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?
Quantum Mechanics – Understanding the Meaning of an Operator
operatorsquantum mechanics
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I think you might try approaching this in the Heisenberg picture.
The time derivative of the position operator is:
$$\dfrac{d \hat x}{dt} = \dfrac{i}{\hbar}[\hat H, \hat x]$$
which is a reasonable velocity operator. The time derivative of the velocity operator is then:
$$\dfrac{d^2 \hat x}{dt^2} = \dfrac{i}{\hbar}[\hat H, \dfrac{d \hat x}{dt}]$$
For example, consider a free particle so that $\hat H = \frac{\hat P^2}{2m}$. The velocity operator would then be $\frac{\hat P}{m}$. This certainly looks reasonable as it is of the form of the classical $\vec v = \frac{\vec p}{m}$ relationship.
But, note that the velocity operator commutes with this Hamiltonian so the commutator in the definition of the acceleration operator is 0. But that is what it must be since we're assuming the Hamiltonian of a free particle which means there is no force acting on it.
Now, consider a particle in a potential so that $\hat H = \frac{\hat P^2}{2m} + \hat U$. The velocity operator, for this system, is then $\frac{\hat P}{m} + \frac{i}{\hbar}[\hat U, \hat x]$.
Assuming the potential is not a function of momentum, the commutator is zero and the velocity operator is the same as for the free particle.
The acceleration operator is then $\dfrac{i}{\hbar}[\hat U, \frac{\hat P}{m}]$.
In the position basis, this operator is just $\frac{-\nabla U(\vec x)}{m} $ which looks like the acceleration of a classical particle of mass m in a potential given by $U(\vec x)$.
The wave function isn't an operator; the word "operator" in quantum mechanics means something more precise than "function". You might say that the momentum operator is "something to put in an integral to get the expectation value of momentum"; let me explain.
We know that for a particle in a state $\psi$,
$$ \langle x \rangle = \int_{-\infty}^{+\infty} x |\psi(x,t)|^2 \mathrm{d}x $$
because $|\psi(x,t)|^2\mathrm{d}x$ is the probability that the particle will be found in the small interval $(x,x+\mathrm{d}x)$ at time $t$. Let's differentiate this and push the derivative into the integral
$$ \frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = \int_{-\infty}^{+\infty} x \frac{\partial|\psi(x,t)|^2}{\partial t} \mathrm{d}x $$
Now we use Schrodinger's equation to replace the time derivative with space derivatives
$$ \frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = \frac{i\hbar}{2m}\int x \frac{\partial}{\partial x} \left(\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial \psi^*}{\partial x}\right) \mathrm{d}x $$
Use integration by parts to eliminate integrate away the outer $\frac{\partial}{\partial x}$ and differentiate the $x$
$$ \frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = -\frac{i\hbar}{2m}\int \left(\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial \psi^*}{\partial x}\right) \mathrm{d}x $$
Perform another integration by parts on the second term
$$ \frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = -\frac{i\hbar}{m}\int \left(\psi^*\frac{\partial\psi}{\partial x}\right) \mathrm{d}x $$
Multiply by $m$
$$ \langle p \rangle = \int \psi^* \left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right) \psi \mathrm{d}x $$
Notice that this is the same form as the integral for $\langle x \rangle$, except we use $\frac{\hbar}{i}\frac{\partial}{\partial x}$ in place of $x$. That's why they're called the momentum and position operators respectively; they are the operators you place between $\psi^*$ and $\psi$ in the integral to obtain the expectation value of that variable. There are also operators for angular momentum, energy, etc.
Of course, there are other more useful definitions of an operator. For example, notice that if we write $\psi$ as a sum of sinusoidal functions, each sinusoidal function is an eigenfunction of the momentum operator, and the integral becomes very simple to evaluate; hence we could think of the momentum operator as an operator that returns the sum of the component sinusoidal functions (momentum eigenstates) of $\psi$, weighted by momentum, which is how I like to think of it as.
Best Answer
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.)