1
$\begingroup$

In the Heisenberg uncertainty principle,

$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$

The values of $\Delta x$ and $\Delta p$ are the standard deviations which we get from the probability distribution function of the particle and I heard that it has nothing to do with the measuring instrument.

Actually while measuring, the probability distribution function of a particle also changes, Does this means that the measuring instrument has some effect?

$\endgroup$
  • $\begingroup$ Some effect on what? The standard deviations are defined without any measurement process. Why are you talking about a measurement instrument at all? $\endgroup$ – ACuriousMind Jan 15 '16 at 13:52
  • $\begingroup$ In English, only use capital letters for the first work of each sentence, for proper nouns (i.e. names of people and places), and for the word "I". It is improper to capitalize the words "Uncertainty Principle", for example. $\endgroup$ – DanielSank Jan 16 '16 at 16:56
  • $\begingroup$ Are you trying to ask what is asked in this other post? We're having a hard time figuring out what you mean here. $\endgroup$ – DanielSank Jan 16 '16 at 17:22
4
$\begingroup$

Actually while measuring, the Probability Distribution function of a particle also changes, Does this means that the measuring instrument has some effect ?

The measuring process may change the boundary conditions of the solutions of the quantum mechanical equations of the system under measurement, so the complex conjugate square of the wave function ( the probability) may change. Good measurements are done with a minimal disturbance of the process under study by the measuring method.

$\endgroup$
1
$\begingroup$

Yes, indeed any measurement where you learn something will disturb the state. This is also known as "no information without disturbance". Recently, there has been a lot of flurry of activity trying to pin this down quantitatively.

As you said, this is not what the Heisenberg relation you write down is about, but the effect still certainly exists. Here is a number of questions you could ask (let's stick to position and momentum):

  • Given a quantum state (i.e. a preparation procedure, which is to say a number of steps you do in a lab), what are the fundamental limits on the variance of position and momentum if I measure them (separately)?
  • Given a quantum state, what happens if I measure the position of the state and then the momentum of the state and compare the momentum distribution with the distribution I obtain when I only measure the state?
  • Given a quantum state, what happens if I measure the position only very coarsly and then measure the momentum? Will anything change from the previous measurement?
  • What is the best way to jointly measure position and momentum? I.e. what measurement can I do that outputs two values every time I measure, one for position the other for momentum. The "best" meaning that the distributions are as close as possible to the distributions where I just measure position or just momentum.

The first is as far as I understand it pretty much what the Heisenberg uncertainty principle describes. The second and third scenario describe a qualitative and quantitative analysis of how much a measurement actually "disturbs" a state. The fourth asks about doing the measurement at the same time (something that you also often hear in connection with the HUP).

As I understand it (but I am by no means an expert), people are getting a much better understanding of the scenarios two to four in various flavour. Note for example that you always have to compare probability distributions for which there are many ways to do it.

Let me close by giving one link to a paper that tries to distinguish a number of different relations (operationally). It might not be the best to start with, but it seems reasonable and if you are interested, you can get a lot from there, since it is a pretty recent paper: http://arxiv.org/abs/1402.6711

$\endgroup$
0
$\begingroup$

These are standard deviations of a probability distribution indeed. The probability distribution is that of getting a particular value while the system is in the state prior to measurement. So we imagine measuring a state - it collapses to some value, and then somehow resetting time back before measurement and measuring again. If we repeat this process, rewinding the clock and measuring, this is what is meant by the probability. Equivalently, and perhaps more accurately you could think of preparing a bunch of identical quantum systems, and the distribution would be of the values you'd get by measuring all of them. So THIS is what the standard deviations apply to - these probability distributions. So if you measure x with complete certainty, then the standard deviation for the momentum distribution becomes infinite. It has nothing to do with the accuracy of measurement or something.

Furthermore measuring does have an effect - causing the system to randomly select a value from the previous probability distribution and making it choose that exact value to be.

Hope this helped!

$\endgroup$
0
$\begingroup$

The Heisenberg uncertainty principle establishes a lower bound on the uncertainty (hence the $\lt$ sign). It follows that if you have additional error (due to your measurement instrument) it can only increase the uncertainty over this quantum mechanical limit.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.