When I was first introduced to the quantum operators, I thought the terminology was just a convention, like the momentum operator is just a convention and it's not really momentum per se, because momentum is defined for classical particles in motion. But then I found the expectation values of those operators behaving just like their classical analogues. The momentum operator is derived from the position operator in the same classical way, and the derivative of the momentum operator with respect to time is a carbon copy of Newton's Second Law. Can someone explain this confusion to me?


2 Answers 2


In all of physics there is a difference between the mathematical concepts and the physical goings-on. For example, in classical physics a velocity in a physical sense is the rate of change of the physical position of something, and in a mathematical abstract sense it can be represented as a mathematical vector $\bf v$. In classical physics we often say that this $\bf v$ 'is' the velocity and we don't need to worry about the distinction between the abstraction and the physical happening. This is because the mapping between the maths and the physics is fairly direct.

In quantum physics, on the other hand, the mapping is not quite so direct, and for this reason it helps to keep distinct in one's mind the mathematical concepts and the physical happenings. The word 'represent' is helpful here. We may say that the operator $\hat{\bf p}$ represents the momentum and the operator $\hat{\bf H}$ represents the energy, and so on.

The state of motion of some physical system is represented by a vector in Hilbert space, often called a state vector or a ket, written thus: $| \psi \rangle$. (The reader may know this but it helps to mention it to see where I am going with the logic).

The result of some physical observation is neither a ket nor an operator but a scalar, that is, a number (with units). The number could be written for example $$ \langle \psi | \hat{\bf p} | \psi \rangle . $$ This example gives the average of the momentum observations if they were to be carried out on a large sample of systems all in the state represented by $|\psi\rangle$. So you see neither the state vector nor the operator on its own tells us what will be observed, but rather both together yield a statement which can be compared with physical observations.

For this reason it is not best to say, of a quantum system, that $\hat{\bf p}$ 'is' it's momentum, but one may say that $\hat{\bf p}$ is the momentum operator and it represents (in certain respects) the system's momentum.

To bring out the fact that the two parts of the description (ket and operator) work together, one can notice that there are many ways to track the dynamics. The two most standard ways are named after Schrodinger and Heisenberg. In the Schrodinger approach the state of the system evolves with time and the basic operators do not. In the Heisenberg approach the state of the system stays fixed and the basic operators evolve.

Let $| \psi(0) \rangle$ be the state of the system at time zero. Both approaches can agree on this. Let $\hat{\bf p}$ be the momentum operator in the Schrodinger representation. Let $\hat{U(t)}$ be the time-evolution operator, also called propagator. Then the Schrodinger state at time $t$ is $$ | \psi(t) \rangle = U(t) | \psi(0) \rangle $$ and the expectation value of momentum at time $t$ is $$ \langle \psi(t) | \hat{\bf p} | \psi(t)\rangle = \langle \psi(0) | U^\dagger(t) \hat{\bf p} U(t)| \psi(0) \rangle. $$

Ok, now let's look at the Heisenberg picture. In this approach the state vector stays equal to $| \psi(0) \rangle$ so the state at time $t$ is $$ | \psi_H(t) \rangle = | \psi_H(0) \rangle = | \psi(0) \rangle. $$ (I am putting a subscript $H$ for Heisenberg here.) This seems rather odd when you first meet it, because one would think the system state evolves, as the Schrodinger picture says it does. But in the Heisenberg picture the momentum is not represented by $\hat{\bf p}$, it is represented by $$ \hat{\bf p}_H = U^\dagger(t) \hat{\bf p} U(t). $$ The expectation value of momentum at time $t$ is $$ \langle \psi_H(t) | \hat{\bf p}_H | \psi_H(t)\rangle = \langle \psi_H(0) | U^\dagger(t) \hat{\bf p} U(t)| \psi_H(0) \rangle = \langle \psi(0) | U^\dagger(t) \hat{\bf p} U(t)| \psi(0) \rangle. $$ This is the same answer as in the Schrodinger picture, so the two calculations agree, and more generally the two approaches agree about all predictions concerning physical observations.

I wrote all this in order to bring out the fact that operators do not represent the physical quantities in a very direct way; and the state vector does not represent the physical state in a very direct way. Rather the physical stuff is captured in the mathematics by the combination of operators and kets (or state vectors).

(And by the way, expectation values behave somewhat like but not exactly like their classical counterparts.)


Not quite sure where your confusion lies, since you seem to have grasped the concept. There is a linear operator corresponding to each observable of a quantum system. The eigenvalues of the operator tell you what values the observable can take (and the physical requirement that these eigenvalues be real means that each operator is Hermitian). The algebra of operators tells you how the measurement of one observable will affect the subsequent measurement of another observable, and gives rise to constraints such as the Heisenberg uncertainty principle.

It Is not surprising that the expectation values of quantum operators behave like their classical analogues. Indeed, it would be surprising if that were not true, since it would imply that there was a disconnect between quantum and classical viewpoints.

Whether the momentum operator actually is momentum or is just a representation of momentum etc. is a question of semantics which probably worries philosophers more than physicists.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.