# Operators Conceptual Meanings in Quantum Mechanics

I'm confused about the meaning of various operators in quantum mechanics. I know that $$X|\psi\rangle = x\psi(x),$$ where $$X$$ is the position operator, and the projection is made onto the position basis. Additionally, $$P|\psi\rangle = -i\hbar\frac{d}{dx}\psi(x),$$ where $$P$$ is the momentum operator and the projection is made onto the position basis. I believe that applying a Hermitian Operator (like $$P$$ or $$X$$) is intended for making measurements. What does such measurements physically represent? Additionally, it seems that the unitary operator $$e^{\frac{-iaP}{\hbar}}$$ modifies $$|x\rangle$$ to $$|x + a\rangle$$ unlike $$P|\psi\rangle.$$ Why are they different, and how does $$P$$ generate a unitary transformation in $$e^{\frac{-iaP}{\hbar}}$$? I believe there's a connection to the Lagrangian (i.e. spacial transformation symmetry), but I'm not sure how it gives that formula. I also don't understand why the unitary operation shifts $$|x\rangle$$ to $$|x + a\rangle$$ but the operation $$P|\psi\rangle = -i\hbar\frac{d}{dx}\psi(x),$$ which seems to retrieve information from the wave.

• One way to deal with the disorganization is to single out one question and ask that. Right now you have a lot of text and it is even difficult to see what exactly you are asking! Even if no one answers, the work spent in trying to organize and single out a question is going to help you understand more, if not everything. Oct 14, 2023 at 22:11
• I apologize for my question's disorganization. I shortened my question to make it easier to read. If it's still difficult to understand, I will edit it again.
– Vikram Kumar
Oct 14, 2023 at 22:40
• Measurements physically represent... measurements ! I.e. the ones you make with instruments $-$ we sometimes tend to forget that physics is an experimental science, even quantum physics, because of the way it is taught. Oct 15, 2023 at 6:41
• $e^{-ia\hat{p}/\hbar} = e^{a\partial_x}$ is nothing else than the translation operator (in the position basis), which you have certainly encountered in your calculus course and whose effect translates $x$, such that $x \mapsto x+a$, hence $e^{a\partial_x}f(x) = f(x+a)$. Oct 15, 2023 at 6:44
• When you write a formula, put dollar signs around the whole formula: $P|\psi\rangle = -i\hbar\frac{d}{dx}\psi(x)$, not just around every single symbol that requires them as you did: P|$\psi$> = -i$\hbar$$\frac{d}{dx}$$\psi$(x). When you are talking about a symbol, like $P$, put dollar signs around it: $P$. See my edits. Oct 15, 2023 at 6:58

## 2 Answers

A measurement of the position in quantum mechanics is to have a small detector with the capability to catch a fish in the river in a small volume $$V(x)$$. There is only one fish in the river for each experiment and the waiting time for detection is unlimited.

If it is a classical fish, the detector is measuring the integral of the density of the static position probability distribution $$\rho(x)$$ multiplied by the characteristic function $$\chi_V(x)$$ over one dimension

$$P(X \in V) = \int_{\mathbb R} \rho(x) \ \chi_V(x) \ dx$$

$$E[X] = \int_{\mathbb R} \ (X \rho(x) ) \ \chi(x) \ dx = \int_{\mathbb R} \ (x \rho(x) ) \ \chi(x) \ dx$$

So the position operator in P-theory is defined by the multiplication of the density with its argument.

If it is a very small quantum fish, we replace each operator X acting on the probabilty density of position by the same operator acting on an element of the Hilbert space

$$X \rho(x) = x \rho(x) \longrightarrow \psi(x)^* (X \psi(x) )= x |\psi(x)|^2$$

The second fundamental operation is motion. The translation operation on analytic functions is readily provided by the Taylor expansion:

$$\psi(x + dx) =\sum_n \frac{dx^n \partial_x^n}{n!} \psi(x) = e^{dx \partial_x} \ \psi(x)$$

Here the exponential can be understood as a formal operator exopnential, that has to coincide with the limit of repeated translation by refinement of the partition of an interval according to

$$\lim_{n\to \infty} (1 + x/n)^n = e^x$$

So its natural to define $$\left(1 + i (- i \partial_x ) \psi(x)\right) - \psi(x)$$ as the canonical displacement differential in Hilbert space. The factor $$i$$ is invented in order to make $$-i\partial_x$$ symmetric with respect to partial integration over all of space and acting in the left factor.

$$\int_{\mathbb R} \psi(x)^* (- i \phi' (x)) dx = \int_{\mathbb R} (-i \psi(x)')^* \phi (x)) dx$$ for all vectors in Hilbert space with images of the derivative inHilbert space, too.

This means that the probabilty distribution of the momentum of that single fish in the experiment is always taken as an integral over the whole space. As always, the probability in quantum mechanics is event counting of independent, single particle experiments with destructive, terminating measurements over times long with respect to the dynamics of the system.

There is a book by Olver called "Applications of Lie Groups to Differential Equations" and I would recommend you read the first chapter to get an intuition as to is happening. This is a Springer Book so you can download the PDF for Free: (https://link.springer.com/book/10.1007/978-1-4684-0274-2)

All of this has to do with infinitesimal generators and the exponential map.

Let's start with the simplest and most fundamental example consider a vector field V, i.e. an object that assigns a vector to each point in your space. Take your space to be a Manifold, i.e. a space that looks locally like Euclidean space.

The important question now is, where do these vectors live?

The answer is that at each point of your space, your manifold M, there is a Vector space called the tangent space at x, $$T_xM$$, where the vector assigned to x lives.

Now consider the following:

Suppose you have a vector field on M, V, and a curve on M whose derivative at each point is given by the tangent vector assgined by V.

That curve is called an integral curve of the vector field. Now consider the maximal integral curve, or flow of a vector field going though one point, that is, (uniequeness by ODE theory) pick some point x on M, then the flow is just the longest integral curve that passes through x.

This flow is precisely what a commutation relation in quantum mechanics describes, or what the momentum generates.

You mentioned the example of $$exp(-iε\hat{p})$$, well here you see what this actually is:

Consider simply $$exp(ε\frac{\partial}{\partial x})$$ acting on a point x.

What is the result, what happens? (Try this out, will give good intuition as to what is happening in general)

You get a translation, $$exp(ε\frac{\partial}{\partial x})= x+ε$$.

In other words the vector field $$v = d/dx$$ generated a translation.(The basis of the vector are the basis of the tangent space)

Momentum is given by $$-i\frac{d}{dx}$$, so $$exp(-iεp)=exp(ε\frac{d}{dx})$$, which as we saw generates translations.

This is extended far beyond translations, as for exampl time evolution is generated by the Energy, rotations by angular momentum etc.

This is Lie theory, where the power lies locally, in the infinitesimal generators!

By exponentiating the infinitesimal generators you construct one-parameter groups of transformations that can describe symmetries.

These infinitesimal generators form what is called a Lie Algebra, which tells you how the flow of one infinitesimal generator affects the other. This is called the Lie Derivative, it is the true form of the "looking-and-comparing-neighboring-points" derivative, and is based on the pullback, but that is not that important right now.

When talking about vector fields, this Lie Derivative, and the Lie algebra is given by the Commutator!

So now let's put all the pieces together:

The way we think of uncertainty in observables is in terms of their infinitesimal generators and how one affects the other!

In other words, one can replace complicated, and perhaps non linear conditions for the invariance and symmetries of a subset, or functions or any object with a linear condition of infinitesimal invariance or condition for the infinitesimal generators that generate the group whose action we originally cared about!

I hope this helps clear things out a bit. Feel free to ask more question or point out anything I missed!