# Tag Info

1

I'm not sure what "Pauli's argument" precisely is because he refers to pages in the first edition of Dirac's Principles of Quantum Mechanics which contain nothing of evident relevance in my fourth edition, but the more common thing to say is that it is the boundedness of the energy from below that forbids a naive time operator, not the discreteness. However,...

0

Commutation relations between position and momentum operators, in quantum mechanics, are valid only when the actions of either of the two operators is contained in the domain of the remaining one. In particular we have: $$[x,p]\psi = (xp)\psi-(px)\psi.$$ In order for the above to be defined $\psi$ must be in the domain of definition of both operators and, ...

-1

(In my experience this tends to be a rather controversial subject, so I think this answer might start some arguments!) First of all, given a specific action, it is a purely mathematical result whether or not there exists a local transformation that depends on an arbitrary smooth function $\lambda(x)$ on spacetime and leaves the action invariant. The ...

5

General wave functions can be expressed in terms of any set of eigenfunctions. But for bound systems, the energy eigenfunctions have a couple of appealing properties that make them popular: The energy eigenfunctions are the solutions to the time-independent problem, so you can work on a steady-state system. This often makes the math a lot easier. The ...

4

You can work in the position representation. It is not difficult: \langle \psi \mid[x,p] \mid \psi \rangle = \langle \psi \mid (xp-px)\mid \psi \rangle =\\=\int dx \ \psi^*(x) \left(-i \hbar x \ \partial_x \psi(x) + i \hbar \ \partial_x (x \ \psi(x)) \right) = \\ =-i \hbar \int dx \ \psi^* \left(x \ \psi'-(\psi+x \ \psi')\right) = \\ =i \hbar \int dx \...

2

Observables are operators, in particular they are of the self-adjoint type (with discrete spectrum spanning the entire Hilbert space). Given a normalised state $|\psi\rangle$, the expectation value of an operator $A$ thereupon is defined as $\langle A \rangle = \langle\psi |\, A\, |\psi\rangle$; equivalently, one can prove that the uncertainty on the ...

8

There doesn't exist any procedure to uniquely associate a Hermitian operator $L$ to a function of the phase space $f(x,p)$. Quantum mechanics is a theory that exists independently of classical physics. Quantum mechanics is not just a cherry on a classical pie that needs the classical theory to exist at every moment. If we want to define a quantum theory, we ...

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