Spatial Translation and the Momentum Operator

Introduction

In our last two lessons, we explored the deep connection between symmetry and the mathematical structure of quantum mechanics, culminating in Wigner's theorem. You've seen that any symmetry transformation must be represented by either a unitary or an anti-unitary operator. Now, we will transition from this general and abstract framework to a concrete and fundamental example: spatial translation.

Our goal in this lesson is to construct the operator that implements a spatial shift on a quantum state. We will start from the physical requirement of what a translation does and use this to derive the operator's form. By analyzing an infinitesimal translation, we will identify its generator and rigorously establish that this generator is none other than the momentum operator. This process will reveal the canonical commutation relation, $[X, P] = i ℏ$ , not as an ad-hoc postulate, but as a direct consequence of momentum's role as the generator of spatial translations.

Defining the Translation Operator

Let's begin by formalizing what we mean by a spatial translation in quantum mechanics. A translation by a vector $a$ is a symmetry of space itself. Following Wigner's theorem, it must be represented by a unitary operator, which we'll denote as $U (a)$ .

What is the defining property of this operator? If we have a state $∣ ψ ⟩$ for which the expectation value of position is $⟨ X ⟩$ , then for the translated state $∣ ψ^{'} ⟩ = U (a) ∣ ψ ⟩$ , the new expectation value must be $⟨ X ⟩ + a$ . Let's write this out:

⟨ ψ^{'} ∣ X ∣ ψ^{'} ⟩ = ⟨ ψ ∣ U^{†} (a) X U (a) ∣ ψ ⟩ = ⟨ ψ ∣ X ∣ ψ ⟩ + a

Since this must hold for any state $∣ ψ ⟩$ , it forces a condition on the operators themselves. The constant vector $a$ can be written as an expectation value of the operator $a I$ , where $I$ is the identity. This leads us to the fundamental operator equation that defines the action of a spatial translation:

U^{†} (a) X U (a) = X + a I

Since $U$ is unitary, we can also write this as $U^{- 1} (a) X U (a) = X + a I$ . This equation is our starting point. It states that conjugating the position operator $X$ by the translation operator $U (a)$ is equivalent to adding the displacement vector $a$ to it.

The following resource provides a clear and rigorous derivation based on this principle.

4 Transformations and Symmetries

This document provides an excellent, rigorous treatment of transformations in quantum mechanics. We will use it to guide our derivation of the translation operator.

Please read the beginning of Section 4.2, "Translations". Focus on the argument leading from the physical requirement of a shifted expectation value to the operator equation that defines the translation operator. Read from the start of the section up to equation (4.18).

The Generator of Infinitesimal Translations

Now we arrive at the core of the argument. Translations form a continuous group. This means we can consider an infinitesimal translation by a vector $δ a$ . Any unitary operator representing such a transformation can be expanded around the identity. By convention, we write this as:

U (δ a) = I - \frac{i}{ℏ} P \cdot δ a + O (δ a^{2})

Here, $P$ is a vector of three Hermitian operators called the generators of the transformation. The factor of $- i /ℏ$ is conventional but crucial; it ensures $P$ is Hermitian (as $U$ is unitary) and will ultimately give it the units of momentum. At this stage, $P$ is simply the name we give to the generator; our task is to prove it is indeed the momentum operator.

Let's substitute this infinitesimal form into the defining relation $U^{†} (δ a) X U (δ a) = X + δ a$ . The adjoint is $U^{†} (δ a) = I + \frac{i}{ℏ} P \cdot δ a + \dots$ . Expanding to first order in $δ a$ :

(I + \frac{i}{ℏ} P \cdot δ a) X (I - \frac{i}{ℏ} P \cdot δ a) \approx X + \frac{i}{ℏ} [(P \cdot δ a), X]

Equating this with $X + δ a$ , we find:

\frac{i}{ℏ} [(P \cdot δ a), X] = δ a

Let's analyze this in component form. The $k$ -th component of the position vector is $X_{k}$ . The equation becomes:

\frac{i}{ℏ} [j \sum P_{j} δ a_{j}, X_{k}] = δ a_{k}

Since this must hold for any arbitrary infinitesimal vector $δ a$ , we can equate the coefficients of $δ a_{j}$ on both sides. This leads to:

\frac{i}{ℏ} [P_{j}, X_{k}] = δ_{j k} ⟹ [X_{k}, P_{j}] = i ℏ δ_{j k}

This is the canonical commutation relation. We have not postulated it; we have derived it from the premise that $P$ is the generator of spatial translations. This provides a profound insight into its physical meaning. The non-commutativity of position and momentum is a direct mathematical consequence of the geometry of translations.

This derivation is central to modern quantum mechanics. The following video provides an excellent visual walkthrough of the same logic.

The Symmetry at the Heart of the Canonical Commutation Relation

The video "The Symmetry at the Heart of the Canonical Commutation Relation" from the channel Physics with Elliot provides a clear, step-by-step derivation.

Watch the segment from the definition to the result. The presenter starts by defining the translation operator in the same way we have and then performs the infinitesimal expansion to arrive at the canonical commutation relation.

The Finite Translation Operator

Having identified the infinitesimal generator, we can construct the operator for a finite translation $a$ . We do this by composing a large number $N$ of infinitesimal translations of size $a / N$ and taking the limit as $N \to \infty$ :

U (a) = N \to \infty lim (I - \frac{i}{ℏ} P \cdot \frac{a}{N})^{N} = exp (- \frac{i}{ℏ} a \cdot P)

Thus, we have successfully constructed the spatial translation operator. It is the exponential of the momentum operator. This operator is often denoted by $T (a)$ .

The following image provides a concise summary of the operator and its effect on a wave function in one dimension.

This image summarizes the mathematical form of the spatial translation operator $T(\alpha)$ in one dimension and its effect on a wave function $\Psi(x)$. It graphically depicts how the wave function is shifted from $x=0$ to $x=\alpha$. For comparison, it also shows the analogous momentum translation operator.

Verifying the Action

We've built the operator from first principles. Now, let's verify that it behaves as expected when acting on states. The wave function of a translated state $∣ ψ^{'} ⟩ = U (a) ∣ ψ ⟩$ is given by $ψ^{'} (x) = ⟨ x ∣ U (a) ∣ ψ ⟩$ . Using the unitarity of $U$ , we have $U (a) = (U (a)^{†})^{- 1} = U (- a)^{- 1}$ . So we can write:

ψ^{'} (x) = ⟨ x ∣ U (a) ∣ ψ ⟩ = ⟨ U^{†} (a) x ∣ ψ ⟩ = ⟨ U (- a) x ∣ ψ ⟩

Since $∣ x ⟩$ is an eigenstate of the position operator $X$ , we can show that $U (- a) ∣ x ⟩$ is an eigenstate of $X$ with eigenvalue $x - a$ . Therefore, $U (- a) ∣ x ⟩$ must be proportional to $∣ x - a ⟩$ . With a standard choice of phase, we find $U (- a) ∣ x ⟩ = ∣ x - a ⟩$ . This gives the final result:

ψ^{'} (x) = ⟨ x - a ∣ ψ ⟩ = ψ (x - a)

The new wave function at position $x$ has the value the old wave function had at $x - a$ , which is exactly what we expect from a translation.

The following video takes an alternative and complementary approach. It starts by defining the translation operator as the exponential of the momentum operator and then proves that it correctly translates position eigenstates. This confirms the logical consistency of our framework.

The translation operator in quantum mechanics

The video "The translation operator in quantum mechanics" from Professor M does Science provides a clear demonstration of the operator's properties.

Watch the segment from the commutator calculation. The presenter first calculates the commutator [X, T(\alpha)] and then uses this result to show that if |x\rangle is an eigenstate of X with eigenvalue x, then T(\alpha)|x\rangle is an eigenstate with eigenvalue x+\alpha.

Connection to Classical Mechanics

This deep connection between momentum and translation is not unique to quantum mechanics. It is, in fact, a direct parallel to the structure of Hamiltonian mechanics. Your studies in advanced physics will have familiarized you with Noether's theorem, which connects symmetries to conserved quantities via the Poisson bracket formalism. In classical mechanics, the momentum $p_{i}$ is the generator of translations in the $q_{i}$ direction in the sense that for any function $f (q, p)$ on phase space, the infinitesimal change under translation is given by the Poisson bracket $δ f = ϵ {f, p_{i}}$ .

The transition to quantum mechanics, canonical quantization, famously replaces the Poisson bracket with the commutator:

{A, B}_{PB} \to \frac{1}{i ℏ} [A^, B^]

Seen in this light, our derivation that momentum is the generator of translations is the quantum mechanical analogue of the corresponding theorem in classical dynamics.

[PDF] Chapter 9 Symmetries and transformation groups

This text provides the formal bridge between the classical and quantum pictures.

Skim the subsection "Hamiltonian version" (p. 164) to refresh your memory of how a generator Q creates a canonical transformation via the Poisson bracket: \hat{\delta}q_i = \{Q, q_i\}_{\text{PB}}. Then, read the subsection "Quantization" (p. 165) and the first paragraph of "Energy, momentum and angular momentum" (p. 166). The text explicitly shows how canonical quantization maps the classical generator relationship to the quantum one: the mapping, and then identifies the momentum as the generator of translations.

Conclusion

In this lesson, we have moved from the abstract concept of symmetry to the concrete construction of the spatial translation operator.

Key Takeaways:

The unitary operator for spatial translation, $U (a)$ , is defined by its action on the position operator: $U^{†} (a) X U (a) = X + a$ .
By analyzing an infinitesimal translation, we identified the operator's generator and derived the canonical commutation relation $[X_{k}, P_{j}] = i ℏ δ_{j k}$ . This shows that momentum's status as the generator of translation is the fundamental origin of this relation.
The finite translation operator is constructed by exponentiating the generator: $U (a) = exp (- \frac{i}{ℏ} a \cdot P)$ .
This operator acts on wave functions as expected, transforming $ψ (x)$ into $ψ (x - a)$ .

We have established a crucial link in the chain connecting symmetries to physical operators. In the next lesson, we will complete this chain by proving Noether's theorem in the quantum context: if the Hamiltonian is invariant under spatial translations, then momentum is conserved.

Recommended resources