I understand that electron spin on different axis is an example of complementary properties, which cannot be measured accurately at the same time. I also understand that if we have an electron is pointing up along some axis, it is meaningless to say it is pointing up or down along another axis, until we make the measurement.

Same principle applies in case of photon polarization. A photon polarised at 45° to the horizontal is neither horizontally or vertically polarised. it is in a "superposition" of both states. The question ‘is it horizontally or vertically polarised?’ is meaningless and unanswerable.

From these two examples we could say:

If A and B are complementary properties of a particle, and if we know that this particle has some value of A; Then it is meaningless to ask which values of B does this particle possess.

The problem arises when we talk about position and momentum (or velocity). According to the last statement we may say: "If we know that this particle has definite velocity or momentum; then it is meaningless to ask which position does this particle occupy."

I cannot understand that there is a prticle (or even a wave) without position. Also as we know, velocity is the rate of change of the position!! So how could this particle has a definite velocity without occupying a definite position?

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    $\begingroup$ Simple enough it doesn't have a definite velocity, unless it is in a momentum eigenstate, which is also a „velocity” eigenstate if you define $\hat v = \hat p / m$ BTW in Hamiltonian formalism you don't define velocity as the rate of change of the position, rather you say that $v = \{x,H\}_{ps}$, where $\{\cdot\}_{ps}$ denotes Poisson brackets , which is equivalent to $\dot x$. However this representation have a nice transition to QM, which is that $\hat v = -i/\hbar [\hat x, H]$. $\endgroup$ – Gonenc Apr 8 '15 at 21:06
  • $\begingroup$ Thank you @gonenc for your reply. I'm trying to understand all that maths but to be honest ; Hamiltonian and Hermitian and all that quantum maths is a way difficult for me. I want a physical meaning for an electron moving with known momentum but does not have a position at all! $\endgroup$ – Osama Azzam Apr 9 '15 at 6:27

The short answer is in this paragraph, with more detail in other paragraphs. The particle acts a bit like a blob of mist spread out over position and momentum space. The intensity of the mist in a given region tells your the probability of finding it there. If you squash the blob too much in position space then it will be more spread out in momentum space and vice versa. the velocity tells you something about how fast the blob will spread out in position space. This doesn't require that the particle has a single well defined position.

The particle has some state $$|\psi(0)\rangle = \sum\sqrt{p_a}|R_a\rangle,$$ where the $|R_a\rangle$ means that the particle is in the region $R_a$. There is no state for a particle being at a single point since no measurement can tell whether a particle is at a single point. Rather, the relevant states are for a particle to be in a region and if you measure more precisely the regions are smaller. More precise measurements are said to be more fine grained.

If you measure the particle then you couple the particle to a measuring instrument, the resulting state is $$|\psi(1)\rangle = \sum\sqrt{p_a}|R_a\rangle|M_a\rangle.$$ where the $|M_a\rangle$ means that the particle has been measured as being in the region $R_a$.

What this means is that there are multiple versions of the particle. Different versions of the particle can sometimes interact with one another in interference experiments, for a discussion see "The Fabric of Reality" by David Deutsch, Chapter 2. There are also multiple versions of the measuring instrument and each version of the instrument sees a particular version of the particle. The different versions of the measuring instrument can't interact with one another as a result of a phenomenon called decoherence:


Decoherence also explains why you don't see other versions of yourself although their existence is a consequence of quantum theory.

The probability for finding the particle in $R_a$ is $p_a$ and this means that the proportion of versions of the particle with value $R_a$ is $p_a$. For an explanation of why this is the appropriate measure see


Now, you could write down the state in a different way as $$|\psi(0)\rangle = \sum\sqrt{q_b}|P_b\rangle,$$ using states $|P_b\rangle$ in which the particle has momentum in a particular range and the $q_b$ are the probabilities for each of those states.

The $q_b$ and $p_a$ are not independent and if you have a suitably fine grained set of position states, then if the $p_a$ are peaked near some particular value, the $q_b$ will have multiple non-zero values, and vice versa.


From the comment above I think that it would be the best if I answered your question with an answer. Furthermore I think you seek for the intuition behind QM and not the beautiful maths behind it. Sadly the answer is there is no good intuition! As Feynman says:

I think I can safely say that nobody understands quantum mechanics.

If you think you understand quantum mechanics, you don't understand quantum mechanics.

However you can perhaps satisfy yourself by the following description of QM, although it is not too accurate and should not be taken too seriously.

Don't think of a quantum object either as a particle or as a wave. Accept it as it is as a quantum particle. You can however think of it as a jelly or a mush. This is the way I usually go around with it. You can see that some mush is moving with velocity $v$ but you can't* really say where it is because well... it is a mush! Furthermore imagine that this mush is made from a very special material, which has the following properties

  • If it goes slow it „contracts“ and accumulates around a certain point.
  • If it goes fast it „relaxes“ and spread out.

so that you can include the uncertainty relations. I hope this helps.

*Of course you can say that the CoM of the mush is where it is but that is not the point.

  • $\begingroup$ Thank you for your answer. I did not vote down BTW :) . $\endgroup$ – Osama Azzam Apr 9 '15 at 12:07
  • $\begingroup$ If you received an answer that solved your problem, you can mark it as accepted. See How does accepting work? for details. $\endgroup$ – Gonenc Apr 9 '15 at 12:17
  • $\begingroup$ Yes I know. I'm still thinking and trying to convince my poor mind to accept the idea that there is something can move with a velocity but without position. It is easy to accept that the uncertainty relations describe our knowledge. It is also easy to understand the case of complementary properties which are independent of each other as spin or polarization on different axis. However, when it comes to momentum (i.e. velocity) and position my mind stop working :) . Velocity is depend on position or at least related to it. That's why I cannot accept it here. $\endgroup$ – Osama Azzam Apr 9 '15 at 14:09
  • $\begingroup$ @OsamaAbbas don't worry you are not the only one having problems with QM. Just accept it as it is and trust the maths behind QM. That is what I do mostly. $\endgroup$ – Gonenc Apr 9 '15 at 14:14

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