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Position and Momentum

In the previous section, we applied the state vector formalism to spin-1/2 particles. Notably, it has a discrete spectrum, and the state vector is a two-component column vector.

We shall now apply the state vector formalism to the description of the position and momentum of a particle. We shall see that the position and momentum operators are the generators of translations in space and time, respectively. Furthermore, we will discuss the canonical commutation relations between the position and momentum operators.

Table of Contents

Position Operator

The position observable has a continuous spectrum; its values can be any real number. Furthermore, we know that there are three spatial dimensions, so the position operator must be a vector operator. We assume that the position eigenkets form a complete basis for the Hilbert space of the particle (an assumption in non-relativistic quantum mechanics).

For a single dimension, the state vector can be expanded as:

In three dimensions, the state vector can be expanded as:

In this case, then, is an eigenket of all three components of the position operator. From what we know about commutators, this means that the position operators in different directions have a commutator of zero:

Translation Operators

Suppose we have a state for a particle at position as . Let's define an operator that translates the state by an infinitesimal distance . We follow Sakurai's notation and denote this operator as :

To see how this operator acts on any state , we can expand in terms of the position eigenkets:

Then, we can apply the translation operator to :

If we make the substitution , we get:

This substitution is valid because the integral is over all space, and the substitution is a simple relabeling of the integration variable. Hence, we have:

Recall that the wavefunction for a state is where is an eigenket of an observable. In our case, the position wavefunction is . When acts on , it changes the wavefunctions from to . This is equivalent to a translation of the wavefunction by .

Properties of Translation Operators

From physical intuition, we expect the translation operator to have the following properties:

  • As , because a translation by zero is the identity operator.

  • If we translate by and then by , it should be the same as translating by :

  • If we translate by , it should be the inverse of translating by :

  • The translation operator should preserve the total probability of finding the particle in a region of space. This means that the translation operator should preserve the inner product of two states:

From the last property, after shifting the operators around a bit, we get:

Thus, the translation operator is unitary.

Explicit Form of Translation Operators

To derive the explicit form of the translation operator, we start with the Taylor expansion of the translation operator:

For the sake of brevity, we will write as :

Recall from the properties of the operator that . Expanding each operator with the Taylor series, we get:

Canceling the terms and the terms, we get:

In other words, . This means that is an anti-Hermitian operator. We can show that is Hermitian:

Hence, is Hermitian. Denote this new Hermitian operator as , with the minus sign by convention. Plugging this back into the Taylor expansion, we get:

For an infinitesimal translation, the terms are negligible, so we can write:

Operator Identity

We can derive a fundamental identity for the operator and the position operator . Consider the commutator of the translation operator with the position operator, :

The first part of the commutator is:

Recall that the position operator acts on the position eigenket as because the eigenket is an eigenstate of the position operator. Hence, the first part of the commutator is:

The second part of the commutator is:

is a constant (it is the eigenvalue of the position operator), so we can pull it out of the translation operator:

Then, we can expand the translation operator:

Thus, the commutator is:

and differ by the second order in , so we can make the approximation:

As such:

This is an operator identity that relates the position operator to the translation operator. This is an operator identity because it holds for any state and the position eigenkets form a complete basis for the Hilbert space. The differential is a number times the identity operator.

If we use the explicit form of the translation operator, we get:

As such, we can write:

If we set , then Equation becomes:

Since , we can write:

is a number, so we can pull it out of the commutator. Then dividing by yields:

Momenta

We will now discuss the physical significance of the operator. From classical mechanics, we know that momentum is the generator of translations in space (see the appendix for a brief review).

From Equation , we have:

Applying both sides to a state , we get:

Rearranging gives:

Since , we have:

The left-hand side is the change in the position of the particle, . Dividing both sides over gives:

The left-hand side becomes a derivative as :

This bears a striking resemblance to the following relation in classical mechanics:

in which we can make the following correspondence:

  • is the Lagrangian, which is analogous to the state vector .
  • is the position, which is analogous to the position operator .
  • is the momentum, which is analogous to the operator .

We thus say that the operator is somehow related to the momentum operator . The dimensions of are (you can see it from the commutation relation ). The dimensions of momentum are . To make the dimensions of match those of momentum, we need to multiply by a constant with dimensions of inverse action:

Since macroscopic physics was constructed before quantum mechanics, our formulation of quantum mechanics necessitates this constant. If we were to start from scratch, we would certainly create a set of units in which this constant is unity (1). In fact, this is what we do in a system called natural units (more precisely, natural units are a set of different systems).

Sakurai provides a useful analogy to understand this constant. The energy of two particles with a charge and a separation is proportional to . In SI units, the constant of proportionality is , but in unrationalized Gaussian cgs units, this constant is simply .

This constant appears in de Broglie's relation for the wavelength of matter waves:

(where is the wavenumber and is our constant with dimensions of action). We finally have:

or,

The infinitesimal translation operator is then:

The commutation relation in can be written as:

Pullling out the term gives:

This is the canonical commutation relation between the position and momentum operators.

Position-Momentum Uncertainty Relation

Recall previously that the uncertainty of the expectation value of two operators and is given by:

Now that we have the canonical commutation relation between the position and momentum operators, we can apply this to the position and momentum operators:

Finite Translation Operators

We have derived the infinitesimal translation operator in Equation . To find the finite translation operator, we simply need to keep applying the infinitesimal translation operator:

If we have infinitesimal translations, then the infinitesimal length is . The right-hand side then becomes:

In the -direction, for example, we have:

As , the right-hand side becomes:

Recall that the exponential function can be written as:

In our case, the "something" is . Thus, we can write:

In vector form, we have:

Note that it would have been possible to derive this result directly from the properties of the translation operator had we not ignored the terms in the Taylor expansion. The infinitesimal expression is a special case of the finite expression - the first term in the Taylor expansion of the exponential function.

In the future, we will see that the operator is equal to in the position eigenbasis. It is amusing to see that we have the exponential of a differential operator, . More generally speaking, Taylor's formula states that the exponential of a differential operator is the translation operator:

Commutation Relations for Finite Translations

Intuitively, it seems apparent that finite translations in different directions should commute. This is because it does not matter which direction we translate first; the final position is the same.

It turns out that they do indeed commute:

We can expand each operator in terms of the exponential function:

Their commutator is then:

By asserting that the commutator of and is zero, we get:

More generally, for all and . Since momentum is the generator of translations, this means that translations in different directions commute. In mathematical terms, the group of translations in three dimensions is an Abelian group.

Since the momentum operators commute, we can simultaneously measure the momentum in different directions. This means we can construct a simultaneous eigenket of the momentum operators:

Since they are eigenkets of the momentum operators, we have:

where , , and are the eigenvalues of the momentum operators , , and respectively.

Dirac's Rule for Canonical Commutation Relations

The canonical commutation relations are the fundamental commutation relations between the position and momentum operators. We have derived all the commutation relations we need:

Dirac proposed a rule for creating the commutation relations between operators by observing their analogue in classical mechanics. More specifically, he proposed that the commutator of two operators is equal to the Poisson bracket of their classical counterparts multiplied by :

where and are the classical counterparts of the operators and respectively. Recall that the Poisson bracket is defined as:

The Poisson bracket of the position and momentum in classical mechanics is:

By Dirac's rule, we have then:

which matches the canonical commutation relation we derived earlier. One justification for this rule is that both the commutator and the Poisson bracket satisfy a similar set of properties, even if they have different dimensions (which is why is needed) and one is real while the other is anti-Hermitian (which is why the is needed). More fundamentally, they are both related to the Lie bracket in mathematics.

Summary and Next Steps

In this section, we applied the principles of quantum mechanics to the position and momentum operators.

Here are the key things to remember:

  • The three position operators commute with each other. This means that we can simultaneously measure the position in different directions:

  • The translation operator generates translations in space. The infinitesimal translation operator is:

    which is derived from the properties that we asserted for the translation operator.

    The finite translation operator is:

  • The group of translations in three dimensions is an Abelian group. This means that translations in different directions commute. The momentum operators also commute with each other:

    Hence, we can simultaneously measure the momentum in different directions:

  • The canonical commutation relations are the fundamental commutation relations between the position and momentum operators. They are:

  • Dirac's rule for canonical commutation relations states that the commutator of two operators is equal to the Poisson bracket of their classical counterparts multiplied by : .

In the next section, we study wavefunctions in quantum mechanics.

Appendix: Generators in Hamiltonian Mechanics

In classical mechanics, we have the concept of a generator. A generator is a function that generates a transformation of the phase space.

Recall that the Hamiltonian of a system is a function of the generalized coordinates and the generalized momenta . For simple cases, imagine we just have a Cartesian coordinate and its conjugate momentum . In most simple cases, the Hamiltonian is simply the total energy of the system:

The Lagrangian of the system is given by the difference between the kinetic and potential energies:

Observe what happens if we take the derivative of the Lagrangian with respect to time:

The Euler-Lagrange equation tells us that the term . Thus, . Plugging this back into the equation above, we get:

Recall that is the momentum . Thus:

Rearranging yields:

Since , we have:

In other words, the time derivative of the Hamiltonian is equal to the negative of the time derivative of the Lagrangian.

We say that the Hamiltonian generates time translations.

Here are a few more examples of generators:

(where is the angular momentum, in a different notation to avoid confusion with the Lagrangian ).