# Showing that position times momentum and energy times time have the same dimensions

I've been asked to show that both the position-momentum uncertainty principle and the energy-time uncertainty principle have the same units.

I've never see a question of this type, so am I allowed to substitute the units into the expressions and then treat them as variables?

If so, here's my attempt. Forgive me if I've done something silly, as I'm no physicist.

Starting with the position-momentum uncertainty principle:

$$\Delta{}x\Delta{}p \geq h / 4\pi$$

Substituting the units into the expression (at this point, diving by $4\pi$ won't necessarily matter):

$$(m)\left(kg \cdot \frac{m}{s}\right) \geq J \cdot s$$

Combining $m$ and bringing $s$ to the other side:

$$\frac{kg \cdot m^2}{s^2} \geq J$$

Knowing that $J = kg \cdot m^2/s^2$:

$$J \geq J$$

Now, for the energy-time uncertainty principle:

$$\Delta{}E\Delta{}t \geq h / 4\pi$$

Substituting the units into the expression (again, diving by $4\pi$ won't necessarily matter):

$$J \cdot s \geq J \cdot s$$

Diving by $s$:

$$J \geq J$$

Is this valid? Or could I not be more wrong?

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Things look good, you should listen to the answers. Also, another common notation is to use "length," $L$, "mass," $M$, and "time," $T$, instead of actually putting in the subjective choice of units (kilogram etc . . .) So velocity has units of length per time: $L/T$. – kηives Jan 8 '13 at 22:56
You have to be careful not to run into multiplication that is really something else in disguise, like a cross product. For instance, informally, "force times distance" could refer to torque (force at an angle, with distance representing displacement from the point of rotation or fulcrum) or "force times distance" could be work (pushing against a resistance, resulting in a displacement across a distance). In the imperial system, foot-pounds is work, but pound-feet are torque. :) – Kaz May 1 '13 at 4:41

I think you are on the right track. There are a couple of bits of advice you may follow:

1. You may simply note that if $A \geq B$, then it follows that $A = B$ is a valid solution, thus $A$ and $B$ must have the same units.

Therefore $\Delta{p}\Delta{x}$ has the same units as $h$ which has the same units as $\Delta{E}\Delta{t}$.

2. The method you used is called Dimensional Analysis and it's perfectly correct to use it.

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It's just that magnitudes can't be ordered, I mean: things like $meter>second$ don't make sense.

In every physical equation, or inequation, both sides must be in the same unit, so what you're doing is actually checking that the units at the right are the same as the ones in the left. That must be true, as you're saying a numerical product of momentum and time (a real value) will be higher than another product of momentum and time, that is, the units of $h$. The $>$ sign doesn't mean that units must be higher, such a thing doesn't exist, it means that quantities measured in the same units (quantities of the same kind of thing) will behave like that, one will bi bigger than the other.

I don't know if I'm explaining it properly.

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Sounds right to me. – Michael Brown Jan 9 '13 at 9:01