the action S is the time integral of the Lagrangian along a path.

If you consider the limit of a very small time interval between $t_i$ and $t_f$, then the integrand of the path integral becomes:

$$ \exp(i(p\dot{x}-H)\Delta t) = \exp(ip\Delta x)\exp(-iH\Delta t) $$

where $\Delta t = t_f - t_i$ and $\Delta x = x_f - x_i$

The lefthand side demonstrates the Legendre conjugacy relationship between x and p, while the righthand side demonstrates the Fourier conjugacy relationship between x and p. If you start from the assumption that we are just rewriting the action in terms of the Hamiltonian, by performing a Legendre transformation, then you end up with a product of two factors. The $\exp(iH\Delta t)$ factor can be interpreted as the Schrodinger Equation, while the $\exp(ip\Delta x)$ factor can be interpreted as a Fourier transform which tells you how momentum states relate to position states. The fact that a Fourier transform is required to change from x to p basis while summing over states is equivalent to the canonical commutation relations between x and p; so both of these, as well as the Schrodinger Equation, can be considered to have been derived from the starting assumption of the path integral.

In summary, it looks like the mathematical relationship between Fourier and Legendre conjugates is somewhat analogous to the relationship between Lie Algebras and Lie Groups. The Schrodinger Equation in terms of the Hamiltonian, where x and p are Fourier conjugates, gives you the infinitesimal rule for time evolution. While the Feynman path integral, where x and p are Legendre conjugates, gives you the exponentiated version of this rule which tells you how to connect an initial state to a final state after some finite amount of time. One tells you about the local properties while the other tells you about global properties, but they both say the same thing.

This seems like it's probably related to @Qmechanic's answer, but I'm not sure exactly how.


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