Return to Question

To see where this question comes from, consider a time independent Hamiltonian $H$ and an initial wave function $\psi(t=0,x)$. We can express time dependant wave function $\psi(t,x) = \sum_j e^{-iE_jt/\hbar} \phi_j(x) c_j$ where $\{\phi_j(x)\}$ is the set of eigen functions of H and $c_j=\int \psi(t=0,y) \phi_j^*(y) dy$.

Let's say we strongly measure energy at some arbitrary time. This causes wave function to collapse to $\psi(t,x)=e^{-iE_kt/\hbar}\phi_k(x)$, and we get $E_k$ as the energy of the system, where k is a random integer. Since $[H(t), H(t')] = 0$, measuring energy over and over again at any time always result in $E_k$ and wave function stays untouched. This is of course only true by assuming that the very act of measurement is not associated with a time dependant Hamiltonian $H_m(t)$ such that $[H_m(t), H] \neq 0$. If it does, then it would imply that measuring eigenvalues of $H$ with arbitrary precision in principle is impossible. There is always some noise coming from measurement itself, and the energy we are measuring corresponds to $H_m(t) + H$, not $H$ itself.

Let's assume otherwise and $H_m(t)=0$. At first, it may seem odd that, on one hand we can measure energy over and over and expect to get $E_k$ and indeed get $E_k$, so $\Delta E = 0$ but then we would have $\Delta E \Delta t= 0$. However, I see no issue, since t is not well defined, and in this context, the time between measurement is not necessarily $\Delta t$ anyway. Perhaps $\Delta t$ refers to inverse rate of changes of some operators that goes to infinity.

So either I am to believe that "measuring energy with arbitrary precision inherently impossible", which seems weird to say the least. Consider spin operator, as long as I measure spin on a fixed axis (say, z) over and over again, I always get the same value no problem, this is the case for all time independent operators, leaving $H$ out is a bit odd. For example, imagine a two state energy system as follows

$$|\psi\rangle = \frac{1}{\sqrt 2} (|e_0 \uparrow \rangle + |e_1 \downarrow \rangle)$$

Measuring spin gives full information about energy. So any limitation set on $H$ measurement should be also set on $S$ measurement. Or accept that $\Delta t = \infty$ somehow and I should not ask why. Or perhaps there is better explanation that I didn't think about, please let me know.

Edit:

Consider an ideal single-photon light emitter that uses a battery as it source. We can monitor the battery and measure its energy with arbitrary precision. Let say we measure battery's energy with 100% precision at arbitrarily small time intervals $\Delta t$. Now as soon as this emitter emits a single photon, we see a jump in battery's energy, indicating that how much energy was used for producing that photon. Since we know photon's energy, we know its exact momentum. On the other hand, we roughly know when the photon was fired too by simply looking at the time of jump in the energy of the battery. Since a time interval $\Delta t$ exists between each battery's energy measurement, the uncertainty in the location of photon is $\Delta x = c\Delta t$ but given $\Delta p = 0$, we have violated uncertainty principle. Again, one of these should be true

1. All measurement devices are noisy, we can never truly measure energy without noise.

2. Or $\Delta t$ DOES refer to the time interval between energy measurements, and it goes to infinity for measuring energy value exactly.

The problem is, neither of this conclusions comes from standard texts in QM. Measurement devices can collapse wave function and measure energy exactly. $\Delta t$ is not introduced as time between measurements either, it can be anything, really.

Some related topics:

Interpretation of the energy-time uncertainty

What is $\Delta t$ in the time-energy uncertainty principle?

Energy-time uncertainty and pair creation

To see where this question comes from, consider a time independent Hamiltonian $H$ and an initial wave function $\psi(t=0,x)$. We can express time dependant wave function $\psi(t,x) = \sum_j e^{-iE_jt/\hbar} \phi_j(x) c_j$ where $\{\phi_j(x)\}$ is the set of eigen functions of H and $c_j=\int \psi(t=0,y) \phi_j^*(y) dy$.

Let's say we strongly measure energy at some arbitrary time. This causes wave function to collapse to $\psi(t,x)=e^{-iE_kt/\hbar}\phi_k(x)$, and we get $E_k$ as the energy of the system, where k is a random integer. Since $[H(t), H(t')] = 0$, measuring energy over and over again at any time always result in $E_k$ and wave function stays untouched. This is of course only true by assuming that the very act of measurement is not associated with a time dependant Hamiltonian $H_m(t)$ such that $[H_m(t), H] \neq 0$. If it does, then it would imply that measuring eigenvalues of $H$ with arbitrary precision in principle is impossible. There is always some noise coming from measurement itself, and the energy we are measuring corresponds to $H_m(t) + H$, not $H$ itself.

Let's assume otherwise and $H_m(t)=0$. At first, it may seem odd that, on one hand we can measure energy over and over and expect to get $E_k$ and indeed get $E_k$, so $\Delta E = 0$ but then we would have $\Delta E \Delta t= 0$. However, I see no issue, since t is not well defined, and in this context, the time between measurement is not necessarily $\Delta t$ anyway. Perhaps $\Delta t$ refers to inverse rate of changes of some operators that goes to infinity.

So either I am to believe that "measuring energy with arbitrary precision inherently impossible", which seems weird to say the least. Consider spin operator, as long as I measure spin on a fixed axis (say, z) over and over again, I always get the same value no problem, this is the case for all time independent operators, leaving $H$ out is a bit odd. For example, imagine a two state energy system as follows

$$|\psi\rangle = \frac{1}{\sqrt 2} (|e_0 \uparrow \rangle + |e_1 \downarrow \rangle)$$

Measuring spin gives full information about energy. So any limitation set on $H$ measurement should be also set on $S$ measurement. Or accept that $\Delta t = \infty$ somehow and I should not ask why. Or perhaps there is better explanation that I didn't think about, please let me know.

Edit:

Consider an ideal single-photon light emitter that uses a battery as it source. We can monitor the battery and measure its energy with arbitrary precision. Let say we measure battery's energy with 100% precision at arbitrarily small time intervals $\Delta t$. Now as soon as this emitter emits a single photon, we see a jump in battery's energy, indicating that how much energy was used for producing that photon. Since we know photon's energy, we know its exact momentum (the emitter emits photons in 1d). On the other hand, we roughly know when the photon was fired too by simply looking at the time of jump in the energy of the battery. Since a time interval $\Delta t$ exists between each battery's energy measurement, the uncertainty in the location of photon is $\Delta x = c\Delta t$ but given $\Delta p = 0$, we have violated uncertainty principle. Again, one of these should be true

1. All measurement devices are noisy, we can never truly measure energy without noise.

2. Or $\Delta t$ DOES refer to the time interval between energy measurements, and it goes to infinity for measuring energy value exactly.

The problem is, neither of this conclusions comes from standard texts in QM. Measurement devices can collapse wave function and measure energy exactly. $\Delta t$ is not introduced as time between measurements either, it can be anything, really.

Some related topics:

Interpretation of the energy-time uncertainty

What is $\Delta t$ in the time-energy uncertainty principle?

Energy-time uncertainty and pair creation

To see where this question comes from, consider a time independent Hamiltonian $H$ and an initial wave function $\psi(t=0,x)$. We can express time dependant wave function $\psi(t,x) = \sum_j e^{-iE_jt/\hbar} \phi_j(x) c_j$ where $\{\phi_j(x)\}$ is the set of eigen functions of H and $c_j=\int \psi(t=0,y) \phi_j^*(y) dy$.

Let's say we strongly measure energy at some arbitrary time. This causes wave function to collapse to $\psi(t,x)=e^{-iE_kt/\hbar}\phi_k(x)$, and we get $E_k$ as the energy of the system, where k is a random integer. Since $[H(t), H(t')] = 0$, measuring energy over and over again at any time always result in $E_k$ and wave function stays untouched. This is of course only true by assuming that the very act of measurement is not associated with a time dependant Hamiltonian $H_m(t)$ such that $[H_m(t), H] \neq 0$. If it does, then it would imply that measuring eigenvalues of $H$ with arbitrary precision in principle is impossible. There is always some noise coming from measurement itself, and the energy we are measuring corresponds to $H_m(t) + H$, not $H$ itself.

Let's assume otherwise and $H_m(t)=0$. At first, it may seem odd that, on one hand we can measure energy over and over and expect to get $E_k$ and indeed get $E_k$, so $\Delta E = 0$ but then we would have $\Delta E \Delta t= 0$. However, I see no issue, since t is not well defined, and in this context, the time between measurement is not necessarily $\Delta t$ anyway. Perhaps $\Delta t$ refers to rate of changes of some operators that goes to infinity.

So either I am to believe that "measuring energy with arbitrary precision inherently impossible", which seems weird to say the least. Consider spin operator, as long as I measure spin on a fixed axis (say, z) over and over again, I always get the same value no problem, this is the case for all time independent operators, leaving $H$ out is a bit odd. For example, imagine a two state energy system as follows

$$|\psi\rangle = \frac{1}{\sqrt 2} (|e_0 \uparrow \rangle + |e_1 \downarrow \rangle)$$

Measuring spin gives full information about energy. So any limitation set on $H$ measurement should be also set on $S$ measurement.

  Or accept that $\Delta t = \infty$ somehow and I should not ask why. Or perhaps there is better explanation that I didn't think about, please let me know.

Some related topics:

Interpretation of the energy-time uncertainty

What is $\Delta t$ in the time-energy uncertainty principle?

Energy-time uncertainty and pair creation

To see where this question comes from, consider a time independent Hamiltonian $H$ and an initial wave function $\psi(t=0,x)$. We can express time dependant wave function $\psi(t,x) = \sum_j e^{-iE_jt/\hbar} \phi_j(x) c_j$ where $\{\phi_j(x)\}$ is the set of eigen functions of H and $c_j=\int \psi(t=0,y) \phi_j^*(y) dy$.

Let's say we strongly measure energy at some arbitrary time. This causes wave function to collapse to $\psi(t,x)=e^{-iE_kt/\hbar}\phi_k(x)$, and we get $E_k$ as the energy of the system, where k is a random integer. Since $[H(t), H(t')] = 0$, measuring energy over and over again at any time always result in $E_k$ and wave function stays untouched. This is of course only true by assuming that the very act of measurement is not associated with a time dependant Hamiltonian $H_m(t)$ such that $[H_m(t), H] \neq 0$. If it does, then it would imply that measuring eigenvalues of $H$ with arbitrary precision in principle is impossible. There is always some noise coming from measurement itself, and the energy we are measuring corresponds to $H_m(t) + H$, not $H$ itself.

Let's assume otherwise and $H_m(t)=0$. At first, it may seem odd that, on one hand we can measure energy over and over and expect to get $E_k$ and indeed get $E_k$, so $\Delta E = 0$ but then we would have $\Delta E \Delta t= 0$. However, I see no issue, since t is not well defined, and in this context, the time between measurement is not necessarily $\Delta t$ anyway. Perhaps $\Delta t$ refers to inverse rate of changes of some operators that goes to infinity.

So either I am to believe that "measuring energy with arbitrary precision inherently impossible", which seems weird to say the least. Consider spin operator, as long as I measure spin on a fixed axis (say, z) over and over again, I always get the same value no problem, this is the case for all time independent operators, leaving $H$ out is a bit odd. For example, imagine a two state energy system as follows

$$|\psi\rangle = \frac{1}{\sqrt 2} (|e_0 \uparrow \rangle + |e_1 \downarrow \rangle)$$

Measuring spin gives full information about energy. So any limitation set on $H$ measurement should be also set on $S$ measurement. Or accept that $\Delta t = \infty$ somehow and I should not ask why. Or perhaps there is better explanation that I didn't think about, please let me know.

Edit:

Consider an ideal single-photon light emitter that uses a battery as it source. We can monitor the battery and measure its energy with arbitrary precision. Let say we measure battery's energy with 100% precision at arbitrarily small time intervals $\Delta t$. Now as soon as this emitter emits a single photon, we see a jump in battery's energy, indicating that how much energy was used for producing that photon. Since we know photon's energy, we know its exact momentum. On the other hand, we roughly know when the photon was fired too by simply looking at the time of jump in the energy of the battery. Since a time interval $\Delta t$ exists between each battery's energy measurement, the uncertainty in the location of photon is $\Delta x = c\Delta t$ but given $\Delta p = 0$, we have violated uncertainty principle. Again, one of these should be true

1. All measurement devices are noisy, we can never truly measure energy without noise.

2. Or $\Delta t$ DOES refer to the time interval between energy measurements, and it goes to infinity for measuring energy value exactly.

The problem is, neither of this conclusions comes from standard texts in QM. Measurement devices can collapse wave function and measure energy exactly. $\Delta t$ is not introduced as time between measurements either, it can be anything, really.

Some related topics:

Interpretation of the energy-time uncertainty

What is $\Delta t$ in the time-energy uncertainty principle?

Energy-time uncertainty and pair creation

To see where this question comes from, consider a time independent Hamiltonian $H$ and an initial wave function $\psi(t=0,x)$. We can express time dependant wave function $\psi(t,x) = \sum_j e^{-iE_jt/\hbar} \phi_j(x) c_j$ where $\{\phi_j(x)\}$ is the set of eigen functions of H and $c_j=\int \psi(t=0,y) \phi_j^*(y) dy$.

Let's say we strongly measure energy at some arbitrary time. This causes wave function to collapse to $\psi(t,x)=e^{-iE_kt/\hbar}\phi_k(x)$, and we get $E_k$ as the energy of the system, where k is a random integer. Since $[H(t), H(t')] = 0$, measuring energy over and over again at any time always result in $E_k$ and wave function stays untouched. This is of course only true by assuming that the very act of measurement is not associated with a time dependant Hamiltonian $H_m(t)$ such that $[H_m(t), H] \neq 0$. If it does, then it would imply that measuring eigenvalues of $H$ with arbitrary precision in principle is impossible. There is always some noise coming from measurement itself, and the energy we are measuring corresponds to $H_m(t) + H$, not $H$ itself.

Let's assume otherwise and $H_m(t)=0$. At first, it may seem odd that, on one hand we can measure energy over and over and expect to get $E_k$ and indeed get $E_k$, so $\Delta E = 0$ but then we would have $\Delta E \Delta t= 0$. However, I see no issue, since t is not well defined, and in this context, the time between measurement is not necessarily $\Delta t$ anyway. Perhaps $\Delta t$ refers to rate of changes of some operators that goes to infinity.

So either I am to believe that "measuring energy with arbitrary precision inherently impossible", which seems weird to say the least. Consider spin operator, as long as I measure spin on a fixed axis (say, z) over and over again, I always get the same value no problem, this is the case for all time independent operators, leaving $H$ out is a bit odd. Or accept that $\Delta t = \infty$ somehow and I should not ask why. Or perhaps there is better explanation that I didn't think about, please let me know.

Some related topics:

Interpretation of the energy-time uncertainty

What is $\Delta t$ in the time-energy uncertainty principle?

Energy-time uncertainty and pair creation

To see where this question comes from, consider a time independent Hamiltonian $H$ and an initial wave function $\psi(t=0,x)$. We can express time dependant wave function $\psi(t,x) = \sum_j e^{-iE_jt/\hbar} \phi_j(x) c_j$ where $\{\phi_j(x)\}$ is the set of eigen functions of H and $c_j=\int \psi(t=0,y) \phi_j^*(y) dy$.

Let's say we strongly measure energy at some arbitrary time. This causes wave function to collapse to $\psi(t,x)=e^{-iE_kt/\hbar}\phi_k(x)$, and we get $E_k$ as the energy of the system, where k is a random integer. Since $[H(t), H(t')] = 0$, measuring energy over and over again at any time always result in $E_k$ and wave function stays untouched. This is of course only true by assuming that the very act of measurement is not associated with a time dependant Hamiltonian $H_m(t)$ such that $[H_m(t), H] \neq 0$. If it does, then it would imply that measuring eigenvalues of $H$ with arbitrary precision in principle is impossible. There is always some noise coming from measurement itself, and the energy we are measuring corresponds to $H_m(t) + H$, not $H$ itself.

Let's assume otherwise and $H_m(t)=0$. At first, it may seem odd that, on one hand we can measure energy over and over and expect to get $E_k$ and indeed get $E_k$, so $\Delta E = 0$ but then we would have $\Delta E \Delta t= 0$. However, I see no issue, since t is not well defined, and in this context, the time between measurement is not necessarily $\Delta t$ anyway. Perhaps $\Delta t$ refers to rate of changes of some operators that goes to infinity.

So either I am to believe that "measuring energy with arbitrary precision inherently impossible", which seems weird to say the least. Consider spin operator, as long as I measure spin on a fixed axis (say, z) over and over again, I always get the same value no problem, this is the case for all time independent operators, leaving $H$ out is a bit odd. For example, imagine a two state energy system as follows

$$|\psi\rangle = \frac{1}{\sqrt 2} (|e_0 \uparrow \rangle + |e_1 \downarrow \rangle)$$

Measuring spin gives full information about energy. So any limitation set on $H$ measurement should be also set on $S$ measurement.

Or accept that $\Delta t = \infty$ somehow and I should not ask why. Or perhaps there is better explanation that I didn't think about, please let me know.

Some related topics:

Interpretation of the energy-time uncertainty

What is $\Delta t$ in the time-energy uncertainty principle?

Energy-time uncertainty and pair creation