# B Heisenberg's uncertainty principle - does it always apply?

#### DannyTr

Summary
A possible experiment in which the uncertainty principle seems not to apply?
I am not too knowledgable about QM, so please forgive me if this is a dumb question. I have outlined below an experiment setup for which Heisenberg's uncertainty principle seems not to apply:
1. Imagine a particle for which we wish to collect the exact position and velocity. We have a detector d1 at which the particle is fired. The detector is at an acute angle to the path of the particle so little velocity is lost in the collision. Detector d1 collects the exact position of the particle (but not velocity).
2. After the particle has collided with detector d1, it is deflected and hits detector d2. Detector d2 collects the exact velocity of the particle (but not position).
3. As the angle between detector d1 and the particle’s initial course is acute, we therefore have an approximate velocity for the particle at detector d1 - the velocity measured at detector d2. Further, we can make the angle between detector d1 and the particle’s initial course as acute as we wish leading to a corresponding increase in the accuracy of the velocity estimate.
4. So in summary, by making the deflection angle arbitrarily small, the velocity of the particle can be determined to arbitrarily precision. The particle’s position is known to arbitrary precision, so it seems the uncertainty principle does not always hold.
This seems too obvious... I assume I must be missing something somewhere?

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#### PeterDonis

Mentor
After the particle has collided with detector d1, it is deflected and hits detector d2.
How do you deflect the particle? You're not measuring its velocity at d1, so you don't know where it's going to go after it hits d1.

#### DannyTr

I'm imagining detector 1 as a beam of photons at an acute angle to the particle being measured. So a photon would be deflected and detected to give the position at d1. The deflection angle is tiny so the particle would only be perturbed a little - so we would be able to pick it up in detector d2 which measures the velocity.

#### PeterDonis

Mentor
The deflection angle is tiny so the particle would only be perturbed a little
You can't make the perturbation arbitrarily small; the uncertainty principle limits you. The more accurately you measure the position, the more you have to perturb the velocity.

we would be able to pick it up in detector d2 which measures the velocity
It measures the velocity after it has been perturbed at d1 (assuming that the perturbation doesn't stop the particle from reaching d2). But that does not tell you what the velocity was before it was perturbed at d1. The only way to measure that would be to not measure position at d1, but to measure velocity instead (and then the position would be perturbed at least as much as the uncertainty principle requires).

#### DannyTr

Yes but by making the perturbation angle smaller, the velocity at d2 becomes closer to the velocity at d1. By making the deflection angle arbitrary small, the velocity at d1 can therefore be know to arbitrary precision.

#### PeterDonis

Mentor
by making the perturbation angle smaller, the velocity at d2 becomes closer to the velocity at d1
The velocity at d1 after the measurement (assuming, as I said, that the perturbation doesn't stop the particle from reaching d2). But because of the perturbation, that's not the same as the velocity at d1 before the measurement. You can't control the perturbation, so there is some inherent uncertainty in what the velocity was before the measurement. And the amount of that uncertainty times the amount of uncertainty in the position will always be at least as large as the uncertainty principle requires.

#### DannyTr

You can control the perturbation - it is reduced in size by using a more acute impact angle between the particle and d1. So by using a deflection angle of say 0.01º the perturbation is reduced to almost nothing and the velocity measured at d2 will be almost the velocity the particle had before measurement at d1.

#### Nugatory

Mentor
......so little velocity is lost in the collision..... we therefore have an approximate velocity for the particle at detector d1......
Exactly how little velocity is lost? Exactly how approximate is the velocity measurement? You need to do real calculations with real numbers to see whether the results fall within the bounds given by the uncertainty principle.

#### DannyTr

My maths is not great I'm afraid. I guess the amount of velocity is lost is a function of the deflection angle:

1º - a little is lost
0.01º - hardly any is lost
0.000001º - even less velocity is lost

So as the deflection angle approaches zero, the velocity at the second detector approaches the velocity at the first detector.

So in principle it appears that velocity is known to arbitrary precision, IE more precisely that the uncertainty principle allows.

#### PeterDonis

Mentor
You can control the perturbation - it is reduced in size by using a more acute impact angle between the particle and d1.
And the more you do this, the more you will increase the uncertainty in the position measurement. So the product of the two uncertainties will, as I said, obey the uncertainty principle.

in principle it appears that velocity is known to arbitrary precision, IE more precisely that the uncertainty principle allows.
The uncertainty principle does not say that you can't know the velocity to arbitrary precision. It just says that the product of the uncertainties in position and velocity (more precisely, in position and momentum, but we're assuming the mass of the particle is fixed so it factors out) must be at least a certain minimum value.

#### DannyTr

And the more you do this, the more you will increase the uncertainty in the position measurement. So the product of the two uncertainties will, as I said, obey the uncertainty principle.
The first detector might be a light gun of some sort perhaps. The reflected photon would be captured after the deflection by a measuring device. I would of thought position can be measured to arbitrary position no matter how acute the angle is between the photon and the particle? Why do you say the uncertainty in the position measurement increases in line with the acuteness of angle?

#### Nugatory

Mentor
Why do you say the uncertainty in the position measurement increases in line with the acuteness of angle?
I'm imagining detector 1 as a beam of photons at an acute angle to the particle being measured. So a photon would be deflected and detected to give the position at d1. The deflection angle is tiny so the particle would only be perturbed a little - so we would be able to pick it up in detector d2 which measures the velocity.
We need to detect both the position and the transverse velocity component of the photon to determine the exact location at which it was deflected but these are also constrained by the uncertainty principle. With a bit of trigonometry you will be able to see that the for any given uncertainty in the transverse velocity, a smaller deflection allows a larger the volume of space within which the collision between photon and could and particle could have happened - that is, the more acute then angle the greater the uncertainty in the position measurement.
(A math-free way to see this is to consider the limiting case in which the angle is so acute that there is zero deflection - the photon beam is exactly parallel to the trajectory of the particle. The collision could happen anywhere along that trajectory and the photon would end up in the same place).

It’s generally not a good use of time trying to analyze any particular experimental setup to prove that it will respect the uncertainty principle; as the setup becomes more complicated and ingenious the analysis becomes more tedious and error-prone, but we already know what the answer will be. It’s as if I were to give you a number ten million digits long ending with the digit “5” and ask if it could be prime; you’d immediately tell me that it can’t be, that it must be divisible by 5. If I refused to be convinced until you actually did the division and showed me a zero remainder you’d rightly accuse me of wasting your time; the mathematical argument that a number ending with the digit 5 must be divisible by 5 is already compelling. The argument that the product of the uncertainty in position and velocity measurements cannot be zero is no less compelling.

#### DannyTr

We need to detect both the position and the transverse velocity component of the photon to determine the exact location at which it was deflected but these are also constrained by the uncertainty principle. With a bit of trigonometry you will be able to see that the for any given uncertainty in the transverse velocity, a smaller deflection allows a larger the volume of space within which the collision between photon and could and particle could have happened - that is, the more acute then angle the greater the uncertainty in the position measurement.
(A math-free way to see this is to consider the limiting case in which the angle is so acute that there is zero deflection - the photon beam is exactly parallel to the trajectory of the particle. The collision could happen anywhere along that trajectory and the photon would end up in the same place).

It’s generally not a good use of time trying to analyze any particular experimental setup to prove that it will respect the uncertainty principle; as the setup becomes more complicated and ingenious the analysis becomes more tedious and error-prone, but we already know what the answer will be. It’s as if I were to give you a number ten million digits long ending with the digit “5” and ask if it could be prime; you’d immediately tell me that it can’t be, that it must be divisible by 5. If I refused to be convinced until you actually did the division and showed me a zero remainder you’d rightly accuse me of wasting your time; the mathematical argument that a number ending with the digit 5 must be divisible by 5 is already compelling. The argument that the product of the uncertainty in position and velocity measurements cannot be zero is no less compelling.
Thanks. I guess I am still a little confused. Perhaps a simplified experiment would help:

So keeping to one dimension to avoid trigonometry, the particle to be measured starts at the right of the above diagram with momentum p1 (unknown), hits a stationary particle of known mass and position which deflects with momentum p2. The initially stationary particle hits detector d1 which gives us the momentum p2. The particle to be measured deflects with momentum p3 which is measured at detector d2.

So we know exactly the position of the particle to be measured. The momentum of the particle is given by conservation of momentum:

0 + p1 = p2 + p3

As we know p2 and p3 precisely, we know p1 precisely?

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#### DrChinese

Gold Member
... hits a stationary particle of known mass and position ...
So we know both p and q of the "stationary" particle to a high degree of certainty. See anything wrong with this assumption?

#### DannyTr

So we know both p and q of the "stationary" particle to a high degree of certainty. See anything wrong with this assumption?
I had imagined maybe a piece of foil in which the stationary particle is initially positioned. So position would be known and the particle would have no momentum until it is knocked out of the foil by the other particle.

I guess you mean the uncertainty principle applies to the stationary particle too? I had not realised that... Bit of a newbie to QM... apologies...

"Heisenberg's uncertainty principle - does it always apply?"

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