# Equivalent version of sudden approximation's time energy uncertainty principle?

## Background

So let's say I have $$2$$ particles which interact with a potential $$V(\vec r_1- \vec r_2)$$. The Hamiltonian of this system is given by:

$$H \psi( \vec r_1, \vec r_2) = \frac{ \vec p_1 \cdot \vec p_1 }{2m} \psi( \vec r_1, \vec r_2) + \frac{\vec p_2 \cdot \vec p_2}{2m} \psi( \vec r_1, \vec r_2) + V(\vec r_1 - \vec r_2 ) \psi( \vec r_1, \vec r_2)$$

Where $$H$$ is the Hamiltonian, $$p_i$$ is the momentum of the $$i$$'th particle and $$V(\vec r_1 - \vec r_2)$$ is the relative position between both wavefunctions $$\psi(x_1,x_2)$$. Now, let's say I "suddenly" reduce the displacement by $$\vec c$$.

But the sudden approximation I can easily write the new Hamiltonian to be:

$$H' \psi( \vec r_1, \vec r_2) = \frac{ \vec p_1 \cdot \vec p_1 }{2m} \psi( \vec r_1, \vec r_2) + \frac{\vec p_2 \cdot \vec p_2}{2m} \psi( \vec r_1, \vec r_2) + V(\vec r_1 - \vec r_2 -\vec c) \psi( \vec r_1, \vec r_2)$$

Where there is a "sudden" change in Hamiltonian:

$$H \to H'$$

However, since the universe cares only about relative position and not actual position (insert general relativity philosophy here), it should be impossible to distinguish this situation from say if the wavefunctions "suddenly" was translated by the translation operator (for particle $$2$$) $$T_{2}(\vec c)$$

$$\psi( \vec r_1, \vec r_2) \to T_{2}(\vec c) \psi ( \vec r_1, \vec r_2) = \psi( \vec r_1, \vec r_2 + \vec c) = \psi'( \vec r_1, \vec r_2)$$

The energies of both are equivalent as expected:

$$\langle \psi| H'| \psi \rangle = \langle \psi' | H| \psi' \rangle$$

## Question

Is my initial analysis correct? Just as I have a uncertainty principle for the "sudden"(diabatic) approximation for:

$$H \to H'$$

Can I derive an uncertainty principle for the equivalent sudden wave function translation?

$$\psi( \vec r_1, \vec r_2) \to \psi'( \vec r_1, \vec r_2)$$

## 1 Answer

Your analysis looks essentially correct to me. To recast the condition for the validity of the sudden approximation, note that your result can be written $$H' = T_2^\dagger(c)\, H\, T_2(c)\;.$$

The condition for the validity of the sudden approximation given in the wiki page you linked is that $$\zeta \ll 1$$ where \begin{align} \zeta &= \frac{\tau^2}{\hbar^2}\left(\langle \bar{H}^2 \rangle_0 - \langle \bar{H} \rangle_0^2 \right)\\ \bar{H} &= \frac{1}{\tau} \int_0^\tau dt\; H(t) \end{align} and $$\langle \cdot \rangle_0$$ denotes the expectation with respect to the state at $$t=0$$ and $$\tau$$ is the time it takes for the sudden process to happen.

We will now assume that we can write $$H(t) = U^{\dagger}(t) H U(t)$$ for all $$0 \le t \le \tau$$, with $$U(0) = 1$$, $$U(\tau) = T_2(c)$$ and $$U(t)$$ a unitary operator for all $$t$$. For example we might have $$U(t) = T_2\left(c \frac{t}{\tau}\right)$$ in the relevant region, but the exact form will depend on the details of the process you are attempting to approximate away.

We can now define \begin{align} |\psi(t)\rangle &= U(t)|\psi(0)\rangle\\ \langle A \rangle_t &= \langle \psi(t)| A | \psi(t) \rangle \end{align} Putting this all together we find $$\zeta = \frac{1}{\hbar}\int_0^\tau\;d t \left(\langle H^2 \rangle_t - \langle H \rangle_t^2\right)$$

• Umm... In my head I know this the time energy uncertainty principle should be the same. Is $\langle \bar H^2 \rangle_0 = \langle \bar H'^2 \rangle_0$ ? Commented Jun 23, 2021 at 11:23
• I may have misunderstood what you meant by an uncertainty principle in this context (time energy uncertainty is tricky to define). I was going by what was in the Wikipedia link. $\bar{H}$ is the "average" Hamiltonian during the rapid process, so it is somewhere between $H$ and $H'$. This means that the notion of $\bar{H}'$ does not really make sense Commented Jun 23, 2021 at 11:30
• I think my first question was posed poorly. Allow me to rephrase: I have $2$ equivalent ways of defining $H'$: $H' = H - V(\vec r_1 - \vec r_2) + V(\vec r_1 - \vec r_2 - \vec c)$ the other as as you state: $H' = T_2^\dagger(\vec c) H T_2(\vec c)$. Do both these Hamiltonians yield the same uncertainty principle? I see no problem in $\langle \bar H \rangle_t$ but am worried in $\langle \bar H^2 \rangle_t$ Commented Jun 23, 2021 at 11:35
• The key requirement here is that the operator I have called $U$ is invertible (and really should be unitary, otherwise you will have issues with the normalization of states). In this case $H^2(t) = U^{-1}(t) H U(t)U^{-1}(t)H U(t) = U^{-1}(t) H^2 U(t)$ which implies that $\langle H^2(t) \rangle_0 = \langle H^2 \rangle_t$. The unitary of $T_2(c)$ is really the mathematical content of your "general relativity philosophy" about the universe not caring about where you put the shift. Commented Jun 23, 2021 at 11:46
• Actually, looking at it again, it is not enough for $U$ to be invertible. $U$ must be fully unitary. Commented Jun 23, 2021 at 11:50