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Uncertainty principle for pairs $x$ and $p$ or $E$ and $t$ data written for a physical particle has nothing to do with virtual pair production.

The "presence of higher states" in a given state has a limited and certain meaning and it is not due to fluctuations.

Let us consider an exact ground state $\psi_0$. It is often unknown as analytical formula. It is searched by the perturbation theory and is obtained in a spectral form like this one:

$$\Psi_0=e^{-iE_0 t/\hbar}\sum_{n\ge 0}C_{0n}\psi_n^{(0)}\quad (1)$$

This spectral decomposition is not a quantum mechanical superposition of states at all! All higher approximate states $\psi_n^{(0)},\: n>0$ are non observable in the exact state $\psi_0$; they are just dumb numbers to correct inexact value $\psi_0^{(0)}$ to get the exact one, the latter being still the ground state. No experiment can find an excited state, exact or approximate, in the ground state, even virtually (vacuum is a ground state). But in the formula (1) the approximate higher states are "present". This leads to an erroneous confusion that in the ground state one may "find" higher states for short period due to uncertainty relationship. No, they are not virtual states.

Note, the spectral expansions like (1) for other exact states ($n>0$) are involved in real calculations where exist observable exact states $\psi_n,\:n>0$ which bring their own $\psi_{n'}^{0}$ because of being expanded in the spectral series too. In that case, those approximate $\psi_n^{0},\:n>0$ may be called observable since $\psi_n\approx C_{nn}\psi_n^{(0)}$.

Again, in any particular state $n$ $$\psi_n=\sum_{n'} C_{nn'}\psi_n^{(0)}=C_{nn}\psi_n^{(0)}+\sum_{n'\ne n} C_{nn'}\psi_n^{(0)}\quad (2)$$ there are no other observable states $n'\ne n$, it should be clear. It is a state with a certain energy an nothing else can be found in it.

The only observable states in a general state $\Psi$ are those that are involved in the quantum mechanical superposition of exact states with their own energetic exponentials:

$$\Psi=\sum_n A_n\psi_n e^{-iE_n t/\hbar}\quad (3)$$

Often some higher observable states are just forbidden in this superposition by the energy conservation law valid, for example, in collisions. On the other hand, there is no limit on $n$ in the dumb spectral decompositions like (1) and (2). In perturbative caclulations the observable states mix with dumb ones. But if you analyse examples carefully, you will find that the "virtual states" are always the approximate functions $\psi_{n'}^{0}$ (corrections to $\psi_n^{0}$ injected from (2) to (3)) and never the exact states, in the regular perturbation theory or in the Feynman diagrams, whatever. This fact shows their true origin (1)-(2).

P.S. For those who did not get the point: there are no virtual states, as a matter of fact.

Uncertainty principle for pairs $x$ and $p$ or $E$ and $t$ data written for a physical particle has nothing to do with virtual pair production.

The "presence of higher states" in a given state has a limited and certain meaning and it is not due to fluctuations.

Let us consider an exact ground state $\psi_0$. It is often unknown as analytical formula. It is searched by the perturbation theory and is obtained in a spectral form like this one:

$$\Psi_0=e^{-iE_0 t/\hbar}\sum_{n\ge 0}C_{0n}\psi_n^{(0)}\quad (1)$$

This spectral decomposition is not a quantum mechanical superposition of states at all! All higher approximate states $\psi_n^{(0)},\: n>0$ are non observable in the exact state $\psi_0$; they are just dumb numbers to correct inexact value $\psi_0^{(0)}$ to get the exact one, the latter being still the ground state. No experiment can find an excited state, exact or approximate, in the ground state, even virtually (vacuum is a ground state). But in the formula (1) the approximate higher states are "present". This leads to an erroneous confusion that in the ground state one may "find" higher states for short period due to uncertainty relationship. No, they are not virtual states.

Note, the spectral expansions like (1) for other exact states ($n>0$) are involved in real calculations where exist observable exact states $\psi_n,\:n>0$ which bring their own $\psi_{n'}^{(0)}$ because of being expanded in the spectral series too. In that case, those approximate $\psi_n^{(0)},\:n>0$ may be called observable since $\psi_n\approx C_{nn}\psi_n^{(0)}$.

Again, in any particular state $n$ $$\psi_n=\sum_{n'} C_{nn'}\psi_n^{(0)}=C_{nn}\psi_n^{(0)}+\sum_{n'\ne n} C_{nn'}\psi_n^{(0)}\quad (2)$$ there are no other observable states $n'\ne n$, it should be clear. It is a state with a certain energy an nothing else can be found in it.

The only observable states in a general state $\Psi$ are those that are involved in the quantum mechanical superposition of exact states with their own energetic exponentials:

$$\Psi=\sum_n A_n\psi_n e^{-iE_n t/\hbar}\quad (3)$$

Often some higher observable states are just forbidden in this superposition by the energy conservation law, which valid, for example, in collisions. On the other hand, there is no limit on $n$ in the dumb spectral decompositions like (1) and (2). In perturbative calculations the observable states mix with dumb ones. But if you analyze examples carefully, you will find that the all "virtual states" are always the approximate functions $\psi_{n'}^{0}$ (corrections to $\psi_n^{0}$ injected from (2) to (3)) and never the exact states, in the regular perturbation theory or in the Feynman diagrams, whatever. This fact shows their true origin (1)-(2).

P.S. For those who did not get the point: there are no virtual states, as a matter of fact.

Uncertainty principle for pairs $x$ and $p$ or $E$ and $t$ data written for a physical particle has nothing to do with virtual pair production.

The "presence of higher states" in a given state has a limited and certain meaning and it is not due to fluctuations.

Let us consider an exact ground state $\psi_0$. It is often unknown as analytical formula. It is searched by the perturbation theory and is obtained in a spectral form like this one:

$$\Psi_0=e^{-iE_0 t/\hbar}\sum_{n\ge 0}C_{0n}\psi_n^{(0)}\quad (1)$$

This spectral decomposition is not a quantum mechanical superposition of states at all! All higher approximate states $\psi_n^{(0)},\: n>0$ are non observable in the exact state $\psi_0$; they are just dumb numbers to correct inexact value $\psi_0^{(0)}$ to get the exact one, the latter being still the ground state. No experiment can find an excited state, exact or approximate, in the ground state, even virtually (vacuum is a ground state). But in the formula (1) the approximate higher states are "present". This leads to an erroneous confusion that in the ground state one may "find" higher states for short period due to uncertainty relationship. No, they are not virtual states.

Note, the spectral expansions like (1) for other exact states are involved in real calculations where exist exact observable states $\psi_n$ which bring their own $\psi_n^{0}$ because of being expanded in the spectral series too.

Again, in any particular state $$\psi_n=\sum_{n'} C_{nn'}\psi_n^{(0)}\quad (2)$$ there are no higher observable states, it should be clear. It is a state with a certain energy an nothing else can be found in it.

The only observable states in a general state $\Psi$ are those that are involved in the quantum mechanical superposition of exact states with their own energetic exponentials:

$$\Psi=\sum_n A_n\psi_n e^{-iE_n t/\hbar}\quad (3)$$

Often some higher states are just forbidden in this superposition by the energy conservation law valid, for example, in collisions. On the other hand, there is no limit on $n$ in the dumb spectral decompositions like (1). If you analyse examples carefully, you will find that the "virtual states" are always the approximate functions $\psi_{n'}^{0}$ (corrections to $\psi_n^{0}$ injected from (2) to (3)) and never the exact states, in the regular perturbation theory or in the Feynman diagrams, whatever. This fact shows their true origin (1)-(2).

P.S. For those who did not get the point: there are no virtual states, as a matter of fact.

Uncertainty principle for pairs $x$ and $p$ or $E$ and $t$ data written for a physical particle has nothing to do with virtual pair production.

The "presence of higher states" in a given state has a limited and certain meaning and it is not due to fluctuations.

Let us consider an exact ground state $\psi_0$. It is often unknown as analytical formula. It is searched by the perturbation theory and is obtained in a spectral form like this one:

$$\Psi_0=e^{-iE_0 t/\hbar}\sum_{n\ge 0}C_{0n}\psi_n^{(0)}\quad (1)$$

This spectral decomposition is not a quantum mechanical superposition of states at all! All higher approximate states $\psi_n^{(0)},\: n>0$ are non observable in the exact state $\psi_0$; they are just dumb numbers to correct inexact value $\psi_0^{(0)}$ to get the exact one, the latter being still the ground state. No experiment can find an excited state, exact or approximate, in the ground state, even virtually (vacuum is a ground state). But in the formula (1) the approximate higher states are "present". This leads to an erroneous confusion that in the ground state one may "find" higher states for short period due to uncertainty relationship. No, they are not virtual states.

Note, the spectral expansions like (1) for other exact states ($n>0$) are involved in real calculations where exist observable exact states $\psi_n,\:n>0$ which bring their own $\psi_{n'}^{0}$ because of being expanded in the spectral series too. In that case, those approximate $\psi_n^{0},\:n>0$ may be called observable since $\psi_n\approx C_{nn}\psi_n^{(0)}$.

Again, in any particular state $n$ $$\psi_n=\sum_{n'} C_{nn'}\psi_n^{(0)}=C_{nn}\psi_n^{(0)}+\sum_{n'\ne n} C_{nn'}\psi_n^{(0)}\quad (2)$$ there are no other observable states $n'\ne n$, it should be clear. It is a state with a certain energy an nothing else can be found in it.

The only observable states in a general state $\Psi$ are those that are involved in the quantum mechanical superposition of exact states with their own energetic exponentials:

$$\Psi=\sum_n A_n\psi_n e^{-iE_n t/\hbar}\quad (3)$$

Often some higher observable states are just forbidden in this superposition by the energy conservation law valid, for example, in collisions. On the other hand, there is no limit on $n$ in the dumb spectral decompositions like (1) and (2). In perturbative caclulations the observable states mix with dumb ones. But if you analyse examples carefully, you will find that the "virtual states" are always the approximate functions $\psi_{n'}^{0}$ (corrections to $\psi_n^{0}$ injected from (2) to (3)) and never the exact states, in the regular perturbation theory or in the Feynman diagrams, whatever. This fact shows their true origin (1)-(2).

P.S. For those who did not get the point: there are no virtual states, as a matter of fact.

Uncertainty principle for pairs $x$ and $p$ or $E$ and $t$ data written for a physical particle has nothing to do with virtual pair production.

The "presence of higher states" in a given state has a limited and certain meaning and it is not due to fluctuations.

Let us consider an exact ground state $\psi_0$. It is often unknown as analytical formula. It is searched by the perturbation theory and is obtained in a spectral form like this one:

$$\Psi_0=e^{-iE_0 t/\hbar}\sum_{n\ge 0}C_{0n}\psi_n^{(0)}\quad (1)$$

This spectral decomposition is not a quantum mechanical superposition of states at all! All higher approximate states $\psi_n^{(0)},\: n>0$ are non observable in the exact state $\psi_0$; they are just dumb numbers to correct inexact value $\psi_0^{(0)}$ to get the exact one, the latter being still the ground state. No experiment can find an excited state, exact or approximate, in the ground state, even virtually (vacuum is a ground state). But in the formula (1) the approximate higher states are "present". This leads to an erroneous confusion that in the ground state one may "find" higher states for short period due to uncertainty relationship. No, they are not virtual states.

Note, the spectral expansions like (1) for other exact states are involved in real calculations where exist exact observable states $\psi_n$ which bring their own $\psi_n^{0}$ because of being expanded in the spectral series too.

Again, in any particular state $$\psi_n=\sum_{n'} C_{nn'}\psi_n^{(0)}\quad (2)$$ there are no higher observable states, it should be clear. It is a state with a certain energy an nothing else can be found in it.

The only observable states in a general state $\Psi$ are those that are involved in the quantum mechanical superposition of exact states with their own energetic exponentials:

$$\Psi=\sum_n A_n\psi_n e^{-iE_n t/\hbar}\quad (3)$$

Often some higher states are just forbidden in this superposition by the energy conservation law valid, for example, in collisions. On the other hand, there is no limit on $n$ in the dumb spectral decompositions like (1). If you analyse examples carefully, you will find that the "virtual states" are always the approximate functions $\psi_{n'}^{0}$ (corrections to $\psi_n^{0}$ injected from (2) to (3)) and never the exact states, in the regular perturbation theory or in the Feynman diagrams, whatever. This fact shows their true origin (1)-(2).

Uncertainty principle for pairs $x$ and $p$ or $E$ and $t$ data written for a physical particle has nothing to do with virtual pair production.

The "presence of higher states" in a given state has a limited and certain meaning and it is not due to fluctuations.

Let us consider an exact ground state $\psi_0$. It is often unknown as analytical formula. It is searched by the perturbation theory and is obtained in a spectral form like this one:

$$\Psi_0=e^{-iE_0 t/\hbar}\sum_{n\ge 0}C_{0n}\psi_n^{(0)}\quad (1)$$

This spectral decomposition is not a quantum mechanical superposition of states at all! All higher approximate states $\psi_n^{(0)},\: n>0$ are non observable in the exact state $\psi_0$; they are just dumb numbers to correct inexact value $\psi_0^{(0)}$ to get the exact one, the latter being still the ground state. No experiment can find an excited state, exact or approximate, in the ground state, even virtually (vacuum is a ground state). But in the formula (1) the approximate higher states are "present". This leads to an erroneous confusion that in the ground state one may "find" higher states for short period due to uncertainty relationship. No, they are not virtual states.

Note, the spectral expansions like (1) for other exact states are involved in real calculations where exist exact observable states $\psi_n$ which bring their own $\psi_n^{0}$ because of being expanded in the spectral series too.

Again, in any particular state $$\psi_n=\sum_{n'} C_{nn'}\psi_n^{(0)}\quad (2)$$ there are no higher observable states, it should be clear. It is a state with a certain energy an nothing else can be found in it.

The only observable states in a general state $\Psi$ are those that are involved in the quantum mechanical superposition of exact states with their own energetic exponentials:

$$\Psi=\sum_n A_n\psi_n e^{-iE_n t/\hbar}\quad (3)$$

Often some higher states are just forbidden in this superposition by the energy conservation law valid, for example, in collisions. On the other hand, there is no limit on $n$ in the dumb spectral decompositions like (1). If you analyse examples carefully, you will find that the "virtual states" are always the approximate functions $\psi_{n'}^{0}$ (corrections to $\psi_n^{0}$ injected from (2) to (3)) and never the exact states, in the regular perturbation theory or in the Feynman diagrams, whatever. This fact shows their true origin (1)-(2).

P.S. For those who did not get the point: there are no virtual states, as a matter of fact.