Interpreting Unitary Time Evolution

[thatboi]

Hi all,
This should be a simple question but it has been bothering me for a bit:
Consider 2 Hamiltonian terms $H_{1},H_{2}$ that satisfy $[H_{1},H_{2}] = 0$. Suppose we are working in the Heisenberg picture and we time evolve some operator $A$ according to $A(t) = e^{-i(H_{1}+H_{2})t}Ae^{i(H_{1}+H_{2})t}$. This can be interpreted as time-evolving the operator $A$ from time $0$ to time $t$. Now if I broke up the exponential so that $A(t) = e^{-iH_{2}t}e^{-iH_{1}t}Ae^{iH_{1}t}e^{iH_{2}t}$, is the right way to interpret this like: the $e^{-iH_{1}t}$ unitary time-evolves the operator $A$ from time $0$ to time $t$, and then the unitary $e^{-iH_{2}t}$ time evolves this new operator $e^{-iH_{1}t}Ae^{iH_{1}t}$ from some new time $0$ to time $t$? Naively looking at the expression would almost seem to imply that somehow we have progressed the time forward by $2t$ by breaking up the exponential but I know this is wrong. I would appreciate if there were a better interpretation of this.

 

[Demystifier]

Here is a classical analogy. A body moves with a constant velocity $v=v_1+v_2$. Its position as a function of time is
$$x(t)=vt=(v_1+v_2)t$$
But one can define the quantity $x_1(t)=v_1t$ and write the formula above as
$$x(t)=x_1(t)+v_2t$$
How would you interpret the last formula? Would you say that the body travels with velocity $v_1$ from time $0$ to $t$, and then travels with velocity $v_2$ from some new time $0$ to $t$?

When you explain it to yourself by yourself, just apply the same type of reasoning to your original quantum question.

 

[PeterDonis]

I would appreciate if there were a better interpretation of this.

There is: just answer "no" to your question:

is the right way to interpret this like: the $e^{-iH_{1}t}$ unitary time-evolves the operator $A$ from time $0$ to time $t$, and then the unitary $e^{-iH_{2}t}$ time evolves this new operator $e^{-iH_{1}t}Ae^{iH_{1}t}$ from some new time $0$ to time $t$?

 

[thatboi]

Here is a classical analogy. A body moves with a constant velocity $v=v_1+v_2$. Its position as a function of time is
$$x(t)=vt=(v_1+v_2)t$$
But one can define the quantity $x_1(t)=v_1t$ and write the formula above as
$$x(t)=x_1(t)+v_2t$$
How would you interpret the last formula? Would you say that the body travels with velocity $v_1$ from time $0$ to $t$, and then travels with velocity $v_2$ from some new time $0$ to $t$?

When you explain it to yourself by yourself, just apply the same type of reasoning to your original quantum question.

Hi,
From the equation alone, everything has a dependence on $t$ so I couldn't use such an argument. But couldn't I achieve the same result as $(v_1+v_2)t$ by first having the body travel with velocity $v_1$ for a time $t$ and then switch to it travelling with velocity $v_2$ for another time $t$? I know this breaks the continuous evolution in time $t$ but the end result is still the same.

 

[thatboi]

Hi,
From the equation alone, everything has a dependence on $t$ so I couldn't use such an argument. But couldn't I achieve the same result as $(v_1+v_2)t$ by first having the body travel with velocity $v_1$ for a time $t$ and then switch to it travelling with velocity $v_2$ for another time $t$? I know this breaks the continuous evolution in time $t$ but the end result is still the same.

I only bring this up because quantum mechanically, I'm imagining a scenario like a quantum circuit where we have the state first evolve under the unitary gate $e^{-iH_{1}t}$ for time $t$ and then it evolves under the unitary gate $e^{-iH_{2}t}$ for another time $t$.

 

[Demystifier]

But couldn't I achieve the same result as $(v_1+v_2)t$ by first having the body travel with velocity $v_1$ for a time $t$ and then switch to it travelling with velocity $v_2$ for another time $t$? I know this breaks the continuous evolution in time $t$ but the end result is still the same.

Are you sure that the end result is the same? Sure, at the end you arrive at the same place. But how much total time do you spend to arrive, is it the same in both cases?

 

[thatboi]

Are you sure that the end result is the same? Sure, at the end you arrive at the same place. But how much total time do you spend to arrive, is it the same in both cases?

Indeed, the time would take twice as long in such a process, but my initial inspiration for this question was that I was looking at the process of trotterization in quantum circuits, which seems to break up the Hamiltonian in the time-evolution process and I just wanted to reconcile how a trotterized circuit with $n$ terms i.e $n$ layers in the circuit would seem to take $n$ times longer than under generic time-evolution.

 

[Demystifier]

I only bring this up because quantum mechanically, I'm imagining a scenario like a quantum circuit where we have the state first evolve under the unitary gate $e^{-iH_{1}t}$ for time $t$ and then it evolves under the unitary gate $e^{-iH_{2}t}$ for another time $t$.

Ah, I see now what's your motivation. If you don't care how much time you spend to perform the job, then those are equivalent. But if you want to perform the job as fast as possible, then they are not equivalent. Furthermore, if you want to describe it as two gates acting at different times, then the unitary evolution operator $U(t)$ is not $e^{-iH_{2}t}e^{-iH_{1}t}$. Instead, it is
$$U(t)=\chi_1(t)e^{-iH_{1}t}+\chi_2(t)e^{-iH_{2}t}$$
where $\chi_a(t)$ are characteristic functions, equal to 1 during their characteristic time intervals and equal to 0 outside of these intervals. The characteristic time interval of $\chi_1(t)$ is $[0,\Delta t)$, while the characteristic time interval of $\chi_2(t)$ is $[\Delta t,2\Delta t)$. Hence the total process lasts time $2\Delta t$. It's important to distinguish $t$ from $\Delta t$, because $t$ is a variable (taking any value in a continuous interval), while $\Delta t$ is a fixed time interval.

 

[thatboi]

Ah, I see now what's your motivation. If you don't care how much time you spend to perform the job, then those are equivalent. But if you want to perform the job as fast as possible, then they are not equivalent. Furthermore, if you want to describe it as two gates acting at different times, then the unitary evolution operator $U(t)$ is not $e^{-iH_{2}t}e^{-iH_{1}t}$. Instead, it is
$$U(t)=\chi_1(t)e^{-iH_{1}t}+\chi_2(t)e^{-iH_{2}t}$$
where $\chi_a(t)$ are characteristic functions, equal to 1 during their characteristic time intervals and equal to 0 outside of these intervals. The characteristic time interval of $\chi_1(t)$ is $[0,\Delta t)$, while the characteristic time interval of $\chi_2(t)$ is $[\Delta t,2\Delta t)$. Hence the total process lasts time $2\Delta t$. It's important to distinguish $t$ from $\Delta t$, because $t$ is a variable (taking any value in a continuous interval), while $\Delta t$ is a fixed time interval.

Great, I think this clarifies a lot. My other motivation just came from the fact that if we considered in the Heisenberg picture to expand something like $A(t)=e^{-iH_{2}t}e^{-iH_{1}t}Ae^{iH_{1}t}e^{iH_{2}t}$ , we would first apply Baker Campbell Hausdorff to evaluate the inner conjugation by $H_{1}$ and then evaluating conjugation under $H_{2}$, which I imagine as different layers in a quantum circuit but I think I was misunderstanding what exactly the time evolution implemented in each layer physically represented.

 

[Demystifier]

Great, I think this clarifies a lot.

Actually, there is even a better formula for $U(t)$:
$$U(t)=e^{-i\chi_2(t)H_2t} e^{-i\chi_1(t)H_1t}$$
because this is unitary for all $t$, while the one in the previous post was unitary only for $t\in [0,2\Delta t)$.

 

[PeterDonis]

couldn't I achieve the same result as $(v_1+v_2)t$ by first having the body travel with velocity $v_1$ for a time $t$ and then switch to it travelling with velocity $v_2$ for another time $t$?

No. The expression $x = vt = \left( v_1 + v_2 \right) t$ is a function. It applies for all times $t$. It describes a single straight worldline with a single constant speed for all time.

What you describe involves changing speed at some point in time. That's not the same thing and can't be described by the same math.

Similar remarks would apply to your "interpretation" of unitary evolution in your OP.