# Time is not an observable, but....

• I
Gold Member
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
Staff Emeritus
What do you mean by saying that a time intervale is represented by a unitary matrix?

Demystifier
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.

Gold Member
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.

Gold Member
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.

Gold Member
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.

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

Demystifier
Gold Member
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.

PaleMoon
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
Gold Member
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.

PaleMoon
A. Neumaier
reduced to a macroscopic space position of something.

Demystifier
Gold Member

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

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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
Gold Member
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.

dextercioby and Demystifier
Demystifier
Gold Member
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
Gold Member
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}##.

Last edited:
Gold Member
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
Gold Member
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
$$\langle {\bf x}|{\bf x}'\rangle \approx \delta_{{\bf x},{\bf x}'}$$.