I Time-energy uncertainty relation

[PeterDonis]

Each of those parameters can be mapped to an operator which multiplies the function by that parameter

I'm not sure what operators you are talking about. For the position operators, which I think is what you mean by the operators corresponding to $x$, $y$, and $z$, those operators only correspond to "multiply by $x$" (or $y$ or $z$) in the position representation. They don't in other representations. $t$ does not work that way; it is a scalar parameter and the only thing you can do with it is multiply by it, regardless of representation.

Also, $x$, $y$, and $z$ are not really parameters; they are labels for degrees of freedom (roughly speaking, dimensions in the Hilbert space). That is not a property that $t$ has.

 

[vanhees71]

$x,y,z$ are not "dimensions in the Hilbert space" but the eigenvalues of the self-adjoint operators that represent the components of the position vector of the particle in the 1st-quantization formalism of (non-relativistic) QT.

 

[PeterDonis]

$x,y,z$ are not "dimensions in the Hilbert space" but the eigenvalues of the self-adjoint operators that represent the components of the position vector of the particle in the 1st-quantization formalism of (non-relativistic) QT.

Strictly speaking, yes, $x$, $y$, and $z$, or more precisely each possible value for each of those, are eigenvalues. But all of the possible eigenvalues, taken together, form a set of 3 continuous parameters that label the possible range of variation in the Hilbert space for a single spinless particle, in the position representation. "Dimension" might not be precisely the right technical term for this, but this is what I was referring to. And none of these things are the same as what $t$ is.

 

[vanhees71]

It's a "generalized basis of eigenvectors" (to put it in the usual physicists' hand-waving slang). The Hilbert space of the 1st-quantization formalism is the separable Hilbert space, i.e., it has a countable basis, isomorphic to $\text{L}^2(\mathbb{R}^3)$. An example are the eigenfunctions of the 3D harmonic oscillator.

As stressed several times before, in QT $t$ is not an observable but a parameter; $t$ thus does also not denote eigenvalues of any operator.

In relativistic QFT $x=(ct,\vec{x})$ are Minkowski-vector valued parameters and also not eigenvalues of observables.