# what is quantum mechanics?

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No assumption necessary when you use the big green smile. Up to now I've associated with either joking or sarcasm. I've only used it to denote joking. I see now that my association of it with sarcasm is invalid. Thank you for the clarification Phi.

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Quantum Mechanics is a statistical theory at best.

Waves determine the probability of finding the absolute square of these statistics, but in whole you cannot say quantum mechanics is purely or fundmentally a wave theory, because if that was the case, we neglect totally the wave-particle duality of nature.

Again, QM is a statistical theory. Nothing more and nothing really less.

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I disagree. There is more to quantum mechanics than a solely wave description can handle. For example; the state ket

$|\psi> = a|+> + b|->$

represents the super position of a particle which in a superposition of the spin up state and the spin down state. There is no wave function associated with this ket.

I am not sure what you are getting at. I could very easily write down a wavefunction for a superposition of spin up and spin down particles, in whatever basis you choose.

The wavefuction in QM is simply the coefficient sitting in front of the basis vectors in the decomposition of the state in terms of the the basis, e.g.

$|\psi \rangle = \int dx \psi(x) |x \rangle$

where $|x \rangle$ is the position eigenbasis and $\psi(x)$ is the position space wavefunction.

In other words, $\psi(x) = \langle x | \psi \rangle$, so in the example you gave the (position space) wavefunction is

$\psi(x) = a\langle x|+ \rangle + b \langle x|- \rangle$.

My objection to the name Quantum Mechnics is that the 'quantum' nature of QM, i.e. that energy, for example, comes in discrete lumps, comes from the wave properties of matter. Not only that, classical field theories also exhibit the same discretisation without being labled as 'quantum'.

For example, a guitar string can only produce notes with particular frequencies (or energies) because of it's boundary conditions at the end of the string. This is really no different to the quantisation of the electron's energy in the hydrogen atom, where the boundary conditions only allow certain Laguerre function to be solutions, and since each one has a different energy, energy is quantised.

This is generally true - all quantisation effects in QM are a consequence of describing particles as waves with boundary conditions.

The thing which distinguishes QM from classical physics is allowing the operators to effect the system. After measurement of a quantity, the system jumps into that corresponding operator's eigenstate.

Notice that simply the presence of operators isn't enough - the classical heat equation, for example, contains operators. Not even commutation relations are enough, since the "operators" in classical mechanics don't commute either.

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