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nomadreid

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- #1

nomadreid

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What do you mean by saying that a time intervale is represented by a unitary matrix?

- #3

A. Neumaier

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How??But a time interval is (represented by) a unitary matrix

Yes. Most things we measure are not directly Hermitian operators in the technical sense. For example life times, cross sections, line widths of spectra, etc.is the identification of " observable" with "Hermitian" using the former term only in a technical sense but no longer one corresponding to our experience?

- #4

nomadreid

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- #5

A. Neumaier

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Well, this gives an operator depending on a time-difference, but it does not represent a time interval in a similar sense as ##q## represents a position. Instead, it is the operator that transforms a state at time ##t_1## into the corresponding state at time ##t_2##. Nothing is measured here or calculated from measurement.When I wrote that a time interval is represented by a unitary matrix, I was referring to the solution to Schrödinger's equation: U(t_{1},t_{2}≡ exp[-iH(t_{2}-t_{1})/ħ], where H is the Hamiltonian. No?

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nomadreid

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- #7

A. Neumaier

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The time to cook eggs is not different, not more or less shadowy than in quantum mechanics; both are equally real. You look at a watch to check the initial time (as a reading of a pointer or a counter), look again at the time the egg is deemed ready, and check if the time difference computed matches your goal. One does essentially the same when timing quantum events, except that the times are often much shorter and you need better equipment in place of a watch.That is, whereas real time is measured by comparing measurements, using distance between events, the speed of light, etc., this is a kind of shadowy time that is represented by changes in states: a kind of clock for quantum states that is related to real clocks

No quantum physics is needed to comprehend the nature of time. Time is the entity used to model in physics change or the lack of it. It is inferred by relating a change of interest to something that changes in an approximately periodic fashion - the position of the sun, the stars, the shadow of a pole, the pointer of a clock, or the counter of a digital watch.perhaps it is a heuristic to comprehend the nature of time, whatever that is?

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nomadreid

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Your previous remark that time in quantum mechanics is real would mean that one have to qualify "something" in the above quote by "something which can be assigned a real-number value", since otherwise the changes of a quantum state would fit this description. So, on one hand the unitary transformation does not represent any sort of time, yet on the other hand it corresponds to changes in time. Sorry, I am flailing a bit here, but I feel that the unitary transformation does represent something that is akin to time if not actually time, but the concept seems a bit slithery to my intuition.Time is the entity used to model in physics change...of something that changes in an approximately periodic fashion

- #9

A. Neumaier

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It representsthe unitary transformation does represent something

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nomadreid

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- #12

A. Neumaier

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Modern measurements of time intervals in quantum experiments typically do not involve a pointer (what you call needle).the state of clock is given by the space position of a needle.

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It was just an example. Modern instruments of time intervals certainly involve something the measurement outcome of which can be reduced to a macroscopic space position of something.Modern measurements of time intervals in quantum experiments typically do not involve a pointer (what you call needle).

- #14

A. Neumaier

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How about a digital counter?reduced to a macroscopic space position of something.

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How about a digital counter?

Each digit consists of a few black lines, and each black line has a well defined macroscopic position.

- #16

A. Neumaier

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Sure, but each of the pixels of the black line has a position known in advance, hence no position is ever measured. Instead it is measured which pixels have which color!View attachment 232631

Each digit consists of a few black lines, and each black line has a well defined macroscopic position.

- #17

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A pixel is relatively small, but still sufficiently big to consider it macroscopic. I'm not sure how exactly a pixel becomes black, but it must be that something physical (call it Z) comes to the position of that pixel which makes it black. Hence it reduces to the position of Z.Sure, but each of the pixels of the black line has a position known in advance, hence no position is ever measured. Instead it is measured which pixels have which color!

- #18

A. Neumaier

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In a modern LED display, the pixel brightness is controlled by an electric field that influences the degree to which polarized light goes through a liquid crystal. Thus the something is the electric field Z, not a particle. It is everywhere with different intensity at different positions.A pixel is relatively small, but still sufficiently big to consider it macroscopic. I'm not sure how exactly a pixel becomes black, but it must be that something physical (call it Z) comes to the position of that pixel which makes it black. Hence it reduces to the position of Z.

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Which confirms what I claim that it is eventually the position of something that determines the macroscopic outcome of measurement. In this case, the position at which the intensity has this or that value.It is everywhere with different intensity at different positions.

- #20

A. Neumaier

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But this is quite different from the quantum measurement of a position operator of something.Which confirms what I claim that it is eventually the position of something that determines the macroscopic outcome of measurement. In this case, the position at which the intensity has this or that value.

- #21

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It's not different. For simplicity, let as assume that electric field is zero where the pixels are not black and non-zero where the pixels are black. Let ##|{\bf x}\rangle## be a quantum state of EM field such thatBut this is quite different from the quantum measurement of a position operator of something.

$$\langle {\bf x}|\hat{\bf E}({\bf x}')|{\bf x}\rangle \sim \delta^3( {\bf x}-{\bf x}')$$

(The exact meaning of ##\sim## is a mathematical subtlety, which is not so important here.) Then we can define the position operator as

$$\hat{\bf X}=\int d^3x\, {\bf x}|{\bf x}\rangle \langle {\bf x}|$$

Our measurement can then be reduced to measurement of the position operators of the form of ##\hat{\bf X}##.

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- #22

nomadreid

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- #23

A. Neumaier

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I don't see how this can suffice, even assuming a discrete set of points ##x## to eliminate problems with a continuous spectrum. Without further assumptions you don't even get the states ##|x\rangle## to be eigenstates of your position operator.It's not different. For simplicity, let as assume that electric field is zero where the pixels are not black and non-zero where the pixels are black. Let ##|{\bf x}\rangle## be a quantum state of EM field such that

$$\langle {\bf x}|\hat{\bf E}({\bf x}')|{\bf x}\rangle \sim \delta^3( {\bf x}-{\bf x}')$$

(The exact meaning of ##\sim## is a mathematical subtlety, which is not so important here.) Then we can define the position operator as

$$\hat{\bf X}=\int d^3x\, {\bf x}|{\bf x}\rangle \langle {\bf x}|$$

Our measurement can then be reduced to measurement of the position operators of the form of ##\hat{\bf X}##.

- #24

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For simplicity, let us take the discrete case. You can think of ##|{\bf x}\rangle## as a state with a significant (expectation value of) electric field near ##{\bf x}## and negligible electric field elsewhere. ThenI don't see how this can suffice, even assuming a discrete set of points ##x## to eliminate problems with a continuous spectrum. Without further assumptions you don't even get the states ##|x\rangle## to be eigenstates of your position operator.

$$\langle {\bf x}|{\bf x}'\rangle \approx \delta_{{\bf x},{\bf x}'}$$

so the position operator

$$\hat{{\bf X}}=\sum_{{\bf x}} {\bf x} |{\bf x}\rangle \langle {\bf x}|$$

satisfies

$$\hat{{\bf X}} |{\bf x}'\rangle \approx {\bf x}' |{\bf x}'\rangle$$

Hence my states are approximate eigenstates of the position operator.

- #25

A. Neumaier

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I don't see how something like this follows from your assumption.For simplicity, let us take the discrete case. You can think of ##|{\bf x}\rangle## as a state with a significant (expectation value of) electric field near ##{\bf x}## and negligible electric field elsewhere. Then

$$\langle {\bf x}|{\bf x}'\rangle \approx \delta_{{\bf x},{\bf x}'}$$.

- #26

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For instance, coherent states of EM field satisfy an approximate orthogonality of that form.I don't see how something like this follows from your assumption.

- #27

A. Neumaier

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Coherent states of the e/m field are not parameterized by a position, so I don't see how they can be used as your kets. Moreover, t make your argument about measuring, you need to get arbitrary close to orthogonality, which coherent states do not provide - they are well-known to be heavily overcomplete.For instance, coherent states of EM field satisfy an approximate orthogonality of that form.

- #28

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Arghhhh!Coherent states of the e/m field are not parameterized by a position

Coherent states are parameterized by expectation value of electric field (and its canonical conjugate), and I already said that my states have defined expectation values of fields. In my case, those fields are localized around certain point x.

- #29

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I don't take all coherent states. I only take a discrete subset, those the electric fields of which are localized around a discrete set of points. In this way I get near orthogonality. The set is not complete in the full Hilbert space of QED, but I don't need that kind of completeness because I only need those states that can describe the states of discrete pixels.you need to get arbitrary close to orthogonality, which coherent states do not provide - they are well-known to be heavily overcomplete.

- #30

A. Neumaier

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Ok, so you take the subset of coherent states whose electric field is zero at all but one pixel ##x##. But for each ##x## this still leaves an infinity of coherent states with different intensity; which ones do you pick for ##|x\rangle##?Coherent states are parameterized by expectation value of electric field (and its canonical conjugate), and I already said that my states have defined expectation values of fields. In my case, those fields are localized around certain point x.

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