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Why can't two different quantum states evolve into the 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 13 mins ago
knzhou
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asked 18 hours ago
Doubting Thomas
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up vote
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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 13 mins ago
knzhou
35.7k8100173
asked 18 hours ago
Doubting Thomas
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up vote
12
down vote
favorite
up vote
12
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 13 mins ago
knzhou
35.7k8100173
asked 18 hours ago
Doubting Thomas
613
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
quantum-mechanics quantum-information
edited 13 mins ago
knzhou
35.7k8100173
asked 18 hours ago
Doubting Thomas
613
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 13 mins ago
knzhou
35.7k8100173
asked 18 hours ago
Doubting Thomas
613
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edited 13 mins ago
knzhou
35.7k8100173
edited 13 mins ago
knzhou
35.7k8100173
edited 13 mins ago
knzhou
35.7k8100173
35.7k8100173
asked 18 hours ago
Doubting Thomas
613
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 18 hours ago
Doubting Thomas
613
asked 18 hours ago
Doubting Thomas
613
613
New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Doubting Thomas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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4 Answers
4
active
oldest
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up vote
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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 16 hours ago
DanielSank
16.5k44978
I think there is a simpler way to argue that two different states cannot simultaneously evolve into the same state: Time evolution by some time $t$ is given by a unitary operator $U(t)$, unitary implies invertible, implies injective. Hence $U(t)Psi = U(t)Phi$ implies $Psi = Phi$.
– Peter
3 hours ago
@Peter Yep, that's a good argument. Why not edit it in?
– DanielSank
3 hours ago
add a comment |Â
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2
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Excluding wave function collapse, wave functions evolve deterministically, and this determinism goes both ways in time. So if you take $Psi$ and $Phi$ such that there is some $t_0$ for which $Psi(t_0)=Phi(t_0)$, then as long as they evolve under the same transformation, you have $Psi(t)=Phi(t)$ for all $t$. Thus, you can have neither two wavefunctions that are the same at one time evolve into different wavefunctions at a different time, nor two wavefunctions that are different at one time evolve into the same wavefunction at another time.
A state can evolve into what another state was at another time, i.e. $Psi(t_1)=Phi(t_2)$. But if they evolve under a time-constant transformation, then if we define $Delta t= t_2-t_1$, then $Psi(t)=Phi(t+Delta t)$ for all $t$; the two states are simply time-shifted version of each other.
answered 8 hours ago
Acccumulation
1,41719
The first 4 words could be removed; "wave function collapse" is not part of QM, just some magical thinking to smear away the difference between how QM describes nature and how many humans expect it to be.
– R..
3 hours ago
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up vote
1
down vote
Let $Psi$ and $Phi$ be two states that evolve into the same state after some time $t$. Time evolution after time $t$ is given by a unitary operator $U(t)$. In particular, this means that $U(t)$ is invertible, so we have $U(t)^-1U(t) = 1$. Now we have by assumption that $U(t)Psi = U(t)Phi$. Multiplying both sides of this equation by $U(t)^-1$ from the left we obtain $Psi = Phi$. So if two states evolve into the same state after some time $t$, they were the same to begin with.
Time evolution furthermore has the property that $U(t)U(s) = U(t+s)$. Let $tneq 0$. Now suppose that we have some state $Phi$ with the property that $U(s)Phi neq Phi$. Set $Psi = U(s)Phi$. We then get
beginequation
U(t)Psi = U(t)(U(s)Phi) = U(t+s)Phi.
endequation
We then have that the states $Psi$ and $Phi$, (which are different), evolve into the same state $U(t+s)Phi$, but after different times.
In fact, this is the only way this can happen. That is, if there are two different states $Phi neq Psi$ such that $U(t)Phi = U(t')Psi$, then there exists a time $s$ such that $Psi = U(s) Phi$. Maybe finding a proof for this claim is a good exercise.
answered 3 hours ago
Peter
1113
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up vote
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$defket#1#1rangle$
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 14 hours ago
Elio Fabri
1412
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
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 16 hours ago
DanielSank
16.5k44978
I think there is a simpler way to argue that two different states cannot simultaneously evolve into the same state: Time evolution by some time $t$ is given by a unitary operator $U(t)$, unitary implies invertible, implies injective. Hence $U(t)Psi = U(t)Phi$ implies $Psi = Phi$.
– Peter
3 hours ago
@Peter Yep, that's a good argument. Why not edit it in?
– DanielSank
3 hours ago
add a comment |Â
up vote
15
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 16 hours ago
DanielSank
16.5k44978
I think there is a simpler way to argue that two different states cannot simultaneously evolve into the same state: Time evolution by some time $t$ is given by a unitary operator $U(t)$, unitary implies invertible, implies injective. Hence $U(t)Psi = U(t)Phi$ implies $Psi = Phi$.
– Peter
3 hours ago
@Peter Yep, that's a good argument. Why not edit it in?
– DanielSank
3 hours ago
add a comment |Â
up vote
15
down vote
up vote
15
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 16 hours ago
DanielSank
16.5k44978
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 16 hours ago
DanielSank
16.5k44978
answered 16 hours ago
DanielSank
16.5k44978
answered 16 hours ago
DanielSank
16.5k44978
answered 16 hours ago
DanielSank
16.5k44978
16.5k44978
I think there is a simpler way to argue that two different states cannot simultaneously evolve into the same state: Time evolution by some time $t$ is given by a unitary operator $U(t)$, unitary implies invertible, implies injective. Hence $U(t)Psi = U(t)Phi$ implies $Psi = Phi$.
– Peter
3 hours ago
@Peter Yep, that's a good argument. Why not edit it in?
– DanielSank
3 hours ago
add a comment |Â
I think there is a simpler way to argue that two different states cannot simultaneously evolve into the same state: Time evolution by some time $t$ is given by a unitary operator $U(t)$, unitary implies invertible, implies injective. Hence $U(t)Psi = U(t)Phi$ implies $Psi = Phi$.
– Peter
3 hours ago
@Peter Yep, that's a good argument. Why not edit it in?
– DanielSank
3 hours ago
I think there is a simpler way to argue that two different states cannot simultaneously evolve into the same state: Time evolution by some time $t$ is given by a unitary operator $U(t)$, unitary implies invertible, implies injective. Hence $U(t)Psi = U(t)Phi$ implies $Psi = Phi$.
– Peter
3 hours ago
I think there is a simpler way to argue that two different states cannot simultaneously evolve into the same state: Time evolution by some time $t$ is given by a unitary operator $U(t)$, unitary implies invertible, implies injective. Hence $U(t)Psi = U(t)Phi$ implies $Psi = Phi$.
– Peter
3 hours ago
@Peter Yep, that's a good argument. Why not edit it in?
– DanielSank
3 hours ago
@Peter Yep, that's a good argument. Why not edit it in?
– DanielSank
3 hours ago
add a comment |Â
up vote
2
down vote
Excluding wave function collapse, wave functions evolve deterministically, and this determinism goes both ways in time. So if you take $Psi$ and $Phi$ such that there is some $t_0$ for which $Psi(t_0)=Phi(t_0)$, then as long as they evolve under the same transformation, you have $Psi(t)=Phi(t)$ for all $t$. Thus, you can have neither two wavefunctions that are the same at one time evolve into different wavefunctions at a different time, nor two wavefunctions that are different at one time evolve into the same wavefunction at another time.
A state can evolve into what another state was at another time, i.e. $Psi(t_1)=Phi(t_2)$. But if they evolve under a time-constant transformation, then if we define $Delta t= t_2-t_1$, then $Psi(t)=Phi(t+Delta t)$ for all $t$; the two states are simply time-shifted version of each other.
answered 8 hours ago
Acccumulation
1,41719
The first 4 words could be removed; "wave function collapse" is not part of QM, just some magical thinking to smear away the difference between how QM describes nature and how many humans expect it to be.
– R..
3 hours ago
add a comment |Â
up vote
2
down vote
Excluding wave function collapse, wave functions evolve deterministically, and this determinism goes both ways in time. So if you take $Psi$ and $Phi$ such that there is some $t_0$ for which $Psi(t_0)=Phi(t_0)$, then as long as they evolve under the same transformation, you have $Psi(t)=Phi(t)$ for all $t$. Thus, you can have neither two wavefunctions that are the same at one time evolve into different wavefunctions at a different time, nor two wavefunctions that are different at one time evolve into the same wavefunction at another time.
A state can evolve into what another state was at another time, i.e. $Psi(t_1)=Phi(t_2)$. But if they evolve under a time-constant transformation, then if we define $Delta t= t_2-t_1$, then $Psi(t)=Phi(t+Delta t)$ for all $t$; the two states are simply time-shifted version of each other.
answered 8 hours ago
Acccumulation
1,41719
The first 4 words could be removed; "wave function collapse" is not part of QM, just some magical thinking to smear away the difference between how QM describes nature and how many humans expect it to be.
– R..
3 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Excluding wave function collapse, wave functions evolve deterministically, and this determinism goes both ways in time. So if you take $Psi$ and $Phi$ such that there is some $t_0$ for which $Psi(t_0)=Phi(t_0)$, then as long as they evolve under the same transformation, you have $Psi(t)=Phi(t)$ for all $t$. Thus, you can have neither two wavefunctions that are the same at one time evolve into different wavefunctions at a different time, nor two wavefunctions that are different at one time evolve into the same wavefunction at another time.
A state can evolve into what another state was at another time, i.e. $Psi(t_1)=Phi(t_2)$. But if they evolve under a time-constant transformation, then if we define $Delta t= t_2-t_1$, then $Psi(t)=Phi(t+Delta t)$ for all $t$; the two states are simply time-shifted version of each other.
answered 8 hours ago
Acccumulation
1,41719
Excluding wave function collapse, wave functions evolve deterministically, and this determinism goes both ways in time. So if you take $Psi$ and $Phi$ such that there is some $t_0$ for which $Psi(t_0)=Phi(t_0)$, then as long as they evolve under the same transformation, you have $Psi(t)=Phi(t)$ for all $t$. Thus, you can have neither two wavefunctions that are the same at one time evolve into different wavefunctions at a different time, nor two wavefunctions that are different at one time evolve into the same wavefunction at another time.
A state can evolve into what another state was at another time, i.e. $Psi(t_1)=Phi(t_2)$. But if they evolve under a time-constant transformation, then if we define $Delta t= t_2-t_1$, then $Psi(t)=Phi(t+Delta t)$ for all $t$; the two states are simply time-shifted version of each other.
answered 8 hours ago
Acccumulation
1,41719
answered 8 hours ago
Acccumulation
1,41719
answered 8 hours ago
Acccumulation
1,41719
answered 8 hours ago
Acccumulation
1,41719
1,41719
The first 4 words could be removed; "wave function collapse" is not part of QM, just some magical thinking to smear away the difference between how QM describes nature and how many humans expect it to be.
– R..
3 hours ago
add a comment |Â
The first 4 words could be removed; "wave function collapse" is not part of QM, just some magical thinking to smear away the difference between how QM describes nature and how many humans expect it to be.
– R..
3 hours ago
The first 4 words could be removed; "wave function collapse" is not part of QM, just some magical thinking to smear away the difference between how QM describes nature and how many humans expect it to be.
– R..
3 hours ago
The first 4 words could be removed; "wave function collapse" is not part of QM, just some magical thinking to smear away the difference between how QM describes nature and how many humans expect it to be.
– R..
3 hours ago
add a comment |Â
up vote
1
down vote
Let $Psi$ and $Phi$ be two states that evolve into the same state after some time $t$. Time evolution after time $t$ is given by a unitary operator $U(t)$. In particular, this means that $U(t)$ is invertible, so we have $U(t)^-1U(t) = 1$. Now we have by assumption that $U(t)Psi = U(t)Phi$. Multiplying both sides of this equation by $U(t)^-1$ from the left we obtain $Psi = Phi$. So if two states evolve into the same state after some time $t$, they were the same to begin with.
Time evolution furthermore has the property that $U(t)U(s) = U(t+s)$. Let $tneq 0$. Now suppose that we have some state $Phi$ with the property that $U(s)Phi neq Phi$. Set $Psi = U(s)Phi$. We then get
beginequation
U(t)Psi = U(t)(U(s)Phi) = U(t+s)Phi.
endequation
We then have that the states $Psi$ and $Phi$, (which are different), evolve into the same state $U(t+s)Phi$, but after different times.
In fact, this is the only way this can happen. That is, if there are two different states $Phi neq Psi$ such that $U(t)Phi = U(t')Psi$, then there exists a time $s$ such that $Psi = U(s) Phi$. Maybe finding a proof for this claim is a good exercise.
answered 3 hours ago
Peter
1113
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add a comment |Â
up vote
1
down vote
Let $Psi$ and $Phi$ be two states that evolve into the same state after some time $t$. Time evolution after time $t$ is given by a unitary operator $U(t)$. In particular, this means that $U(t)$ is invertible, so we have $U(t)^-1U(t) = 1$. Now we have by assumption that $U(t)Psi = U(t)Phi$. Multiplying both sides of this equation by $U(t)^-1$ from the left we obtain $Psi = Phi$. So if two states evolve into the same state after some time $t$, they were the same to begin with.
Time evolution furthermore has the property that $U(t)U(s) = U(t+s)$. Let $tneq 0$. Now suppose that we have some state $Phi$ with the property that $U(s)Phi neq Phi$. Set $Psi = U(s)Phi$. We then get
beginequation
U(t)Psi = U(t)(U(s)Phi) = U(t+s)Phi.
endequation
We then have that the states $Psi$ and $Phi$, (which are different), evolve into the same state $U(t+s)Phi$, but after different times.
In fact, this is the only way this can happen. That is, if there are two different states $Phi neq Psi$ such that $U(t)Phi = U(t')Psi$, then there exists a time $s$ such that $Psi = U(s) Phi$. Maybe finding a proof for this claim is a good exercise.
answered 3 hours ago
Peter
1113
New contributor
Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $Psi$ and $Phi$ be two states that evolve into the same state after some time $t$. Time evolution after time $t$ is given by a unitary operator $U(t)$. In particular, this means that $U(t)$ is invertible, so we have $U(t)^-1U(t) = 1$. Now we have by assumption that $U(t)Psi = U(t)Phi$. Multiplying both sides of this equation by $U(t)^-1$ from the left we obtain $Psi = Phi$. So if two states evolve into the same state after some time $t$, they were the same to begin with.
Time evolution furthermore has the property that $U(t)U(s) = U(t+s)$. Let $tneq 0$. Now suppose that we have some state $Phi$ with the property that $U(s)Phi neq Phi$. Set $Psi = U(s)Phi$. We then get
beginequation
U(t)Psi = U(t)(U(s)Phi) = U(t+s)Phi.
endequation
We then have that the states $Psi$ and $Phi$, (which are different), evolve into the same state $U(t+s)Phi$, but after different times.
In fact, this is the only way this can happen. That is, if there are two different states $Phi neq Psi$ such that $U(t)Phi = U(t')Psi$, then there exists a time $s$ such that $Psi = U(s) Phi$. Maybe finding a proof for this claim is a good exercise.
answered 3 hours ago
Peter
1113
New contributor
Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $Psi$ and $Phi$ be two states that evolve into the same state after some time $t$. Time evolution after time $t$ is given by a unitary operator $U(t)$. In particular, this means that $U(t)$ is invertible, so we have $U(t)^-1U(t) = 1$. Now we have by assumption that $U(t)Psi = U(t)Phi$. Multiplying both sides of this equation by $U(t)^-1$ from the left we obtain $Psi = Phi$. So if two states evolve into the same state after some time $t$, they were the same to begin with.
Time evolution furthermore has the property that $U(t)U(s) = U(t+s)$. Let $tneq 0$. Now suppose that we have some state $Phi$ with the property that $U(s)Phi neq Phi$. Set $Psi = U(s)Phi$. We then get
beginequation
U(t)Psi = U(t)(U(s)Phi) = U(t+s)Phi.
endequation
We then have that the states $Psi$ and $Phi$, (which are different), evolve into the same state $U(t+s)Phi$, but after different times.
In fact, this is the only way this can happen. That is, if there are two different states $Phi neq Psi$ such that $U(t)Phi = U(t')Psi$, then there exists a time $s$ such that $Psi = U(s) Phi$. Maybe finding a proof for this claim is a good exercise.
answered 3 hours ago
Peter
1113
New contributor
Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Peter
1113
New contributor
Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Peter
1113
answered 3 hours ago
Peter
1113
1113
New contributor
Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
0
down vote
$defket#1#1rangle$
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 14 hours ago
Elio Fabri
1412
add a comment |Â
up vote
0
down vote
$defket#1#1rangle$
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 14 hours ago
Elio Fabri
1412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$defket#1#1rangle$
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 14 hours ago
Elio Fabri
1412
$defket#1#1rangle$
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 14 hours ago
Elio Fabri
1412
answered 14 hours ago
Elio Fabri
1412
answered 14 hours ago
Elio Fabri
1412
answered 14 hours ago
Elio Fabri
1412
1412
add a comment |Â
add a comment |Â
Doubting Thomas is a new contributor. Be nice, and check out our Code of Conduct.
Doubting Thomas is a new contributor. Be nice, and check out our Code of Conduct.
Doubting Thomas is a new contributor. Be nice, and check out our Code of Conduct.
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