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?










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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?










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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?










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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






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      edited 13 mins ago









      knzhou

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          4 Answers
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          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.






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          • 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


















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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.






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          • 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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          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.






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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.






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              4 Answers
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              4 Answers
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              active

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              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.






              share|cite|improve this answer




















              • 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















              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.






              share|cite|improve this answer




















              • 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













              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.






              share|cite|improve this answer













              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.







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              share|cite|improve this answer



              share|cite|improve this answer










              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

















              • 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











              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.






              share|cite|improve this answer




















              • 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














              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.






              share|cite|improve this answer




















              • 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












              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.






              share|cite|improve this answer












              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.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              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
















              • 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










              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.






              share|cite|improve this answer








              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.





















                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.






                share|cite|improve this answer








                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.



















                  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.






                  share|cite|improve this answer








                  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.







                  share|cite|improve this answer








                  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.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  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




                  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.




















                      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.






                      share|cite|improve this answer
























                        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.






                        share|cite|improve this answer






















                          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.






                          share|cite|improve this answer












                          $defket#1#1rangle$
                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 14 hours ago









                          Elio Fabri

                          1412




                          1412




















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