Time is not an observable, but....

[nomadreid]

It is one of the most repeated responses whenever the subject of time comes up. OK, point taken. But a time interval is (represented by) a unitary matrix, and therefore can be written as exp(iK) for K being Hermitian. So it sounds like although a time (interval) is not directly measurable, it is (theoretically) directly calculable from measurements, and is a real number. But many measurements have at least one step in their derivation which is actually a calculation, such as when we measure velocity. So is the identification of " observable" with "Hermitian" using the former term only in a technical sense but no longer one corresponding to our experience?

 

[stevendaryl]

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

 

[A. Neumaier]

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

How??

is the identification of " observable" with "Hermitian" using the former term only in a technical sense but no longer one corresponding to our experience?

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.

 

[nomadreid]

When I wrote that a time interval is represented by a unitary matrix, I was referring to the solution to Schrödinger's equation: U(t₁,t₂) ≡ exp[-iH(t₂-t₁)/ħ], where H is the Hamiltonian. No?

 

[A. Neumaier]

When I wrote that a time interval is represented by a unitary matrix, I was referring to the solution to Schrödinger's equation: U(t₁,t₂ ≡ exp[-iH(t₂-t₁)/ħ], where H is the Hamiltonian. No?

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.

 

[nomadreid]

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. Not to be used to cook eggs, but perhaps it is a heuristic to comprehend the nature of time, whatever that is?

 

[A. Neumaier]

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

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.

perhaps it is a heuristic to comprehend the nature of time, whatever that is?

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.

 

[nomadreid]

Time is the entity used to model in physics change...of something that changes in an approximately periodic fashion

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.

 

[A. Neumaier]

the unitary transformation does represent something

It represents the act of changing in a given time interval the state of an arbitrary system of the kind modeled, in the same way as a 2 by 2 rotation matrix in the Euclidean plane (or the wall of a gallery, say) represents the act of changing an arbitrary figure in the plane by rotating it around the origin by a given angle. But the two times are external to the first act in the same way as the angle is external to the second act. They are just parameters in the description.

 

[nomadreid]

Thanks, A.Neumaier. That is an enlightening analogy. This gives me something to mull over; also thanks for your patience.

 

[Demystifier]

Time is not an observable, in the sense that there is no time operator. But it doesn't mean that time is not measurable. Time is measured by a clock, and the state of clock is given by the space position of a needle. The space position is an observable.

 

[A. Neumaier]

the state of clock is given by the space position of a needle.

Modern measurements of time intervals in quantum experiments typically do not involve a pointer (what you call needle).

 

[Demystifier]

Modern measurements of time intervals in quantum experiments typically do not involve a pointer (what you call needle).

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.

 

[A. Neumaier]

reduced to a macroscopic space position of something.

How about a digital counter?

 

[Demystifier]

How about a digital counter?

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

 

[A. Neumaier]

View attachment 232631
Each digit consists of a few black lines, and each black line has a well defined macroscopic position.

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!

 

[Demystifier]

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!

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.

 

[A. Neumaier]

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.

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.

 

[Demystifier]

It is everywhere with different intensity at different positions.

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.

 

[A. Neumaier]

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.

But this is quite different from the quantum measurement of a position operator of something.

 

[Demystifier]

But this is quite different from the quantum measurement of a position operator of something.

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 $\hat{\bf X}$.

 

[nomadreid]

I presume I am overlooking some fundamental point here, because it seems to me that saying that the time can be measured via distances is a bit circular, since the distances alone are not sufficient: that is, the distance between two events can be the same in two different measurements with two different time intervals. If we then say that we just need to throw in the speed of light ( and curvature and all that) to calculate the time once given a distance measurement, we then are implicitly assuming a time measurement in place so that the speed of light has a meaning.

 

[A. Neumaier]

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 $\hat{\bf X}$.

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.

 

[Demystifier]

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.

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

 

[A. Neumaier]

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

I don't see how something like this follows from your assumption.

 

[Demystifier]

I don't see how something like this follows from your assumption.

For instance, coherent states of EM field satisfy an approximate orthogonality of that form.

 

[A. Neumaier]

For instance, coherent states of EM field satisfy an approximate orthogonality of that form.

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.

 

[Demystifier]

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

Arghhhh!
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.

 

[Demystifier]

you need to get arbitrary close to orthogonality, which coherent states do not provide - they are well-known to be heavily overcomplete.

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.

 

[A. Neumaier]

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.

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 $|x\rangle$?

 

[Demystifier]

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 $|x\rangle$?

It is an experimental question. Those the expectation values of fields best describe the measured "classical" fields in the LCD screen.

 

[A. Neumaier]

It is an experimental question. Those the expectation values of fields best describe the measured "classical" fields in the LCD screen.

Ok, so you pick one coherent state per pixel with the experimental intensity. Now coherent states corresponding to neighboring pixels will substantially overlap. Thus the eigenstates of your position operator will have significant support in a number of pixels close to the intended one. This means that measuring $X$ will produce a superposition of pixels with probabilities that are maximal at the intended pixel but significantly less than 1. This means that you always get blurred measurements of the pixels, not true position measurements. This is sufficient for recognizing whether the digit 3 or 4 appeared, but not for a reduction of the digital time measurement process to one with true position measurements.

 

[Demystifier]

Now coherent states corresponding to neighboring pixels will substantially overlap. Thus the eigenstates of your position operator will have significant support in a number of pixels close to the intended one.

My intuition is that the overlap will be small, but I guess we are both guilty of not quantifying our intuitions.

But when I think again about all this, I realize that we are actually discussing irrelevant issues. What is relevant is that even though there is no time observable, there is an observable associated with a clock. In the case of LCD screen with digits, a relevant observable is the electric field in the screen. I believe you would agree with that.

 

[A. Neumaier]

a relevant observable is the electric field in the screen. I believe you would agree with that.

Yes, and the electric field changes with time. Thus time is not an observable but a parameter that tells which value $E(t)$ of the electric field applies in a particular instant. Similar for position. To talk about position you need to say at which time you mean it.

A measured time interval is usually the difference of the times at which two particular events happen. For example, you may ask about the time it takes between two adjacent sign changes of some component of the electric field. There is no operator corresponding to this measurement, only operators corresponding to the triggering events.