Dimensions of physical quantities in quantum mechanics

Question

In most introductory quantum mechanics classes, we are introduced to the Dirac notation, concept of the 'state' of the system being represented as an abstract vector in the Hilbert space associated with it, and we are told that measurements of physical quantities involve the action of a Hermitian operator associated with the respective quantity on the wavefunction. Then, we are told that the result of the measurement is one of the eigenvalues of the operator, etc.

However, these measurements are supposed to be of physical observable quantities. Thus, whenever a measurement is done in classical physics, it must give rise to a physical quantity, with dimensions (by dimensions, I mean length, Energy, etc.).

Where exactly does the dimension of the quantity being measured come into the picture while discussing QM? For example, if I say that $|x_0\rangle$ is an eigenstate of the position operator $X$, then the action of the position operator in the ket is written as follows: \begin{equation} X|x_0\rangle=x_0|x_0\rangle \end{equation} Here, what is the quantity $x_0$? Is it just a pure number? Or does $x_0$ have units of length?

If $x_0$ is just a pure number, then where does the length dimension come into picture? If the value $x_0$ is has the dimensions of length, then can the operator $X$ operate on a wavevector multiplied by a quantity with physical dimensions?

Also, are operators of the form $L^2+L_z$, or $P+X$ (which refer to physical quantities with different dimensions; there are no constants being multipled with them) valid operators, and what is the justification (for them either existing or not existing)?

Answer 1

There is nothing special about the treatment of dimensionful quantities in quantum mechanics. In the specific example of the position operator you mentioned,

$$ \hat{x}|x_0\rangle = x_0 |x_0\rangle, $$

the number $x_0$ has dimension of length, since it's one of the possible outcomes of a position measurement.

The addition of operators makes sense only if the operators have the same units. For example, in the ladder operator approach to the quantum harmonic oscillator, the annihilation operator is defined,

$$ \hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} + \frac{\imath}{m\omega}\hat{p}\right) $$

The factor of $\frac{\imath}{m\omega}$ ensures that both operators have the same units. Similarly, the expression $\hat{L}^2 + \hbar \hat{L}_z$ is dimensionally correct, because $\hbar$ has units of angular momentum. (Note that if you're using natural units, then $\hbar = 1$, and you might as well write $\hat{L}^2 + \hat{L}_z$.)

Answer 2

The dimensions of a Hilbert space are countable, i.e. each dimension can be assigned a whole number and thereby all dimensions are referenced in a unique way with 1, 2, 3, .... A vector space that is a Hilbert space has some additional properties.......

In the case of the hydrogen atom, we had to use three qunatum numbers n, l, and m to characterize the energy eigenfunctions completely. Without proof, the indices l and m characterize the eigenvalues of the square of the angular momentum operator L 2, and of the z-component of the angular moment Lz with eigenvalues l(l + 1) ~2 and m~, respectively.

One can show, that the Hamilton operator of the hydrogen atom, the square of the angular momentum operator and the z-component of the angular moment operator commute with each other and build a complete system of commuting operators (CSCO), whose eigenvalues enable a unique characterization of the energy eigenstates of the hydrogen atom>

The above quote is from http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-974-fundamentals-of-photonics-quantum-electronics-spring-2006/lecture-notes/chapter5.pdf

he above quote was only to prepare a ground for how things/observables are being represented in Hilbert space.

You are raising the issue of relationship of measured observable's Units and Dimensions of measurement (as is available in classical mechanics/newtonian mechanics) and its relation to dimensions of vector spaces

I think when the formalism of classical mechanics moves to canonical representations and the governing equations are say Hamilton's equations the normal coordinates are replaced with 'generalized coordinates' and energy, momenta are also taken on an equivalent footing leading to some advantage in solving problems.

when one moves over to quantum mechanics the dimensions of an observable gets defined by the number of states in which the system is expected to stay.

When one is measuring say position, momenta or energy the expected value of result of measurement are given in their proper units rather than some abstract number. e.g. say neutron beam energy is measured in MeV though its an Expectation value of Hamiltonian taken in a particular eigen state. similarly the ground state of Hydrogen atom...... or the ground state of Deuteron....