# What is 'definite' variable in QM?

I have gone through a few of the questions on the website regarding this particular query, but I have not understood what they meant.

When a question says that a particle has definite momentum, are they saying it neglecting the approximations implied by uncertainty principle?

Because in QM, as far as I know, there can not be anything known beyond a certain limit.

P.S. I have a high school math background.

• I'm afraid you're not precise enough about what the problem here is: Is it: a) that you think that no state in QM can ever have zero uncertainty for any observable, b) that you think no state can ever have zero uncertainty for momentum because that would violate the uncertainty principle? – ACuriousMind Dec 30 '18 at 16:48
• Are you sure the term you've seen is "definite" and not "well-defined"? The latter is part of the standard terminology but the former isn't (much). If it's the former, you should provide explicit links to where you've seen it being used. – Emilio Pisanty Dec 30 '18 at 17:40
• @ACuriousMind: (a), and so (b) – KV18 Dec 30 '18 at 18:19
• @EmilioPisanty I've seen "definite". I heard my professor use it explaining the square well problem - a particle with definite momentum. But I guess now that he really meant 'almost' definite momentum, with the uncertainties included. – KV18 Dec 30 '18 at 19:20

I have understood your question as two separate questions.

1. What does it mean that a state has definite momentum?

A general state $$|\psi \rangle$$in quantum mechanics is a superposition of different eigenstates $$|p\rangle$$ of the momentum operator:

$$|\psi \rangle=\int \psi(p)|p \rangle dp$$

I have written this for 1 dimension, since the concept does not change in 3d.

However, you can consider the case that $$|\psi \rangle=|p \rangle$$. Actually there are some issues with this, as $$|p\rangle$$ is not normalizable, but it works to some approximation. The state $$|p \rangle$$ is a state with definite momentum in the sense that it has a 100% probability to give momentum $$p$$, and 0% probability to give any other momentum.

1. Doesn't such a state violate the uncertainty principle?

Good question, and I see why you'd think so, but it does not! We wish for

$$\sigma_x \sigma_p \geq \hbar/2$$

Where $$\sigma_x , \sigma_p$$ are the standard deviations in the x and p probability distributions, respectively.

And if there is no momentum uncertainty, as in the case $$|\psi \rangle = |p \rangle$$, then $$\sigma _p =0$$, so you'd think that the product is also 0. However, for a momentum eigenstate, the position wave function is proportional to $$e^{ipx}$$. If you try to calculate $$\sigma_x$$, you will get infinity. So you have

$$\sigma_x \sigma _p = \infty * 0 \geq \hbar/2 ?$$

This is an indefinite form and a priori can equal any number. However you can prove that the inequality still holds (maybe lacking a bit of mathematical rigor): $$\psi (p)$$ will be a delta function $$\delta (p'-p)$$, and you can write this as a limit of gaussian functions. For any of the gaussians in the limit the uncertainty principle holds.

Lastly: Since (according to your edit) you're looking for an answer with high school math background, I'll just summarize without any technical stuff.

Summary (no math background needed)

Mathematically, in quantum mechanics you can conceive of a particle with definite momentum, i.e. with 100% chance that the momentum takes a particular value $$p$$. But that on its own does not violate the Heisenberg uncertainty principle: The HUP says that the error in position and momentum cannot both be arbitrarily small at the same time. However, having 0 error in one of the variables does not mean the HUP is violated, it just means you have to have large (actually infinite) uncertainty in the other variable.

There is a caveat to this. Mathematically you can conceive of a state that has a definite momentum, and this state is very useful in quantum mechanics. But there are reasons to believe that such a state would never exist physically: you could get states which approximate this state very well, but you'd never quite get there. So if you like, you can think of the definite momentum state as an almost definite momentum state. A state with almost definite momentum has almost a 100% chance to give a momentum very close to $$p$$, and that chance is so close to 100% that you can just consider it exactly 100% for most purposes.

• It is also worth mentioning that you can have as precise a measurement as you want with momentum if you aren't also trying to measure the position as well. – Aaron Stevens Dec 30 '18 at 17:13
• Actually an assumption of the uncertainty relation is that the state is normalizable; states of definite momentum are not so the HUR should be used with extreme care in this case, i.e. not valid. – ZeroTheHero Dec 30 '18 at 18:30
• @user27084 do you mean to say that the statement, say, "definite position" mean that if we did an experiment, we will find the particle in there somewhere? – KV18 Dec 30 '18 at 18:45
• But if that were the case, then uncertainties would be non zero right? As we are calculating the uncertainty (standard deviation) after doing a series of experiments? – KV18 Dec 30 '18 at 18:50
• "definite position" would not just mean that we will find the particle somewhere, it is even more information than that: If a state has definite position $x=2$ (in some units), it means there is a 100% chance that it will be found at $x=2$ if we measure its position. That leaves no uncertainty. – doublefelix Dec 30 '18 at 18:59