# Are quantum operators dimensionless?

I'm slightly confused as to whether quantum (hermitian) operators, which we get by promoting observables to operators, are dimensionless or not?

Clearly the Hamiltonian of the system, say of the harmonic oscillator, has units of energy. Does the hamiltonian of the corresponding quantum mechanical system also have units of energy?

I suspect this is the case because only then would the procedure of making the Hamiltonian dimensionless by introducing creation and annihilation operators make sense, as done on: http://en.wikipedia.org/wiki/Creation_and_annihilation_operators, starting from the position representation of the SE, though. But the idea of operators having dimensions doesn't strike me intuitively.

It'd be great if someone could explain this.

The creation and annihilation operators are not observables. They are obviously not hermitean because $a \neq a^\dagger$. But, regarding your question, consider the number operator $( N_k = a_k a^\dagger_k )$. As the eigenvalue of the number operator is a dimensionless number, the creation and annihilation operators must be dimensionless as well.

The full Hamiltonian has the dimension of energy, because $[\hbar] = \mathrm{J s / rad}$ and $[\omega = \mathrm{rad / s}$, so $[\hbar \omega] = \mathrm{J}$.

• That's exactly what I had/have in mind; however, just thinking about operators having a dimension doesn't make intuitive sense. I'm trying to visualize it through the matrix representation of operators. The only subtle point I'm missing is that: hey, here is a n*n matrix; multiply it with, for example, h bar and you'll get a dimensional operator. Right? – Junaid Aftab Jun 9 '16 at 16:51
• I think a dimensional operator is not a problem per se. Dimensional matrices are encountered in classical mechanics as well (think of the inertia tensor or the force field tensor). To satisfy the mathematically more rigorous requirements for matrix operations: You usually deal with equations, so you can rewrite everything in a dimensionless way. – magnon2020 Jun 10 '16 at 9:25

The raising and lowering operators are dimensionless. The position and momentum operators are written according to $$x = \sqrt{\frac{\hbar}{m\omega}}q,~\frac{\partial}{\partial x} = \sqrt{\frac{m\omega}{\hbar}}\frac{\partial}{\partial q}$$ with $p = -i\hbar\partial/\partial x$ we then write the raising and lowering operators according to these dimensionless operators $$a = \frac{1}{\sqrt{2}}\left(q - \frac{\partial}{\partial q}\right),~a^\dagger = \frac{1}{\sqrt{2}}\left(q + \frac{\partial}{\partial q}\right)$$ The Hamiltonian is $H = \frac{1}{2}\hbar\omega a^\dagger a$ and the dimensions are restored.

• Since this is a question about dimensions, you might want to restore the $\hbar$ in the Hamiltonian, since otherwise it has frequency rather than energy units. – zeldredge Jun 9 '16 at 13:07

Two ways that might help you see that operators in general must have units:

1. The quantum Hamiltonian must have units of energy, because $\exp\left(i\,\frac{H}{\hbar}\,t\right)$ is the time evolution operator, so that the exponent is dimensionless; otherwise put: Schrödinger's equation is $i\,\hbar\,\partial_t\,\psi = H\,\psi$, so that $H$ must have the same units as $\hbar/t$;

2. The operators you speak of aside from the ladder operators are all also observables: their real eigenvalues represent possible real outcomes from measurements and, as such, the units of the eigenvalues must match those of the potential measurements. In the case of a finite dimensional operator, one could factorize the operator as $A = P\,\Lambda\,P^{-1}$ and any units of $P$ and $P^{-1}$ cancel.

There are already good answers, here are a few more ways to help see it:

1. If you already accept that vectors can have units (like in $\mathbf{F} = m \mathbf{a}$), then you already have accepted that algebraic constructs that aren't just numbers can have units. The operator case isn't any more complicated.
2. You've already seen examples of operators with units in classical mechanics. For example, consider the equation $L = I \omega$. For a general 3D object, the generalization is $\mathbf{L} = \hat{I} \boldsymbol{\omega}$, where $\hat{I}$ is the "moment of inertia tensor". That's a fancy way of saying $\hat{I}$ is a linear operator, and it clearly must have the same units as the scalar $I$ did.

## protected by Qmechanic♦Jun 9 '16 at 12:03

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