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In classical mechanics momentum is the generator of spatial translations. This remains true in quantum mechanics. The way we define the momentum operator in one-dimension, for example, already shows that

$$\langle x |P |\psi\rangle =-i\hbar \dfrac{\partial}{\partial x}\langle x|\psi\rangle \, . $$

Now if we have a particle in one dimension and this particle is acted upon by a spatial translation, its position will change. In other words, its $x$ coordinate will change.

We have thus, two observables: the momentum $P$ and the position $X$. The momentum is the generator of translations. In that way, momentum generates transformations which directly affect the position $X$.

On the other hand we know that the canonical commutation relation (CCR) $$[X,P]=i\hbar$$ is enough to characterize $X$ and $P$.

Following my reasoning, is there a deeper connection between momentum being the generator of spatial translations and the CCR? How the two things relate? Can we interpret and make sense of the CCR by thinking about this point of view that momentum generates spatial translations?

DanielSank
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Gold
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3 Answers3

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Generically, given two self-adjoint operators $A,B$, the transformation of $A$ under the unitary transformation $U_B(t) := \mathrm{e}^{\mathrm{i}Bt}$ with parameter $t$ generated by $B$ through Stone's theorem is given for central $[X,Y]$ by a form of the BCH formula: $$ U_B(t)AU_B(t)^\dagger = \mathrm{e}^{\mathrm{i}Bt}A\mathrm{e}^{-\mathrm{i}Bt} = A + [B,A]t$$ that is, transforming $A$ by $B$ just shifts $A$ by the commutator of the two, and even if the commutator is not central, this still holds for infinitesimal $t$. This is the quantum version of the classical statement that the Poisson bracket of two functions on the phase space gives the infinitesimal shift of one by the transformation generated by the other, cf. this answer of mine.

In the case of $x$ and $p$, the commutator is unity, so transforming with parameter $t$ just shifts one observable by $t$. That is, the commutation relation is indeed the quantum mechanical version of the statement that the position operator generates translations in momentum, and the momentum generates translations in position.

Conversely, knowing that the translation operator is infinitesimally given by the momentum operator allows one to deduce the form of the momentum operator itself if the representation of the position operator is fixed, see this question. It is essentially the content of the Stone-von Neumann theorem that the commutation relations between position and momentum (or rather their exponentiated form, the Weyl relation between the translation in position and the translation in momentum) uniquely (up to commutation relation preserving unitary isomorphism) fixes the operators themselves.

ACuriousMind
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Yes there is a deep connection. Suppose you just tell me that $P$ is the generator of translation. Then I know that all position states $|r\rangle$ can be obtained as $$|x\rangle=\exp(iP\cdot x)|0\rangle$$ This can be used to determine the action of $P$ on a position eigenstate. $$P|x\rangle=P\exp(iP\cdot x)|0\rangle=(-i\nabla_x)\exp(iP\cdot x)|0\rangle=(-i\nabla_x)|x\rangle$$ With this I can determine the commutation relation for each position state $$[X,P]|x\rangle=XP|x\rangle-PX|x\rangle=X(-i\nabla_x)|x\rangle-(-i\nabla_x)X|x\rangle=i|x\rangle$$ Finally we know that any state can be expressed as a superposition of position states so $$[X,P]|\psi\rangle=[X,P]\sum_x \psi(x)|x\rangle=i|\psi\rangle$$ true for any state $|\psi\rangle$ with wavefunction $\psi(x)$. You can restore the $\hbar$ everywhere to make units work out.

James Rowland
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  • This answer begs the question. How did $P$ magically turn into $-i \nabla_x$? – DanielSank Mar 06 '16 at 02:36
  • I hope the edit is clear. – James Rowland Mar 06 '16 at 02:55
  • Yeah, that's way more clear :) – DanielSank Mar 06 '16 at 03:35
  • I only might add (implied in several of these answers, but not quite stated directly) that the natural field of study to learn more about this connection is that of Lie groups and Lie algebra. In this language, one would say that all the spatial translations together form a Lie group, and the commutation relation between X and P is the corresponding Lie algebra, which is called the Heisenberg algebra. Study of the Lie groups in general will indeed allow you to understand this connection and how it relates to, for example, the angular momentum commutation relations. – Rococo Mar 06 '16 at 22:10
  • Your sign is wrong, which you may verify by complex-conjugating the first formula in the question! In fact, acting on kets, P translates you in the opposite direction to its action on bras, the canonical one. See this answer. You should have, instead, $|x\rangle=\exp(-iP\cdot x)|0\rangle$, however counterintuitive this might look. – Cosmas Zachos Feb 28 '20 at 22:46
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Suppose there is a normalized state $|s>$ where $<s|s>=1$. The expectation value of its position is

$$ x_0=<s|X|s> $$

If the object |s> is translated by $x$, the new state is $e^{-\frac{ixP}{\hbar}}|s>$, and the expectation value $x_{new}$ of its position is $$ x_{new}=<s|e^{\frac{ixP}{\hbar}}Xe^{-\frac{ixP}{\hbar}}|s> $$ Then using the BCH formula (which is just expanding the exponentials and collecting terms) $$ x_{new}=<s|X+[\frac{ixP}{\hbar},X]+\frac{1}{2!}[\frac{ixP}{\hbar},[\frac{ixP}{\hbar},X]]+\frac{1}{3!}[\frac{ixP}{\hbar},[\frac{ixP}{\hbar},[\frac{ixP}{\hbar},X]]]+...|s> $$ Now substitute in $[P,X]=-i\hbar$ and notice the higher order multiple commutators are with a constant and are therefore zero. $$ x_{new}=<s|X+x+0+0+...|s> $$ $$ x_{new}=x_0+x $$ As advertised, the translation operator $e^{-\frac{ixP}{\hbar}}$ with the canonical comutation relation $[X,P]=i\hbar$ moved the state so that its new expectation value of position increased by $x$. The generator of the transformation (ie: the operator in the exponent) is P, which is therefore called the generator of translation.

Gary Godfrey
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