I do not understand the idea behind Uncertainty principle: the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa.

But as I know, momentum is mass times velocity, and because velocity is derivative of distance function and also we know mass of particle, why can't we determine momentum? What is problem there? Suppose that we can determine distance-time function and therefore we can determine its position at time $t_0$, for example, and we know mass of this particle, why can't we determine its momentum?


If you measure the position of a particle with uncertainty delta-x, you have to use another particle with a wavelength smaller then that uncertainty.

So, if you try to measure the position of a particle as precise as possible, you'd need for example a photon with a very high frequency (small wavelength) but energy is = plancks constant * frequency. Energy squared of a photon is the moment squared (einsteins formula). This implies that you'd have to bombard the particle you try to measure the position of with a very high momentum photon.

But, if you bombard that particle with a high momentum particle, they will be both knocked in a random direction and the original momentum of that particle will have been changed. Measuring the position of a particle necesarely results in an uncertain chance of momentum of that particle.

By using the distance time function, you'll be able to calculate the momentum the particle had between the two measuring points but you won't know the current momentum of that particle as measuring its position once again changed its momentum.

This shows intuitively that by measuring the position of a particle, you cannot measure the momentum at the same time with the same precision.

The uncertainty principle is more fundamental then the intuitive explanation above. It states that you cannot measure the momentum and position of a particle at the same time with very high precision. Measuring one changes the other.

There are many more relationships between variables of a physical system that we can't precisely measure at the same time.

  • $\begingroup$ thanks very much,so it means that this principle is more or less related to probability measurment $\endgroup$ – dato datuashvili Jan 4 '14 at 12:50
  • $\begingroup$ yes, the result of a measurement a quantum-mechanical system is a probability distribution. $\endgroup$ – camel Jan 4 '14 at 12:58

In classical mechanics there is no uncertainty principle.

However in QM, you represent the state of a system as an element $\left|\psi\right> = \left|\psi(x,t)\right>$ of a Hilbert space with norm $1$. Observations (i.e. measuring properties of the state) are done by applying self-adjoint linear operators to the state, and you always get back an eigenvalue of the operator as a result. Now operators don't usually commute (think e.g. of matrices...), and thus we get the following inequality $$\left<(\Delta A)^2\right>_\psi\left<(\Delta B)^2\right>_\psi\ge\frac{1}{2}\left|\left<[A,B]\right>_\psi\right|^2$$ where $\left<A\right>_\psi=\left<\psi|A|\psi\right>_\psi$ for an operator $A$ and $(\Delta A)^2=(A-\left<A\right>_\psi)^2$. The left hand side represents the product of the "uncertainties of measurement" of the two operators, when they are applied together to the same state.

In particular, we have the operators measuring position $X=x$ and momentum $P = \frac{\hbar}{i}\frac{\partial}{\partial x}$ which give us $$\left<(\Delta X)^2\right>_\psi\left<(\Delta P)^2\right>_\psi\ge\frac{\hbar}{4}$$

My exposition is obviously lacking a lot of details, and very unprecise at places. I would advise you to read any introductory book on QM to get a better understanding of what I'm trying to explain here.

  • $\begingroup$ ok thanks very much,no just this terminology is used also in DSP,that why i was interested $\endgroup$ – dato datuashvili Jan 4 '14 at 12:53
  • $\begingroup$ @datodatuashvili Ok, good :) Anyway, what you should take away from my answer is that the reason of the uncertainty principle is mainly that in QM the result of measuring different things can change if you interchange the order of the measures (applying an operator will get the system in an eigenstate of the operator). A special case is explained in the answer of marteen. $\endgroup$ – Daniel Robert-Nicoud Jan 4 '14 at 12:57
  • $\begingroup$ thanks very much for helping,happy new year $\endgroup$ – dato datuashvili Jan 4 '14 at 13:01

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.