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13

It will be the latter case, $m^2/s^3m$ which is just $m^3/s^3$. Remember that the integral is the sum of all the products $f(x)\;\text{ times } \;dx$. $dx$ is a tiny piece of the path from $0$ to $x$, so it is in units of $m$ as well. Each of the products $f(x)dx$ have units $m^3/s^3$, and the sum of all these products keeps those units.


8

The dimensions of the integral are simply those of $f(x)dx$, so in this case they would be $m^2/s^3 \times m = m^3/s^3$.


3

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 ...


3

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 ...


2

No this is not a physically valid equation. It is a mathematical description of the pendulum in which the variables do not have units. The corresponding physical equation would be: $$I \frac{d^2\theta}{dt^2} + b \frac{d\theta}{dt} + c \sin(\theta) = T \sin(2\pi ft),$$ where $$[I]=kgm^2, [b] = Nms, [c] = [T] = Nm$$ For small angles $$ sin(\theta) = \...


2

For the purposes of dimensions (units), you can treat a derivative like a division. So when you apply $\frac{{\rm d}}{{\rm d}t}$ to a function you divide the dimensions of the function by a unit of time. In your example I get: $$\frac{{\rm d}S}{{\rm d}t}\left[{\rm m}\,{\rm s}^{-1}\right] = v \left[{\rm m}\,{\rm s}^{-1}\right] + a\,\left[{\rm m}\,{\rm s}^{-2}...


2

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


2

At the time of Isaac Newton there was no defined unit for force. This gave him the freedom of choosing any unit which was convenient.So he defined a unit of force such that the proportionality constant $k$ becomes 1 which simplified the formula.


1

The most obvious answer is you measure the total sound energy that you released and is carried away with the sound and the unit is Joule. However, this does depend on intensity, the higher the intensity, the more energy per second and the more energy/sound is released in the 10 minutes. If you want your measurement to be independent of the intensity, you ...



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