Unitary time evolution - products of consecutive evolutions

In summary: But using the exponential form is definitely easier and more intuitive. Thanks for the suggestion.In summary, the conversation discusses the unitary time evolution of quantum states and how to determine the state of a system at a later time by evolving it from its initial configuration. The question of whether this evolution is equivalent to immediately evolving from the initial to the final state is raised. The participants also discuss the proof that time evolution is a unitary operation through the preservation of the inner product of states. It is noted that the exponential form of the unitary time evolution operator for the Schrodinger equation makes the proof simpler.
  • #1
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Hi all,

I was wondering if anyone could clarify my understanding of unitary time evolution of quantum states, in particular for products of time evolution's:

Suppose we know state of a quantum system at [itex] t=t_{0}[/itex], given by [itex] \vert\psi\left(t_{0}\right)\rangle[/itex], then to determine its state at a later time [itex]t=t_{2}[/itex] we can first evolve the state from its configuration at [itex] t=t_{0}[/itex] to its configuration at [itex] t=t_{1}[/itex], i.e [tex]\vert\psi\left(t_{1}\right)\rangle= U\left(t_{1},t_{0}\right)\vert\psi\left(t_{0}\right)\rangle[/tex]
and then subsequently evolve this to its final state at [itex] t=t_{2}[/itex] via a further time evolution, [tex]\vert\psi\left(t_{2}\right)\rangle= U\left(t_{2},t_{1}\right)\vert\psi \left(t_{1}\right)\rangle= U\left(t_{2},t_{1}\right)U \left(t_{1},t_{0}\right)\vert\psi\left(t_{0}\right)\rangle[/tex]
Alternatively, we could immediately evolve the state from its initial configuration at [itex]t=t_{0}[/itex] to its final state at [itex]t=t_{2}[/itex] in the following manner, [tex]\vert\psi \left(t_{2}\right)\rangle= U\left(t_{2},t_{0}\right)\vert\psi \left(t_{0}\right)\rangle[/tex]
Given that time is homogeneous (as far as we know), the results of these two evolution's of the system should be equivalent (as the path we take through time in evolving the state should not affect its initial and final configurations), this implies that [tex]U\left(t_{2},t_{0}\right)= U\left(t_{2},t_{1}\right)U\left(t_{1},t_{0}\right)[/tex]

Is this a correct description?
 
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  • #2
Yes.
 
  • #3
Ok, thanks.

On a related note, to prove that time evolution is a unitary operation is it enough to consider a inner product of a quantum state [itex]\vert \psi\left(t\right)\rangle[/itex] that satisfies the Schrodinger equation (with [itex]\hbar=c=1[/itex]) [tex]i \frac{\partial}{\partial t} \left(\vert \psi\left(t\right)\rangle\right) = \hat{H}\vert \psi\left(t\right)\rangle \rightarrow \frac{\partial}{\partial t} \left(\vert \psi\left(t\right)\rangle\right)= -i\hat{H}\vert \psi\left(t\right)\rangle [/tex] and then differentiate it with respect to time, such that [tex]\frac {\partial}{\partial t}\left(\langle \psi\left(t\right) \vert \psi\left(t\right)\rangle \right)=\frac {\partial}{\partial t}\left(\langle \psi\left(t\right) \vert\right) \vert \psi\left(t\right)\rangle + \langle \psi\left(t\right) \vert\frac {\partial}{\partial t}\left(\vert \psi\left(t\right)\rangle \right)= i\langle\psi \left(t\right)\vert\hat{H}^{\dagger} \vert\psi \left(t\right) \rangle - i\langle\psi \left(t\right)\vert\hat{H} \vert\psi \left(t\right) \rangle = 0 [/tex]
as [itex]\hat{H}^{\dagger} = \hat{H}[/itex]. Hence, the inner product is preserved for all times, i.e. it is preserved under finite time evolution. As a result of this, if we consider a state at time [itex]t=t_{0}[/itex], [itex]\vert\psi \left(t_{0}\right)\rangle [/itex] and its subsequent evolved state at some later time [itex] t[/itex], [itex]\vert\psi \left(t\right)\rangle=U \left(t,t_{0}\right) \vert\psi \left(t_{0}\right)\rangle [/itex], we then have that, [tex] \langle \psi \left(t\right)\vert\psi \left(t\right)\rangle= \langle \psi \left(t_{0}\right)\vert U^{\dagger} \left(t,t_{0}\right) U \left(t,t_{0}\right) \vert\psi \left(t_{0}\right)\rangle= \langle \psi \left(t_{0}\right)\vert\psi \left(t_{0}\right)\rangle [/tex] which is true iff [itex] U^{\dagger} \left(t,t_{0}\right) U \left(t,t_{0}\right) = 1 [/itex], i.e. [itex]U\left( t,t_{0}\right) [/itex] must be a unitary operator?!

Sorry, I know this should probably be in a new thread, but I thought I'd try and combine the two questions as they are related.
 
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  • #4
To prove than an operation is unitary you need to show that the inner product of any two states is preserved. It's not enough to show that the norm of each individual state is preserved. There are operations that preserve the norm of all states but are not unitary.

So you need to consider a general inner product ##\langle \phi | \psi \rangle## in your proof. But actually it should go through in pretty much the same way.
 
  • #5
Ah ok, so would it be fair to consider two different states, [itex] \vert\psi\left( t\right)\rangle[/itex] and [itex] \vert\phi\left( t\right)\rangle [/itex] which both satisfy the schrodinger equation, i.e. [tex] i\frac{ \partial}{\partial t}\left( \vert\psi\left( t\right)\rangle\right)= \hat{H} \vert\psi\left( t\right)\rangle, \qquad i\frac{ \partial}{\partial t}\left( \vert\phi\left( t\right)\rangle\right)= \hat{H} \vert\phi\left( t\right)\rangle [/tex] and then follow the same procedure from here as before?
 
  • #6
Yes.
 
  • #7
Cool. Thanks for the help guys, much appreciated!
 
  • #8
A much easier way to see the same thing perhaps is to notice that the unitary time evolution operator for the Schroedinger equation is simply (for time independent Hamiltonians) ##U(t,t_0)=e^{iH(t-t_0)}##. Since the Hamiltonian is Hermitian, the the time evolution operator is unitary.
 
  • #9
Yes, you're right. I understood it in that case, but was trying to go for a more general approach, including time-dependent Hamiltonians.
 

1. What is unitary time evolution?

Unitary time evolution refers to the mathematical concept in quantum mechanics where the state of a quantum system evolves over time in a reversible and deterministic manner. This means that the probability of finding the system in a particular state at a given time can be calculated using the unitary operator, which is a mathematical representation of the system's evolution.

2. What are products of consecutive evolutions?

Products of consecutive evolutions refer to the result of applying multiple unitary operators to a quantum system in succession. This can be thought of as a series of transformations that the system undergoes over time, each described by a different unitary operator.

3. How are products of consecutive evolutions related to unitary time evolution?

Products of consecutive evolutions are a way of representing the overall time evolution of a quantum system. By applying each individual unitary operator in sequence, the resulting product operator represents the total evolution of the system over a specific time interval.

4. What is the significance of unitary time evolution and products of consecutive evolutions?

Unitary time evolution and products of consecutive evolutions are fundamental concepts in quantum mechanics that allow us to predict and understand the behavior of quantum systems. They provide a mathematical framework for calculating the probabilities of different outcomes and making predictions about the behavior of particles at the quantum level.

5. Are there any limitations to unitary time evolution and products of consecutive evolutions?

Unitary time evolution and products of consecutive evolutions are based on the principles of quantum mechanics, which are still being studied and refined. There are some cases where they may not accurately predict the behavior of a quantum system, such as in certain extreme conditions or when dealing with large numbers of particles. Additionally, the concept of unitarity itself is still a topic of debate among scientists.

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