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Why can't two different quantum states evolve into a same final state?
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Is it true that two different states cannot evolve into the same final state? Can they achieve this state at different times? If yes what is the proof?
quantum-mechanics quantum-information
edited 4 hours ago
DanielSank
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Doubting Thomas
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Is it true that two different states cannot evolve into the same final state? Can they achieve this state at different times? If yes what is the proof?
quantum-mechanics quantum-information
edited 4 hours ago
DanielSank
16.4k44977
asked 4 hours ago
Doubting Thomas
111
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is it true that two different states cannot evolve into the same final state? Can they achieve this state at different times? If yes what is the proof?
quantum-mechanics quantum-information
edited 4 hours ago
DanielSank
16.4k44977
asked 4 hours ago
Doubting Thomas
111
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is it true that two different states cannot evolve into the same final state? Can they achieve this state at different times? If yes what is the proof?
quantum-mechanics quantum-information
quantum-mechanics quantum-information
edited 4 hours ago
DanielSank
16.4k44977
asked 4 hours ago
Doubting Thomas
111
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
DanielSank
16.4k44977
asked 4 hours ago
Doubting Thomas
111
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
DanielSank
16.4k44977
edited 4 hours ago
DanielSank
16.4k44977
edited 4 hours ago
DanielSank
16.4k44977
16.4k44977
asked 4 hours ago
Doubting Thomas
111
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Doubting Thomas
111
asked 4 hours ago
Doubting Thomas
111
111
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
2
active
oldest
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up vote
4
down vote
Is it true that two different states cannot evolve into the same final state?
That depends on exactly what you mean.
If we consider the total state of a closed system, then two different states will never simultaneously evolve into the same state at any later time.
You may have learned that quantum states evolve with a unitary transformation, i.e.
$$ lvert Psi(t) rangle = U(t) lvert Psi(0) rangle$$
where $U(t)$ is unitary, which means that $U(t)^dagger = U(t)^-1$.
That being the case,
beginalign
langle Phi(t)|Psi(t) rangle
&= langle U(t) Phi(0) | U(t) Psi(0) rangle \
text(definition of Heritian conjugate) quad &= langle Phi(0) | U(t)^dagger U(t) | Psi (0) rangle \
text(unitarity) quad &= langle Phi(0) | U(t)^-1 U(t) | Psi (0) rangle \
&= langle Phi(0) | Psi (0) rangle , .
endalign
So you can see, the inner product between two states does not change as time evolves.
Two states that are the same have inner product of 1, but states that are not the same have inner product not 1.
Therefore, two states that are not initially the same cannot become the same later under unitary evolution.
On the other hand, if we allow measurement, it is possible for two initially different states to wind up being the same.
For example, if we have a two level system starting in state $(lvert 0 rangle + e^i phi lvert 1 rangle)/sqrt 2$ for any value of $phi$, it could collapse to $lvert 0 rangle$ after a measurement.
Note, however, that in this case there is randomness, i.e. we cannot make a situation where two intitially different states deterministically evolve to the same final state.
If you could do that, I'm pretty sure you could control the future, communicate faster than light, and destroy the entire universe.
Can they achieve this state at different times?
Yeah, sure.
Consider a two level system with Hamiltonian
$$ H = hbar fracomega2 sigma_x , .$$
The propagator for this system is
$$U(t) = cos(omega t / 2) mathbbI - i sin(omega t / 2) sigma_x =
left(
beginarraycc
cos(omega t / 2) && -i sin(omega t / 2) \
-i sin(omega t / 2) && cos(omega t / 2)
endarray
right)$$
where $mathbbI$ means the identity.
If we start with state $lvert 0 rangle$, then the state at time $t$ is
$$
U(t) lvert 0 rangle = cos(omega t / 2) lvert 0 rangle - i sin(omega t / 2) lvert 1 rangle
$$
Similarly, if we had started with $i lvert 1 rangle$, we'd get
$$U(t) i lvert 1 rangle = sin(omega t / 2) lvert 0 rangle + i cos(omega t / 2) lvert 1 rangle , .
$$
Now look at two particular times:
$$U(t = pi / 2 omega) lvert 0 rangle
= cos(pi / 4) lvert 0 rangle - i sin(pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle)$$
and
$$U(t = 3 pi / 2 omega) i lvert 1 rangle
= sin(3pi/4)lvert 0 rangle + i cos(3 pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle) , .$$
So we can see that two initially different states evolve to the same state, but at different times.
answered 2 hours ago
DanielSank
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0
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$defket#1$
More simply. As to first question: let $keta,0$ be a state vector
taken at time 0, $keta,t$ the same state evolved at time $t$.
Note 1: Schrödinger picture is used, where states evolve in time, observables don't.
Note 2: I'm using a rather different ket notation. I don't write things like $ketPsi(t)$ because I think this is a misinterpretation of Dirac's notation.
We have
$$keta,t = U(t),keta,0$$
where $U(t)$ is unitary. Assume now that another $ketb,0$ exists, such that also
$$keta,t = U(t),ketb,0.$$
Then
$$ketb,0 = U^-1(t),keta,t = U^-1(t),U(t),keta,0 = keta,0.$$
Now for the second question. You're asking if
$$keta,t = ketb,t' qquad(1)$$
could happen, for $t'ne t$. Let's expand eq. (1), using $U$:
$$U(t),keta,0 = U(t'),ketb,0$$
$$ketb,0 = U^-1(t'),U(t),keta,0 = U(-t'),U(t),keta,0 =
U(t-t'),keta,0.$$
This is the condition $keta,0$ and $ketb,0$ are to obey, in order that $keta,t$ and $ketb,t'$ be equal.
answered 33 mins ago
Elio Fabri
1412
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Is it true that two different states cannot evolve into the same final state?
That depends on exactly what you mean.
If we consider the total state of a closed system, then two different states will never simultaneously evolve into the same state at any later time.
You may have learned that quantum states evolve with a unitary transformation, i.e.
$$ lvert Psi(t) rangle = U(t) lvert Psi(0) rangle$$
where $U(t)$ is unitary, which means that $U(t)^dagger = U(t)^-1$.
That being the case,
beginalign
langle Phi(t)|Psi(t) rangle
&= langle U(t) Phi(0) | U(t) Psi(0) rangle \
text(definition of Heritian conjugate) quad &= langle Phi(0) | U(t)^dagger U(t) | Psi (0) rangle \
text(unitarity) quad &= langle Phi(0) | U(t)^-1 U(t) | Psi (0) rangle \
&= langle Phi(0) | Psi (0) rangle , .
endalign
So you can see, the inner product between two states does not change as time evolves.
Two states that are the same have inner product of 1, but states that are not the same have inner product not 1.
Therefore, two states that are not initially the same cannot become the same later under unitary evolution.
On the other hand, if we allow measurement, it is possible for two initially different states to wind up being the same.
For example, if we have a two level system starting in state $(lvert 0 rangle + e^i phi lvert 1 rangle)/sqrt 2$ for any value of $phi$, it could collapse to $lvert 0 rangle$ after a measurement.
Note, however, that in this case there is randomness, i.e. we cannot make a situation where two intitially different states deterministically evolve to the same final state.
If you could do that, I'm pretty sure you could control the future, communicate faster than light, and destroy the entire universe.
Can they achieve this state at different times?
Yeah, sure.
Consider a two level system with Hamiltonian
$$ H = hbar fracomega2 sigma_x , .$$
The propagator for this system is
$$U(t) = cos(omega t / 2) mathbbI - i sin(omega t / 2) sigma_x =
left(
beginarraycc
cos(omega t / 2) && -i sin(omega t / 2) \
-i sin(omega t / 2) && cos(omega t / 2)
endarray
right)$$
where $mathbbI$ means the identity.
If we start with state $lvert 0 rangle$, then the state at time $t$ is
$$
U(t) lvert 0 rangle = cos(omega t / 2) lvert 0 rangle - i sin(omega t / 2) lvert 1 rangle
$$
Similarly, if we had started with $i lvert 1 rangle$, we'd get
$$U(t) i lvert 1 rangle = sin(omega t / 2) lvert 0 rangle + i cos(omega t / 2) lvert 1 rangle , .
$$
Now look at two particular times:
$$U(t = pi / 2 omega) lvert 0 rangle
= cos(pi / 4) lvert 0 rangle - i sin(pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle)$$
and
$$U(t = 3 pi / 2 omega) i lvert 1 rangle
= sin(3pi/4)lvert 0 rangle + i cos(3 pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle) , .$$
So we can see that two initially different states evolve to the same state, but at different times.
answered 2 hours ago
DanielSank
16.4k44977
add a comment |Â
up vote
4
down vote
Is it true that two different states cannot evolve into the same final state?
That depends on exactly what you mean.
If we consider the total state of a closed system, then two different states will never simultaneously evolve into the same state at any later time.
You may have learned that quantum states evolve with a unitary transformation, i.e.
$$ lvert Psi(t) rangle = U(t) lvert Psi(0) rangle$$
where $U(t)$ is unitary, which means that $U(t)^dagger = U(t)^-1$.
That being the case,
beginalign
langle Phi(t)|Psi(t) rangle
&= langle U(t) Phi(0) | U(t) Psi(0) rangle \
text(definition of Heritian conjugate) quad &= langle Phi(0) | U(t)^dagger U(t) | Psi (0) rangle \
text(unitarity) quad &= langle Phi(0) | U(t)^-1 U(t) | Psi (0) rangle \
&= langle Phi(0) | Psi (0) rangle , .
endalign
So you can see, the inner product between two states does not change as time evolves.
Two states that are the same have inner product of 1, but states that are not the same have inner product not 1.
Therefore, two states that are not initially the same cannot become the same later under unitary evolution.
On the other hand, if we allow measurement, it is possible for two initially different states to wind up being the same.
For example, if we have a two level system starting in state $(lvert 0 rangle + e^i phi lvert 1 rangle)/sqrt 2$ for any value of $phi$, it could collapse to $lvert 0 rangle$ after a measurement.
Note, however, that in this case there is randomness, i.e. we cannot make a situation where two intitially different states deterministically evolve to the same final state.
If you could do that, I'm pretty sure you could control the future, communicate faster than light, and destroy the entire universe.
Can they achieve this state at different times?
Yeah, sure.
Consider a two level system with Hamiltonian
$$ H = hbar fracomega2 sigma_x , .$$
The propagator for this system is
$$U(t) = cos(omega t / 2) mathbbI - i sin(omega t / 2) sigma_x =
left(
beginarraycc
cos(omega t / 2) && -i sin(omega t / 2) \
-i sin(omega t / 2) && cos(omega t / 2)
endarray
right)$$
where $mathbbI$ means the identity.
If we start with state $lvert 0 rangle$, then the state at time $t$ is
$$
U(t) lvert 0 rangle = cos(omega t / 2) lvert 0 rangle - i sin(omega t / 2) lvert 1 rangle
$$
Similarly, if we had started with $i lvert 1 rangle$, we'd get
$$U(t) i lvert 1 rangle = sin(omega t / 2) lvert 0 rangle + i cos(omega t / 2) lvert 1 rangle , .
$$
Now look at two particular times:
$$U(t = pi / 2 omega) lvert 0 rangle
= cos(pi / 4) lvert 0 rangle - i sin(pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle)$$
and
$$U(t = 3 pi / 2 omega) i lvert 1 rangle
= sin(3pi/4)lvert 0 rangle + i cos(3 pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle) , .$$
So we can see that two initially different states evolve to the same state, but at different times.
answered 2 hours ago
DanielSank
16.4k44977
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Is it true that two different states cannot evolve into the same final state?
That depends on exactly what you mean.
If we consider the total state of a closed system, then two different states will never simultaneously evolve into the same state at any later time.
You may have learned that quantum states evolve with a unitary transformation, i.e.
$$ lvert Psi(t) rangle = U(t) lvert Psi(0) rangle$$
where $U(t)$ is unitary, which means that $U(t)^dagger = U(t)^-1$.
That being the case,
beginalign
langle Phi(t)|Psi(t) rangle
&= langle U(t) Phi(0) | U(t) Psi(0) rangle \
text(definition of Heritian conjugate) quad &= langle Phi(0) | U(t)^dagger U(t) | Psi (0) rangle \
text(unitarity) quad &= langle Phi(0) | U(t)^-1 U(t) | Psi (0) rangle \
&= langle Phi(0) | Psi (0) rangle , .
endalign
So you can see, the inner product between two states does not change as time evolves.
Two states that are the same have inner product of 1, but states that are not the same have inner product not 1.
Therefore, two states that are not initially the same cannot become the same later under unitary evolution.
On the other hand, if we allow measurement, it is possible for two initially different states to wind up being the same.
For example, if we have a two level system starting in state $(lvert 0 rangle + e^i phi lvert 1 rangle)/sqrt 2$ for any value of $phi$, it could collapse to $lvert 0 rangle$ after a measurement.
Note, however, that in this case there is randomness, i.e. we cannot make a situation where two intitially different states deterministically evolve to the same final state.
If you could do that, I'm pretty sure you could control the future, communicate faster than light, and destroy the entire universe.
Can they achieve this state at different times?
Yeah, sure.
Consider a two level system with Hamiltonian
$$ H = hbar fracomega2 sigma_x , .$$
The propagator for this system is
$$U(t) = cos(omega t / 2) mathbbI - i sin(omega t / 2) sigma_x =
left(
beginarraycc
cos(omega t / 2) && -i sin(omega t / 2) \
-i sin(omega t / 2) && cos(omega t / 2)
endarray
right)$$
where $mathbbI$ means the identity.
If we start with state $lvert 0 rangle$, then the state at time $t$ is
$$
U(t) lvert 0 rangle = cos(omega t / 2) lvert 0 rangle - i sin(omega t / 2) lvert 1 rangle
$$
Similarly, if we had started with $i lvert 1 rangle$, we'd get
$$U(t) i lvert 1 rangle = sin(omega t / 2) lvert 0 rangle + i cos(omega t / 2) lvert 1 rangle , .
$$
Now look at two particular times:
$$U(t = pi / 2 omega) lvert 0 rangle
= cos(pi / 4) lvert 0 rangle - i sin(pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle)$$
and
$$U(t = 3 pi / 2 omega) i lvert 1 rangle
= sin(3pi/4)lvert 0 rangle + i cos(3 pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle) , .$$
So we can see that two initially different states evolve to the same state, but at different times.
answered 2 hours ago
DanielSank
16.4k44977
Is it true that two different states cannot evolve into the same final state?
That depends on exactly what you mean.
If we consider the total state of a closed system, then two different states will never simultaneously evolve into the same state at any later time.
You may have learned that quantum states evolve with a unitary transformation, i.e.
$$ lvert Psi(t) rangle = U(t) lvert Psi(0) rangle$$
where $U(t)$ is unitary, which means that $U(t)^dagger = U(t)^-1$.
That being the case,
beginalign
langle Phi(t)|Psi(t) rangle
&= langle U(t) Phi(0) | U(t) Psi(0) rangle \
text(definition of Heritian conjugate) quad &= langle Phi(0) | U(t)^dagger U(t) | Psi (0) rangle \
text(unitarity) quad &= langle Phi(0) | U(t)^-1 U(t) | Psi (0) rangle \
&= langle Phi(0) | Psi (0) rangle , .
endalign
So you can see, the inner product between two states does not change as time evolves.
Two states that are the same have inner product of 1, but states that are not the same have inner product not 1.
Therefore, two states that are not initially the same cannot become the same later under unitary evolution.
On the other hand, if we allow measurement, it is possible for two initially different states to wind up being the same.
For example, if we have a two level system starting in state $(lvert 0 rangle + e^i phi lvert 1 rangle)/sqrt 2$ for any value of $phi$, it could collapse to $lvert 0 rangle$ after a measurement.
Note, however, that in this case there is randomness, i.e. we cannot make a situation where two intitially different states deterministically evolve to the same final state.
If you could do that, I'm pretty sure you could control the future, communicate faster than light, and destroy the entire universe.
Can they achieve this state at different times?
Yeah, sure.
Consider a two level system with Hamiltonian
$$ H = hbar fracomega2 sigma_x , .$$
The propagator for this system is
$$U(t) = cos(omega t / 2) mathbbI - i sin(omega t / 2) sigma_x =
left(
beginarraycc
cos(omega t / 2) && -i sin(omega t / 2) \
-i sin(omega t / 2) && cos(omega t / 2)
endarray
right)$$
where $mathbbI$ means the identity.
If we start with state $lvert 0 rangle$, then the state at time $t$ is
$$
U(t) lvert 0 rangle = cos(omega t / 2) lvert 0 rangle - i sin(omega t / 2) lvert 1 rangle
$$
Similarly, if we had started with $i lvert 1 rangle$, we'd get
$$U(t) i lvert 1 rangle = sin(omega t / 2) lvert 0 rangle + i cos(omega t / 2) lvert 1 rangle , .
$$
Now look at two particular times:
$$U(t = pi / 2 omega) lvert 0 rangle
= cos(pi / 4) lvert 0 rangle - i sin(pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle)$$
and
$$U(t = 3 pi / 2 omega) i lvert 1 rangle
= sin(3pi/4)lvert 0 rangle + i cos(3 pi / 4) lvert 1 rangle
= frac1sqrt 2(lvert 0 rangle - i lvert 1 rangle) , .$$
So we can see that two initially different states evolve to the same state, but at different times.
answered 2 hours ago
DanielSank
16.4k44977
answered 2 hours ago
DanielSank
16.4k44977
answered 2 hours ago
DanielSank
16.4k44977
answered 2 hours ago
DanielSank
16.4k44977
16.4k44977
add a comment |Â
add a comment |Â
up vote
0
down vote
$defket#1$
More simply. As to first question: let $keta,0$ be a state vector
taken at time 0, $keta,t$ the same state evolved at time $t$.
Note 1: Schrödinger picture is used, where states evolve in time, observables don't.
Note 2: I'm using a rather different ket notation. I don't write things like $ketPsi(t)$ because I think this is a misinterpretation of Dirac's notation.
We have
$$keta,t = U(t),keta,0$$
where $U(t)$ is unitary. Assume now that another $ketb,0$ exists, such that also
$$keta,t = U(t),ketb,0.$$
Then
$$ketb,0 = U^-1(t),keta,t = U^-1(t),U(t),keta,0 = keta,0.$$
Now for the second question. You're asking if
$$keta,t = ketb,t' qquad(1)$$
could happen, for $t'ne t$. Let's expand eq. (1), using $U$:
$$U(t),keta,0 = U(t'),ketb,0$$
$$ketb,0 = U^-1(t'),U(t),keta,0 = U(-t'),U(t),keta,0 =
U(t-t'),keta,0.$$
This is the condition $keta,0$ and $ketb,0$ are to obey, in order that $keta,t$ and $ketb,t'$ be equal.
answered 33 mins ago
Elio Fabri
1412
add a comment |Â
up vote
0
down vote
$defket#1$
More simply. As to first question: let $keta,0$ be a state vector
taken at time 0, $keta,t$ the same state evolved at time $t$.
Note 1: Schrödinger picture is used, where states evolve in time, observables don't.
Note 2: I'm using a rather different ket notation. I don't write things like $ketPsi(t)$ because I think this is a misinterpretation of Dirac's notation.
We have
$$keta,t = U(t),keta,0$$
where $U(t)$ is unitary. Assume now that another $ketb,0$ exists, such that also
$$keta,t = U(t),ketb,0.$$
Then
$$ketb,0 = U^-1(t),keta,t = U^-1(t),U(t),keta,0 = keta,0.$$
Now for the second question. You're asking if
$$keta,t = ketb,t' qquad(1)$$
could happen, for $t'ne t$. Let's expand eq. (1), using $U$:
$$U(t),keta,0 = U(t'),ketb,0$$
$$ketb,0 = U^-1(t'),U(t),keta,0 = U(-t'),U(t),keta,0 =
U(t-t'),keta,0.$$
This is the condition $keta,0$ and $ketb,0$ are to obey, in order that $keta,t$ and $ketb,t'$ be equal.
answered 33 mins ago
Elio Fabri
1412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$defket#1$
More simply. As to first question: let $keta,0$ be a state vector
taken at time 0, $keta,t$ the same state evolved at time $t$.
Note 1: Schrödinger picture is used, where states evolve in time, observables don't.
Note 2: I'm using a rather different ket notation. I don't write things like $ketPsi(t)$ because I think this is a misinterpretation of Dirac's notation.
We have
$$keta,t = U(t),keta,0$$
where $U(t)$ is unitary. Assume now that another $ketb,0$ exists, such that also
$$keta,t = U(t),ketb,0.$$
Then
$$ketb,0 = U^-1(t),keta,t = U^-1(t),U(t),keta,0 = keta,0.$$
Now for the second question. You're asking if
$$keta,t = ketb,t' qquad(1)$$
could happen, for $t'ne t$. Let's expand eq. (1), using $U$:
$$U(t),keta,0 = U(t'),ketb,0$$
$$ketb,0 = U^-1(t'),U(t),keta,0 = U(-t'),U(t),keta,0 =
U(t-t'),keta,0.$$
This is the condition $keta,0$ and $ketb,0$ are to obey, in order that $keta,t$ and $ketb,t'$ be equal.
answered 33 mins ago
Elio Fabri
1412
$defket#1$
More simply. As to first question: let $keta,0$ be a state vector
taken at time 0, $keta,t$ the same state evolved at time $t$.
Note 1: Schrödinger picture is used, where states evolve in time, observables don't.
Note 2: I'm using a rather different ket notation. I don't write things like $ketPsi(t)$ because I think this is a misinterpretation of Dirac's notation.
We have
$$keta,t = U(t),keta,0$$
where $U(t)$ is unitary. Assume now that another $ketb,0$ exists, such that also
$$keta,t = U(t),ketb,0.$$
Then
$$ketb,0 = U^-1(t),keta,t = U^-1(t),U(t),keta,0 = keta,0.$$
Now for the second question. You're asking if
$$keta,t = ketb,t' qquad(1)$$
could happen, for $t'ne t$. Let's expand eq. (1), using $U$:
$$U(t),keta,0 = U(t'),ketb,0$$
$$ketb,0 = U^-1(t'),U(t),keta,0 = U(-t'),U(t),keta,0 =
U(t-t'),keta,0.$$
This is the condition $keta,0$ and $ketb,0$ are to obey, in order that $keta,t$ and $ketb,t'$ be equal.
answered 33 mins ago
Elio Fabri
1412
answered 33 mins ago
Elio Fabri
1412
answered 33 mins ago
Elio Fabri
1412
answered 33 mins ago
Elio Fabri
1412
1412
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