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Commutators and Uncertainty

Uncertainty is one of the most puzzling, yet most fundamental, aspects of quantum mechanics. It initially seems counterintuitive that it is impossible, not practically, but in-principle, to know both the position and momentum of a particle with arbitrary precision. I want to emphasize that this is not a limitation of our measuring devices, but a fundamental property of the universe itself. Even if we had perfect ideal measuring devices, we would still be unable to know both the position and momentum of a particle with arbitrary precision.

There are a few ways to approach the Uncertainty Principle. Here, we will use various concepts from linear algebra to derive the Uncertainty Principle. Sometimes, the Uncertainty Principle is derived from Fourier analysis, which deserves its own discussion.

Table of Contents​

Poisson Brackets in Classical Mechanics​

A bit of background knowledge is required. Hamiltonian mechanics is a reformulation of classical mechanics, and many concepts in Hamiltonian mechanics are directly applicable to quantum mechanics. One of these is the concept of the Poisson bracket.

Hamiltonian mechanics is based around the concept of the Hamiltonian, which is a function that describes the total energy of a system. The Hamiltonian is a function of the generalized coordinates and generalized momenta , which are related to the position and momentum of the particles in the system.

The total mechanical energy of a system is given by the sum of the kinetic and potential energies:

In Hamiltonian mechanics, it is convenient to visualize and as a pair of coordinates on a plane known as the Phase space. The phase space is a 2D space where each point represents a possible state of the system. Over time, the system moves through the phase space, tracing out a path known as the trajectory.

We can define any quantity based on the position and momentum of the system. For instance, for a mass-spring system, we can define the potential as .

The Poisson bracket is a mathematical operation that describes how two quantities evolve over time. For two quantities and , the Poisson bracket is defined as:

The Poisson bracket is a measure of how two quantities change with respect to each other. There are a few reasons we use the Poisson bracket in Hamiltonian mechanics - they reveal deep connections in the system.

For example, consider the poisson bracket :

where we have used the fact that (since is independent of ) and Newton's second law . In fact, is a valid representation of Newton's second law in Hamiltonian mechanics.

Let's take another example: .

More generally, the Poisson bracket is a measure of how the quantity changes over time. To see this, consider the time evolution of . We can use the multivariable chain rule to write:

From our previous results, and . Thus:

This is the first important result of the Poisson bracket - it describes how a quantity changes over time. Another one comes from the conservation of a quantity. If a quantity is conserved, then , and hence . When this happens, we say that commutes with the Hamiltonian .

The Poisson bracket's connection to conserved quantities is important, and it leads to the Hamiltonian version of Noether's theorem.

The Commutator​

In quantum mechanics, we replace the Poisson bracket with the commutator. The commutator of two operators and is defined as:

If both operators commute, then , just like in classical mechanics. If the commutator is nonzero, then the operators do not commute.

If we add instead of subtracting, we get the anti-commutator, another important operation, defined as:

With both the commutator and anti-commutator, we can write a general property of operators:

Eigenvalue Properties​

Suppose two operators and represent two observables in quantum mechanics, and they happen to commute. We can make some interesting observations about the system.

First, suppose is an eigenvector of with eigenvalue . In other words, . Then, we can act on both sides of this equation with :

But since the operators commute, , so we can rewrite the right-hand side as . Thus:

In other words, is also an eigenvector of with the same eigenvalue .

There are two types of eigenvalues: degenerate and non-degenerate. If has a non-degenerate eigenvalue, then that eigenvalue only has one corresponding eigenvector. Otherwise, if has a degenerate eigenvalue, then there are multiple eigenvectors corresponding to that eigenvalue.

In our case, if is non-degenerate, then it only has one eigenvector . But that means is nothing but a scalar multiple of , which means . In other words, is also an eigenvector of !

For degenerate eigenvalues, the situation is a bit more complicated. Suppose is degenerate. The eigenvectors of form a basis for the degenerate eigenspace; acting on any vector in this eigenspace scales it by .

First, suppose lies within this eigenspace. As we know from Equation , . So is also an eigenvector of with eigenvalue . But that means that is inside the degenerate eigenspace. Vectors within the degenerate eigenspace of remain within said eigenspace when acted upon by .

Second, suppose is orthogonal to this eigenspace. If the eigenspace is a plane, for example, is a vector perpendicular to this plane. What happens when we act on ? To find out, we can try taking the inner product . Since is a Hermitian operator, we can write this as . Here's the key insight: since is within the degenerate eigenspace, is also within the degenerate eigenspace (check the previous paragraph). And hence, since is orthogonal to the eigenspace, . Vectors orthogonal to the degenerate eigenspace of remain orthogonal to said eigenspace when acted upon by .

Combining the two, let be any arbitrary eigenvector of . It can be written out as a linear combination of vectors within the eigenspace and orthogonal to it:

Acting on this vector, we get by linearity:

But we know that is an eigenvector of , so . Furthermore, we know that and must stay inside and orthogonal to the degenerate eigenspace of , respectively. As such, the only way for this equation to hold is if both and are scaled by the same factor :

Hence, both and are eigenvectors of with the same eigenvalue . The eigenvectors of are either within the degenerate eigenspace of or orthogonal to it.

In order for 's eigenvectors to span the eigenspace of , they must be within the eigenspace. If they're orthogonal, then they can't span the eigenspace. But if they are within the eigenspace, then by definition, they are eigenvectors of . As such, the eigenvalue property holds even for degenerate eigenvalues.

The conclusion is the following:

If two operators and commute, then they share a common eigenvector basis.

This is a powerful result, and it has deep implications in quantum mechanics. Let and be the momentum and energy operators, and suppose they commute (and hence share a simultaneous eigenbasis). Then, if we make a measurement of the momentum of the particle, the state vector collapses to an eigenstate of . But we know that the eigenvector for is also an eigenvector for . Hence, if we make a measurement for the energy immediately after the momentum measurement, we will still get a definite value. In other words, we can measure the momentum and energy of a particle simultaneously with arbitrary precision:

If two operators and commute, a particle can have a definite value for both observables simultaneously.

Non-Commuting Operators​

Now, let's consider the case where two operators do not commute. Once again, let and be two operators. This time, suppose they do not commute. In this case, they cannot share a common eigenvector basis.

We can make a proof by contradiction.

Suppose they did share a common eigenvector basis . Then, we can write both operators as linear combinations of projectors onto these eigenvectors:

Next, apply to :

Similarly, apply to :

Since and are just scalars, multiplication between them is commutative and we can switch them around. Other than that, since and are just indices, we can switch them around as well. Hence, we can rewrite the second equation as:

Thus, we have shown that . But this contradicts our initial assumption that and do not commute. Hence, our assumption that they share a common eigenvector basis is false. Therefore, if two operators do not commute, they do not share a common eigenvector basis.

If two operators don't commute, then, there exists eigenvectors of one operator that are linear combinations of eigenvectors of the other operator. If and are the position and momentum operators, and if they do not commute, then it means that there exists eigenvectors of that are superpositions of eigenvectors of .

In other words, if one measures the position of a particle, the state vector will collapse onto the eigenvector of the position operator. However, it will still be in a superposition of momentum eigenstates, and hence, will require multiple measurements to determine the momentum of the particle.

If two operators and do not commute, then a particle cannot have a definite value for both observables simultaneously.

This motivates the Uncertainty Principle, which states that the position and momentum of a particle cannot be known with arbitrary precision.

Hermiticity​

Let's consider what happens if we take the Hermitian adjoint of the commutator:

If and are Hermitian operators, then and . Hence, the commutator of two Hermitian operators is:

If an operator is equal to the negative of its Hermitian adjoint, then it is known as an anti-Hermitian operator. As such, the commutator of two Hermitian operators is an anti-Hermitian operator.

The anti-commutator of two Hermitian operators, on the other hand, is a Hermitian operator. This can be seen by taking the Hermitian adjoint of the anti-commutator:

Hence, the anti-commutator of two Hermitian operators is a Hermitian operator.

In conclusion:

If two operators and are Hermitian, then the commutator is anti-Hermitian, and the anti-commutator is Hermitian.

The Uncertainty Principle​

The Uncertainty Principle is a fundamental property of quantum mechanics.

First, let's define the following operator:

This operator represents the deviation of the observable from its expectation value . The expectation value of is known as the variance of . We can show that this indeed matches the definition of variance:

When a state vector collapses onto an eigenvector, then this variance is zero.

The general form of the Uncertainty Principle is given by the following inequality:

Uncertainty Principle: For any two observables with operators and :

We will now derive this inequality.

Derivation​

We follow the derivation given in Sakurai's book. First, recall that the Cauchy-Schwarz inequality states that for any two vectors and :

Using Dirac notation:

(If you read my notes on linear algebra, I told you we would use it to prove the uncertainty principle! 😉)

Sakurai has a very short proof for the Cauchy-Schwarz inequality:

Sakurai's Proof

Let and be two arbitrary vectors, and be an arbitrary complex number. We can see that:

since the dot product of any vector with itself is nonnegative. Since this holds for any complex , we can choose . This yields:

Multiplying both sides by and rearranging gives the Cauchy-Schwarz inequality:

Thus the Cauchy-Schwarz inequality is proven.

Next, we need another result: the expectation value of a Hermitian operator is always real, and the expectation value of an anti-Hermitian operator is always imaginary.

We know that Hermitian operators represent observables in quantum mechanics. As such, they have to have real expectation values by definition. Mathematically, the expectation value of is . By the definition of Hermitian operators, , and hence . If a complex number is equal to its complex conjugate, then it must be real.

Similarly, the expectation value of an anti-Hermitian operator is . Applying the same logic, . If a complex number is equal to the negative of its complex conjugate, then it must be imaginary.

Now, let's prove the Uncertainty Principle.

First, define two kets and as follows:

where can be any arbitrary ket. By the Cauchy-Schwarz inequality applied to these kets:

Since and are Hermitian operators, we can freely interchange the order of the bras and kets. The inner product is then , which is equal to the expectation value of . Similarly, the inner products and are the expectation values of and . Hence, we can rewrite the Cauchy-Schwarz inequality as:

By Equation , we can write as:

Taking the expectation values of both sides gives:

We know from Hermiticity that the commutator of two Hermitian operators is anti-Hermitian, and the anti-commutator is Hermitian. As such, the expectation value of is imaginary, and the expectation value of is real. Hence, the square of the magnitude of the expectation value of the commutator is:

The commutator is simply equal to . (To see this, expand and in terms of and , and then subtract them.) Thus:

Plugging this back into Equation gives:

Finally, we can get rid of the first term on the left - it does not change the inequality. This gives

which is the Uncertainty Principle.

There are actually other "uncertainty principles"; this is not the only one. For example, the Maccone-Pati uncertainty relation describes the sum of variances instead of the product.

Summary and Next Steps​

In this note, we derived the Uncertainty Principle using the commutator of two operators.

Here are the key things to remember:

  • The Poisson bracket in classical mechanics is defined as:

    • The Poisson bracket of any quantity with the Hamiltonian is a measure of how changes over time.
    • If the Poisson bracket of a quantity with the Hamiltonian is zero, then is conserved.

  • The commutator in quantum mechanics is defined as:

  • The anti-commutator in quantum mechanics is defined as:

  • The composition of the commutator and anti-commutator is:

  • An anti-Hermitian operator is equal to the negative of its Hermitian adjoint, and a Hermitian operator is equal to its Hermitian adjoint.

  • The commutator of two Hermitian operators is anti-Hermitian, and the anti-commutator of two Hermitian operators is Hermitian.

  • If two operators commute, then they share a common eigenvector basis, whether degenerate or non-degenerate. As a result, if two operators commute, a particle can have a definite value for both observables simultaneously.

  • If two operators do not commute, then they do not share a common eigenvector basis. As a result, if two operators do not commute, a particle cannot have a definite value for both observables simultaneously. This leads to the Uncertainty Principle.

  • The Uncertainty Principle states:

In the next section, we will explore matrix representations of operators and how they can be used.