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2

The zero-point energy is the energy eigenvalue $\langle H \rangle_\Omega$ of the lowest lying energy eigenstate $\Omega$. $\Delta H_\Omega = 0$, so this has nothing to do with the zero-point energy. This does also not contradict the uncertainty principle, since $\frac{\mathrm{d}\langle A \rangle}{\mathrm{d}t} = 0$ for a energy eigenstate (which are the ...


1

In classical mechanics, each body has an exact position at all times (and hence an exact speed given by its derivative, and the equivalent quantities for rotary motion). To understand the difference to quantum mechanics, think of it as wave mechanics: Each body is described by a wave(function) and properties such as position or speed are only defined to the ...


1

Roughly speaking, and restricting to particles for now, a classical trajectory is a set of exact positions and corresponding velocities (or momenta) of the particle, which (usually) change over time. In quantum mechanics, the uncertainty principle says that it is not possible to know simultaneously the exact position and momentum of the particle, and so it ...


4

The question is so ambiguous that it allows a resounding yes. This is because "balance" is not defined, neither are the dimensions and the material used for the pencil, nor the location of where the "balancing" is to happen. The material and the shape of the surface to balance the pencil on, are not specified, nor the length of time it should stay balanced. ...


2

Let me be the firs to answer 'Yes' (more or less). As the saying goes: In theory there is no difference between theory and practice. In practice there is. What I'm getting at is that there will always be differences between theory and practice, and that it is up to the physicist to decide which assumptions/simplications are suitable and which ones are ...


17

No. The weight of the pencil is roughly 1 Newton, and the area is about 500 square picometer (5 * 10-22) which means the pressure on the tip is around 2 ZettaPascal. That's quite a bit more than what graphite (or diamond) can withstand (that's masured in GigaPascal)


4

No. Firstly, the point of the pencil is generally not sharp enough to have just one single atom. People attempt to make that kind of tip in STM's. Even if you somehow did manage to get it sharp enough, graphite is so soft the the weight of the pencil will crush the tip. It won't stay a single atom wide. So there is no way to balance the pencil on a single ...


78

TL;DR: there are many factors that prevent a pencil remaining perfectly balanced. The most important of these is the uncertainty principle that will make the pencil fall over in less than four seconds. For details, read on... Short answer: NO. The first photon of light that hits it would disturb your perfect equilibrium. The moon's tidal forces (which are ...


81

No. To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time. Momentum and position form a conjugate pair. $\Delta x \Delta p \geq \hbar$. Angular momentum and angular position form one too. $\Delta L \Delta \Theta \geq \hbar$ This doesn't guarantee ...


1

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 ...


-1

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 ...


0

Light speed being invariant is necessary for there to be "quantum effects" (if by that, you mean probabilistic phenomena). Since light speed is invariant, fixed at a rate of 1 lp per 1 tp, and no measurements are possible below the Planck scale, therefore any speed slower than c must be described as a probability at that scale. Even if you plot the ...


1

Your everyday physics intuition is leading you astray. The kinetic energy of the incoming particles can't just disappear, so they can't just come to a stop unless they can transfer the kinetic energy into some other degree of freedom. If you collide two balls of clay they will come to a stop, but only because the kinetic energy can go into deforming and ...



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