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In quantum mechanics, we know that the generator of the infinitesimal time translation operator $$U(dt) = e^{-\frac{i}{\hbar}\hat{H}dt}$$ is the Hamiltonian $\hat{H}$.
Similarly, the generator of the infinitesimal position translation operator $$U(d\vec{x}) = e^{-\frac{i}{\hbar}\hat{\vec{p}} \cdot d\vec{x}}$$ is the momentum operator $\hat{\vec{p}}$.
There are other such relationship between operators in the theory. These relationships can be mathematically shown using group theory.
My question is this : How can we intuitively understand and realize such relationships between two operators?

Qmechanic
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1 Answers1

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The most intuitive explanation behind those formulas that I can think of comes from applying Taylor's expansion.

Say I want to know what the wave function $\psi(x,t)$ looks like when I shift it by $dt$ in time. In other words, I want to calculate $\psi(x, t + dt)$. Since $dt$ is a small increment, I can expand it like

$$\psi(x,t + dt) = \psi(x,t) + dt\frac{\partial\psi}{\partial t} + \frac{1}{2}dt^2\frac{\partial^2 \psi}{\partial t^2} + ... = \left(\sum^\infty_{n=0}\frac{dt^n}{n!}\frac{\partial^n}{\partial t^n}\right)\psi(x,t).$$

If you look at the sum on the right-hand side, it should feel familiar as $$e^z = \sum^{\infty}_{n=0} \frac{z^n}{n!}.$$ So if I take $z := dt\frac{\partial}{\partial t}$, I can easily rewrite our first expression $$\exp\left(dt{\frac{\partial}{\partial t}}\right)\psi(x,t) = \left(\sum^\infty_{n=0}\frac{dt^n}{n!}\frac{\partial^n}{\partial t^n}\right)\psi(x,t) = \psi(x,t + dt).$$ The last step is to get the Hamiltonian in. This is just simple algebra, from the time-dependent Schrödinger's equation we know that $$\hat{H} = i\hbar\frac{\partial}{\partial t} \implies -\frac{i}{\hbar}\hat{H} = \frac{\partial}{\partial t},$$ finally plugging into the expression above $$\psi(x,t + dt) = \exp\left(-\frac{i}{\hbar}\hat{H}dt\right)\psi(x,t).$$ You can go through the same procedure for $p$ and $x$, as $p$ is the operator of spatial derivatives in the coordinate representation ($\vec{p} = -i\hbar\nabla$).