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In each of the uncertainty relations $$\Delta p_x \Delta x \geq \hbar/2$$ $$\Delta p_y \Delta y \geq \hbar/2$$$$\Delta p_z \Delta z \geq \hbar/2$$$$\Delta E \Delta t \geq \hbar/2$$ the second term on the left side is one of the compoments of a position 4-vector, while the first term seems to be the corresponding component of the momentum-energy 4-vector, i.e. the proper time derivative of the position component multiplied by the invariant mass.

Does this have any "deeper" (whatever that means) physical significance or something else is going on here?

I'm perfectly aware that this might be a silly question, but I'm still learning both quantum and spacetime physics and I never had to deal with the intersection between those two, so I would appreciate a simple clarification!

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Note that the position vector $x^j$ is a "contravariant" vector, with upper indices, while momentum vector $p_i$ is fundamentally a "covariant" vector, with lower indices. Transformation laws for contravariant and covariant vectors are different. Of course, in special relativity, the momentum is sometimes written with upper indices like $p^i = m\frac{dx^i}{d \tau}$, but it hides the true nature of the momentum. This said, your expressions may be written in the compact form $\Delta p_i \Delta x^j \geq \delta_i^j \large \frac{\hbar}{2}$ –  Trimok Sep 11 '13 at 11:41
    
I see, that makes sense. –  Schlomo Sep 11 '13 at 11:49

3 Answers 3

up vote 5 down vote accepted

There is a very interesting story behind your question. In the early 1900's (after Special Relativity had been introduced) the solution to wave equation in vacuum:

$a\exp(i(kx-wt)) + b\exp(-i(kx-wt)) $

where $k$ is the wave vector and $w/k=v$ and v is the phase velocity.

De Broglie was the first to notice that the phase factors of the equation at every event remain invariant under Lorentz transformations if $(k,w)$ is considered a 4-vector. This meant invariant amplitude at every event. This is because the scalar product of the two 4-vectors $(k,w)$ and $(x,t)$ remains invariant under Lorentz transformations. From this remarkable piece of insight, he deduced that $(k,w)$ could represent the 4-momentum, $(p,E)$, of a massive particle. This is how wave-particle duality was first discovered. Special Relativity gave birth to quantum mechanics in its proper form!

The commutation relation discovered subsequently

$[p,x]=-i\hbar$

on solving gives (setting the arbitrary phase factor in p to 1): $p=-i\hbar \frac{∂}{∂x} $. The uncertainty relation in momentum is derived from here.

The fundamental form of Schrodinger's equation

$i\hbar \frac{∂T}{∂t} = ET$

Where $E$ is the Hamiltonian and $T(t_0,t)$ is the Unitary time evolution operator has its origins in the relationship between $x$ and $p$ being extended to $t$ and $E$.

When introducing the Schrodinger equation in his book, Dirac points out that its derivation comes mainly from considerations of relativity. In fact Schrodinger's original equation was actually relativistic, where $E$ was the relativistic energy. Schrodinger wasn't sure whether to take the positive or negative root of $E^2$. So he discarded it in favor of its widely known non-relativistic form.

So in fact, other than the discovery of quantized energy levels in blackbody radiation and the photoelectric effect, every breakthrough in QM owes it's existence to special relativity.

EDIT: I earlier said w/k = c for a massive particle, this is incorrect. w/k is equal to the phase velocity, which is proportional to 1/u. u is the group velocity which is equal to the classical velocity of a massive particle. I have fixed the offending sentences.

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You've been on this wi-ru-log for a while already. It would be nice if you'd start using MathJaX now, or at least not put dots in between the variables for multiplication, as it only makes more trouble for the person editingg the post : ) Anyway, +1. –  Dimensio1n0 Sep 11 '13 at 14:50
    
Note however that the I am considering only a single, massive particle, which was historically the first object to be analyzed by Schrodinger's equation. –  dj_mummy Sep 11 '13 at 14:50
    
@DImension10AbhimanyuPS I really do apologize for my poor formatting. :D One day, though, I will surely learn how to use this 'MathJax' one day. –  dj_mummy Sep 11 '13 at 14:52
    
Do these help ? : cs.brown.edu/system/software/latex/doc/symbols.pdf , tobi.oetiker.ch/lshort/lshort.pdf ? LM has the links to these on his blog... There was also this great pdf of LaTeX symbols at "rafflesnus.edu.sg" or something, but it's now dead. –  Dimensio1n0 Sep 11 '13 at 14:56

I saw a similar observation in Griffith's Quantum Mechanics (page 114, 2nd Edition), which I'm liberally borrowing from here.

In a relativistic quantum theory, the last equation would follow from any of the first three since the splitting of spacetime into space-plus-time is dependent on the choice of Lorentz frame.

However, in non-relativistic quantum mechanics, the logic fails because space and time are very distinct; while $p$, $x$ and $E$ are all observables (operators whose spectrum is the set of allowed values, etc etc), $t$ is not. In fact, the Schrodinger equation is first-order in $t$ but second-order in $x$, and Dirac obtained his relativistic wave equation by trying various methods by modifying the Schrodinger equation so that it was first-order in both.

The $\Delta E \Delta t$ uncertainty relation is also rather harder, in my opinion, to interpret than the $\Delta p \Delta x$ ones. The later has a simple interpretation because any quantum state has some $\Delta p$ and some $\Delta x$ associated with it; unfortunately, it has no $\Delta t$ - what does it mean if we say, this particle trapped in a box has $\Delta t = 1\mu s$? One of the many interpretations treats $\Delta t$ as the time is takes for some other observable to change by one standard deviation (p 116), or the oscillation period of a state of energy $E$, or the lifetime of an unstable particle of rest mass energy $E$; of course, we get slightly different bounds each time, and it is interesting that $\Delta E \Delta t$ always be bounded in these cases; it's certainly not obvious to me.

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The Heisenberg uncertainty relations reads

$$\tag{1} \Delta x^{\mu}~\Delta p_{\nu} ~\gtrsim~ \frac{\hbar}{2} \delta^{\mu}_{\nu}, \qquad \mu,\nu\in\{0,1,2,3\}. $$

Clearly the deeper physical principle that OP is alluring to is Poincare symmetry $ISO(3,1)$. Now, the position-momentum uncertainty relations can easily be derived from the canonical commutation relations (CCR)

$$\tag{2} [\hat{x}^j,\hat{p}_k] ~=~i\hbar~ \delta^j_k~{\bf 1}, \qquad j,k\in\{1,2,3\}, $$

see e.g. the Wikipedia page. The time-energy uncertainty relation is more subtle to explain within non-relativistic quantum mechanics, see e.g. this and this Phys.SE posts. But we could consider the analogous simpler question

$$\tag{3} \Delta x^{j}~\Delta p_{k} ~\gtrsim~ \frac{\hbar}{2} \delta^{j}_{k}, \qquad j,k\in\{1,2,3\}, $$

for the Euclidean group $E(3)=ISO(3)$ in the three spatial dimensions, i.e. translations and rotations without the boosts. The latter simpler question follows because (i) the uncertainty relations (3) are a direct consequence of the CCR (2), and (ii) the CCR (2) are invariant under Euclidean symmetries.

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