# Can uncertainity principle be tweaked

1. Jul 27, 2013

### sr241

I am new to quantum mechanics. By hitting a particle with photon in vacuum we find its momentum and then again hit it with another photon to find its position . can we calculate backwards using conservation of momentum the position an momentum of that particle in both two states ( since we know the properties of photon)?

2. Jul 27, 2013

### DimReg

There are two important things when it comes to the uncertainty principle: what it means in principle, and what it means in practice. In my experience, most of the questions people ask when first learning the PRINCIPLE is actually questions about the PRACTICE. So I'll tell you what would actually happen in a lab.

Whenever you measure something, there is some non-zero error on the measurement. However you always get a number, and as you correctly realized, you can get BOTH a momentum and position measurement at the same time. However, to get a handle on the errors, you would repeat the experiment over and over, measuring the position and momentum of the particles (no one takes a single data point seriously). What you would find is that no matter how well you reduce your errors through careful measurements / good statistics, the minimum uncertainties will always be constrained by the uncertainty principle. What this means is that if you look at your table of data, there will always be a spread of data points.

The real issue is that particles have a wave-like nature in quantum mechanics. When you can talk about the momentum of a wave, it's usually difficult to talk about it's position (and the other way around). And in general, waves don't admit a particularly good notion of either. The uncertainty principle is a result of this fact.

3. Jul 27, 2013

### DimReg

Another way to think about it:

Consider a sine wave. This has a single well defined frequency, which means it has a single well defined momentum. But "where" is the wave? Conversely, a spike would have a well defined position, but what would be it's momentum?

When you measure the position of a particle, the wave becomes a spike centered at the result of your measurement. This is because if you made a second measurement, you would have to find it in the same place, so the wave has to be very well localized. This means if you first measure the particle's position, you'll force it to change shape to something that doesn't have a well defined momentum. So there is no guarantee about what you would actually observe to be the particles momentum at that point.

4. Jul 28, 2013

### DrChinese

What you are saying is: "Can I make a reasonable deduction about the state of the particle which violates the HUP? And would the particle now be prepared in that state?" Answer: No, this is not possible in the normal sense of the language. As has been mentioned a number of times, such an inference cannot be supported by experiment. This can be seen by performing experiments on entangled particle pairs which DO follow conservation rules exactly.

Say you have particles Alice and Bob which are entangled such that their non-commuting properties X and Y are conserved. You measure X and then Y on Alice. What do you know about Bob? You only know X about Bob and that can be confirmed by measuring Bob's Y. It will not be correlated to Alice's Y at all. That will demonstrate that Alice's Y was not known - or deducible - in the first place.