$$\Delta x \Delta p_x \geq \frac{\hbar}{2} $$

I understand what does Heisenberg's uncertainty principle states i.e. it's definition and it has been proven experimentally. But, can anyone please explain the reason for whatever this principle states i.e. why it happens?

Stating this principle in reference to the single slit experiment with electron (one particle experiment i.e. electrons are fired one at a time): Assuming the variable width of the slit is along the X-axis. So, if an electron passes through the a thin slit we can say the uncertainty in position is $=\Delta x$ which is equal to the width of the slit. Now, we can also know the $x$ component of the momentum at he time when electron passes through the slit using distance between the slit and screen, time interval and the distance between the centre of the screen and the point where the electron hits.

As, the electron was fired towards the centre of screen we should expect a point on the screen but instead we get a gaussian distribution, and as we decrease the width of the slit (decrease $\Delta x$) the FWHM of the gaussian distribution increases i.e. $\Delta p_x$ increases which means the uncertainty in the measurement or prediction of $p_x$ increases.

Which is exactly what Heisenberg's Uncertainty principle states. But, can anyone please explain how/why this happens?

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    $\begingroup$ possible duplicate of What is the physical meaning of commutators in quantum mechanics? $\endgroup$ Aug 14, 2015 at 9:19
  • $\begingroup$ @JohnRennie: Although commutators are the correct way to answer this question, I don't think that the linked question is a duplicate in any sense. $\endgroup$
    – ACuriousMind
    Aug 14, 2015 at 21:18
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    $\begingroup$ In one of your comments, you state "What I actually want to understand is from the view of diffraction experiment, in which the centre spot widens as we decrease the width of the single slit", could you edit your question to clarify the kind of answer you're looking for? $\endgroup$
    – David
    Aug 14, 2015 at 22:23

3 Answers 3


The uncertainty principle is a simple consequence of the idea that quantum mechanical operators do not necessarily commute.

In quantum mechanics, you find that the state which describes a state of definite value of an observable $A$ is not the state which describes a state of definite value for an observable $B$ if the commutator of both observables $[A,B]$ is not zero. (Formally, the two operators are not simultaneously diagonalizable.)

You just write down the definition of the standard deviation of the operator $A$ on a state $\psi$, $$ \sigma_A(\psi) = \sqrt{\langle A^2\rangle_\psi - \langle A\rangle^2_\psi}$$ where $\langle\dot{}\rangle_\psi$ is the expectation value in the state $\psi$ and with a bit of algebraic manipulation (done e.g. on Wikipedia) we find that $$ \sigma_A(\psi)\sigma_B(\psi)\ge \frac{1}{2}\lvert \langle[A,B]\rangle_\psi\lvert$$

Now, the standard deviation (or "uncertainty") of an observable on a state tells you how much the state "fluctuates" between different values of the observable. The standard deviation is, for instance, zero for eigenstates of the observable, since you always just measure the one eigenvalue that state has.

Plugging in the canonical commutation relation $$ [x,p] = \mathrm{i}\hbar$$ yields the "famous" version of the uncertainty relation, namely $$ \sigma_x\sigma_p\ge \frac{\hbar}{2}$$ but there is nothing special about position and momentum in this respect - every other operator pair likewise fulfills such an uncertainty relation.

It is, in my opinion, crucial to note that the uncertainty principle does not rely on any conception of "particles" or "waves". In particular, it also holds in finite-dimensional quantum systems (like a particle with spin that is somehow confined to a point) for observables like spin or angular momentum which have nothing to do with anything one might call "wavenature". The principle is just a consequence of the basic assumption of quantum mechanics that observables are well-modeled by operators on a Hilbert space.

The reason how "waves" enter is that the uncertainty relation for $x$ and $p$ is precisely that of the "widths" of functions in Fourier conjugate variables, and the Fourier relationship we are most familiar with is that between position and momentum space. That the canonical commutation relations are equivalent to such a description by Fourier conjugate variables is the content of the Stone-von Neumann theorem.

However, it is the description by commutation relations and not that by Fourier conjugacy that generalizes to all quantum states and all operators. Therefore, it is the commutation relation between the operators that should be seen as the origin of their quantum mechanical uncertainty relations.

  • $\begingroup$ So, I suppose the axioms are much fundamental than the principle, right? $\endgroup$
    – user36790
    Aug 15, 2015 at 2:01
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    $\begingroup$ @user36790: I'd indeed say the axioms are more "fundamental" because they imply a more general uncertainty relation then the wave argument, but, as always, "fundamental" is a matter of taste. And I definitely think the other question is related - but it doesn't ask the same thing as this one at all (although they might be answered similarily), so I'd not call it a duplicate. $\endgroup$
    – ACuriousMind
    Aug 15, 2015 at 3:05

This was going to be a comment and then I decided to make an answer by an experimentalist. Too many theoretically based physicists are answering here, and in my opinion give a wrong standing to what physics theories are.

Physics theories are not mathematics. For mathematics axioms are fundamental, because starting from the axioms one can built up self consistently the theory,( the basic and ancient example being plane geometry), using logic and mathematical tools. Physics theories are developed to describe observations and predict new ones. The beauty and consistency of mathematics should not fool us into thinking the mathematics with its axioms and proofs are fundamental . Mathematics is necessary but is irrelevant to physics if there are no postulates to tie up physical observation to the mathematical solutions. The postulates of quantum mechanics are fundamental to the theory of quantum mechanics,

Postulates 1,2 in the link above is the one relevant for the Heisenberg Uncertainty principle.

1) Associated with any particle moving in a conservative field of force is a wave function which determines everything that can be known about the system.

2) With every physical observable q there is associated an operator Q, which when operating upon the wavefunction associated with a definite value of that observable will yield that value times the wavefunction.

This is what makes the correspondence between the mathematical tool of differential wave equations and their properties relevant to physics. Thus the answer with the wave explanation is relevant because the mathematics of waves is the same whether one uses it for probability distributions or for water waves.

The correspondence of observables to operators, also imposed outside mathematical axioms, leads to the commutation relations which are responsible for the Heisenberg uncertainty principle.

Studying the papers of Dirac and Jordan, while in frequent correspondence with Wolfgang Pauli, Heisenberg discovered a problem in the way one could measure basic physical variables appearing in the equations. His analysis showed that uncertainties, or imprecisions, always turned up if one tried to measure the position and the momentum of a particle at the same time. (Similar uncertainties occurred when measuring the energy and the time variables of the particle simultaneously.) These uncertainties or imprecisions in the measurements were not the fault of the experimenter, said Heisenberg, they were inherent in quantum mechanics.

Reading the history, one sees that the uncertainties in experiments finally led to the principle, the adoption of which in physics is consistent with the mathematical framework chosen to model the quantum mechanical nature of observations.

So in my opinion, the why it happens is directly connected to the wave nature of the quantum mechanical framework, probability wave of course. And the choice of wave equations was imposed by the observations.

  • $\begingroup$ @user36790 please read the above, it was going to be a comment in your response to acuriousmind $\endgroup$
    – anna v
    Aug 15, 2015 at 6:12
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    $\begingroup$ The problem with the wave nature explanation is that it doesn't hold for finite-dimensional systems. It doesn't explain why there is an uncertainty relation between the components of angular momentum, for instance. The postulates of "quantum mechanics" you like are actually the postulates of (Schrödinger's) wave mechanics, not of all of quantum theory. One can just as well use the Dirac-von Neumann postulates (and add that time evolution is generated by the Hamiltonian) and see no wave nature at all. $\endgroup$
    – ACuriousMind
    Aug 15, 2015 at 14:05
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    $\begingroup$ @ACuriousMind The equivalence between the two systems ( matrix and schrodinger) has been proven I think. Just different mathematical expression , and the experiments will always show diffraction and interference patterns no matter the formulation, let it be string theory too."see" is what the experiment says, not what the calculations cook. $\endgroup$
    – anna v
    Aug 15, 2015 at 18:55

The standard answer is given ACuriousMind, namely that it is the consequence of the mathematical structure of QM. An alternative answer (approach to QM) begins with the commutator relations and derives the mathematical structure of QM. The Uncertainty Principle (HUP) is then a version of Robertson's Inequality.

Why start with the commutator relations? Mach's Principle: All measurements are relative. Lets assume a measuring device for quantity Q produces a stream of measurements $X=x_1,X= x_2$... A second measuring device $Y = f(X)$ would produce a second measurement stream $Y=y_1,Y= y_2$..., with say $f(x) = x + \epsilon.sin(t)$. Looking at the measurement streams only, which "correctly" measures Q?

Simply speaking it is not possible to tell. One would have to look at the "construction" of the measuring device etc. which is not really satisfactory for various reasons. The quantum equivalent of Mach's statement "If there is only one particle in the Universe, it is meaningless to talk about it's physical properties" would be "If there is only a single measurement measuring device in the Universe, it is meaningless to talk about the measurement produced".

The requirement for repeatability means that the only opportunity to characterise the measurement streams is when the measurement by one device is followed by the measurement using a different measuring device. That leads to commutator relations and ultimately the standard QM mathematical structure.

So which is the correct way to talk about the HUP? You can't go wrong with the explanation given byACuriousMind, but it does depend on a particular set of postulates and other possible reasons 'why' exist.


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