# Is this analysis contradictory with the $2$'nd measurment?

## Summary and Motivation

"The below idea is about making a mathematical statement on system $$2$$ which induces a measurement on system $$1$$ while $$1+2$$ obeys unitary evolution."

Basically, I'm modelling the measurement (occurring at time $$t$$) as an interaction and that I have some constraints based on the conditions before ($$\tilde t^-$$)and after ($$\tilde t^+$$)

## Introduction

There are $$2$$ systems $$1$$ and $$2$$. Let the Hamiltonian of system $$1$$ be $$H_1$$ and let it be in an energy eigenstate:

$$\hat H_1|E_m \rangle = E_m |E_m \rangle$$

Now, a measurement is done (forcing the system to a momentum eigenstate):

$$\hat p |p_j \rangle = p_j |p_j \rangle$$

This measurement must me induced by a another system (see here why I think so: Energy cost of the measurement without perturbing the system? ). Let the Hamiltonian of this system be $$H_2$$. Let us express the net Hamiltonian as:

$$\hat H_{net} = \hat H_1 + \hat H_2 + \hat H'_{\text{int}(1,2)}$$

Where, $$H_{\text{int}(1,2)}$$ is the interaction Hamiltonian between the systems. But let us consider another system before we proceed:

$$\hat H_{\text{non-int}} = \hat H_1 \otimes \hat 1 + \hat 1 \otimes \hat H_2$$

where $$\hat 1$$ is the identity matrix. In the non-interacting case we have a separable wave function:

$$|\psi_{\text{non-int}} \rangle = |\psi_1 \rangle \otimes |\psi_2 \rangle = |\psi_1 ,\psi_2 \rangle$$

where $$|\psi_1 \rangle$$ and $$|\psi_2 \rangle$$ are the wave functions of system $$1$$ and $$2$$, respectively.

Now, I know something about the time-evolution of system $$1$$ and I know the net system $$1+2$$ obeys unitarity. Hence, I should be able to use this to say something about system $$2$$.

After doing some "calculations" I got the following equation (first-order strong summation condition):

$$0 = \sum_{\lambda '} \sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_2|E'_{\lambda'} \rangle \langle \psi^{(1)}_{\lambda '}|\psi^{(0)}_{n} \rangle =\sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_{\text{non-int}} |\psi^{(1)}_{n} \rangle$$

## Questions

Now let's say I want to perform a measurement on $$| \psi_{net} \rangle$$ does the above equation add an additional constraint? If so does it interfere with the usual measurement postulates (and how badly) $$|\psi_{net} \rangle \to |\text{eigenstate}\rangle$$ and the probability of the eigenstate is $$|\langle \psi_{net} |\text{eigenstate} \rangle |^2$$? If it invalidates the model I'm curious to know what was the false assumption?

Also, the current order of logic is: $$\text{Measurement in system 1} \implies \text{first-order strong summation condition}$$. Is the inverse true? $$\text{first-order strong summation condition} \implies \text{Measurement in system 1}$$

## Calculations

To make contact with the interacting case we use perturbation theory (assume $$H'_{12}$$ is small, see detour to justify this assumption):

$$\hat H_{net} = \hat H_{\text{non-int}} + \epsilon \hat H_{\text{int}(1,2)}$$

Using perturbation theory (upto first order correction) in the energy eigenstates:

$$|\psi_{\text{net}-n} \rangle = |\psi^{(0)}_{\text{non-int}} \rangle + \epsilon |\psi^{(1)}_{\text{non-int}} \rangle = |\psi^{(0)}_{n} \rangle + \epsilon |\psi^{(1)}_{n} \rangle$$

Where:

$$\left|\psi_n^{(1)}\right\rangle =\sum _{k\neq n}{\frac {\left\langle \psi^{(0)}_k\right|\hat H_{\text{int}(1,2)} \left|\psi^{(0)}_n\right\rangle }{E_{n}^{(0)}-E_{k}^{(0)}}}\left|\psi^{(0)}_{k}\right\rangle$$

## Case $$\tilde t^-$$:

(When not mentioned the kets are at time $$\tilde t^-$$ where $$\tilde t^- = t- \tilde \epsilon_-$$ and $$\tilde t^+ = t + \tilde \epsilon_+$$)

Let $$\psi_\text{net}$$ be in some superposition of energy eigenstates:

$$| \psi_{net} \rangle = \sum_{n} c_n | \psi_{net-n} \rangle$$

Let, us assume the measurement was done at a time $$t$$. Hence,

$$|\psi_1 (\tilde t^-)\rangle =| E_m \rangle$$

On the other hand, let system $$2$$ be in some superposition of energy eigenstates:

$$|\psi_2 \rangle = \sum_{\lambda '} c_{\lambda '}' |E'_{\lambda '} \rangle$$

Putting things together:

$$| \psi_{net} (\tilde t^-) \rangle = \sum_{n} c_n | \psi_{net-n} (\tilde t^-) \rangle = \sum_{n} c_n (|\psi^{(0)}_{n} (\tilde t^-) \rangle + \epsilon |\psi^{(1)}_{n} (\tilde t^-) \rangle )$$

However, we know,

$$|\psi^{(0)}_{n} (\tilde t^-)\rangle = |E_m , E'_{\lambda'} \rangle_{n(m,\lambda)}$$

where $$n(m,\lambda)$$ is a function which puts $$\tilde E_n = E_m + E'_{\lambda '}$$ ($$\tilde E_n$$ is the energy of the non-interacting Hamiltonian) in ascending order (ignoring degeneracy). Also, to relate coefficients by the below procedure:

$$| \psi_{\text{non-int}} (\tilde t^-) \rangle = |\psi_1 , \psi_2 \rangle = \sum_{\lambda '} c_{\lambda '}' |E_m , E'_{\lambda '} \rangle$$

But we can also throw light on: $$|\psi_{\text{net}-n} \rangle = |\psi^{(0)}_{n} \rangle + \epsilon |\psi^{(1)}_{n} \rangle \implies \sum_{\lambda '} c'_{\lambda '} |\psi_{\text{net}-\lambda} \rangle = | \psi_{\text{non-int}} \rangle +\epsilon \sum_{\lambda '} c'_{\lambda '}|\psi^{(1)}_{\lambda '} \rangle$$

Taking the inner product with $$| \psi_{net} \rangle = \sum_{n} c_n | \psi_{net-n} \rangle = \sum_{n} c_n (|\psi^{(0)}_{n} \rangle + \epsilon |\psi^{(1)}_{n} \rangle )$$ and the right side of the above equation:

$$0 = \epsilon (\sum_{n} c_n \langle \psi_{\text{non-int}} |\psi^{(1)}_{n} \rangle + (\sum_{\lambda '} \bar{c}'_{\lambda '} \langle \psi^{(1)}_{\lambda '}|) (\sum_{n} c_n |\psi^{(0)}_{n} \rangle )) +O(\epsilon^2)$$

Ignoring $$\epsilon^2$$:

$$0=\sum_{n} c_n \langle \psi_{\text{non-int}} |\psi^{(1)}_{n} \rangle + (\sum_{\lambda '} \bar{c}'_{\lambda '} \langle \psi^{(1)}_{\lambda '}|) (\sum_{n} c_n |\psi^{(0)}_{n} \rangle )$$

Let us write the above more generally:

$$0=\sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_{\text{non-int}} |\psi^{(1)}_{n} \rangle + \sum_{\lambda '} \sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_{\text{non-int}} |E_m,E'_{\lambda'} \rangle \langle \psi^{(1)}_{\lambda '}|\psi^{(0)}_{n} \rangle$$

Let, the above equation be called the $$\tilde t^-$$ equation.

## Case $$\tilde t^+$$ :

(When not mentioned the kets are at time $$\tilde t^+$$)

After the measurement on system $$1$$:

$$|\psi_1 (\tilde t^+)\rangle =| p_j \rangle = \sum_k \langle E_k | p_j \rangle |E_k \rangle$$

Let system $$2$$ be in some superposition of eigen-energies:

$$|\psi_2 (\tilde t^+)\rangle = \sum_{\lambda '} d_{\lambda '}' |E'_{\lambda '} \rangle = \sum_\lambda' \langle E'_{\lambda '} |\psi_2 \rangle |E'_{\lambda '} \rangle$$

Again, to relate coefficients by the below procedure:

$$| \psi_{\text{non-int}} (\tilde t^+) \rangle = |\psi_1 , \psi_2 \rangle = \sum_k \sum_{\lambda '} \langle E_k | p_j \rangle \langle E'_{\lambda '} |\psi_2 \rangle |E_k , E'_{\lambda '} \rangle$$

Following the same route as last time from $$|\psi_{\text{net}-n}(\tilde t^+) \rangle = |\psi^{(0)}_{n} \rangle + \epsilon |\psi^{(1)}_{n} \rangle$$:

$$\sum_k \sum_{\lambda '} \langle E_k | p_j \rangle \langle E'_{\lambda '} |\psi_2 \rangle |\psi_{\text{net}-\lambda'} \rangle = | \psi_{\text{non-int}} \rangle +\epsilon \sum_k \sum_{\lambda '} \langle E_k | p_j \rangle \langle E'_{\lambda '} |\psi_2 \rangle |\psi^{(1)}_{\lambda '} \rangle$$

Also:

$$|\psi_{\text{net}}(\tilde t^+) \rangle = \sum_n \langle \psi_{\text{net-n}} | \psi_{\text{net}} \rangle (|\psi^{(0)}_{n} \rangle + \epsilon |\psi^{(1)}_{n} \rangle)$$

Taking the inner-product of the above $$2$$ equations and focusing on the $$1$$'st order $$\epsilon$$: $$0 = \sum_n \sum_k \sum_{\lambda '} \langle p_j | E_k \rangle \langle \psi_2 | E'_{\lambda '} \rangle \langle \psi_{\text{net-n}} | \psi_{\text{net}} \rangle \langle \psi^{(1)}_{\lambda '} | \psi^{(0)}_{n} \rangle + \sum_n \langle \psi_{\text{net-n}} | \psi_{\text{net}} \rangle \langle \psi_{\text{non-int}}|\psi^{(1)}_{n} \rangle$$

Let, the above equation be called the $$\tilde t^+$$ equation.

## Combining the $$\tilde t^-$$ and $$\tilde t^+$$ Cases:

We also know, that $$\psi_{net}$$ undergoes unitary evolution:

$$| \psi_{net} (\tilde t^+)\rangle = U(\tilde t^+,\tilde t^-)|\psi_{net} (\tilde t^-)\rangle= e^{\frac{-iH_{net}(\tilde t^+ - \tilde t^-}{\hbar})} |\psi_{net} (\tilde t^-)\rangle$$

Let us go to the Heisenberg picture (the kets do not depend on time), we do so for a straight forward example (and leave the rest for the reader to work out):

$$\langle \psi_{\text{net-n}} (\tilde t^+) | \psi_{\text{net}} (\tilde t^+) \rangle = \langle \psi_{\text{net-n}}| \underbrace{U^\dagger (\tilde t^+,t') U(\tilde t^+,t')}_{\hat 1}| \psi_{\text{net}} \rangle = \langle \psi_{\text{net-n}}| \psi_{\text{net}} \rangle$$

We do this for all the kets (remove the time dependence). Hence, writing the $$t^+$$ equation minus the $$t^-$$ equation:

$$0 = \sum_n \sum_k \sum_{\lambda '} \langle p_j | E_k \rangle \langle \psi_2 | E'_{\lambda '} \rangle \langle \psi_{\text{net-n}} | \psi_{\text{net}} \rangle \langle \psi^{(1)}_{\lambda '} | \psi^{(0)}_{n} \rangle + \sum_n \langle \psi_{\text{net-n}} | \psi_{\text{net}} \rangle \langle \psi_{\text{non-int}}|\psi^{(1)}_{n} \rangle -\sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_{\text{non-int}} |\psi^{(1)}_{n} \rangle - \sum_{\lambda '} \sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_{\text{non-int}} |E_m,E'_{\lambda'} \rangle \langle \psi^{(1)}_{\lambda '}|\psi^{(0)}_{n} \rangle$$

Cancelling the term $$\sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_{\text{non-int}} |\psi^{(1)}_{n} \rangle$$:

$$0 = \sum_n \sum_k \sum_{\lambda '} \langle p_j | E_k \rangle \langle \psi_2 | E'_{\lambda '} \rangle \langle \psi_{\text{net-n}} | \psi_{\text{net}} \rangle \langle \psi^{(1)}_{\lambda '} | \psi^{(0)}_{n} \rangle - \sum_{\lambda '} \sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_{\text{non-int}} |E_m,E'_{\lambda'} \rangle \langle \psi^{(1)}_{\lambda '}|\psi^{(0)}_{n} \rangle$$

Note: tracing back the calculations we notice: $$\langle \psi_{\text{non-int}} |E_m,E'_{\lambda'} \rangle = \langle \psi_2 , E_m |E_m,E'_{\lambda'} \rangle = \langle \psi_2|E'_{\lambda'} \rangle$$. Now taking the summation common:

$$0 = \Big(\sum_{\lambda '} \sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_2|E'_{\lambda'} \rangle \langle \psi^{(1)}_{\lambda '}|\psi^{(0)}_{n} \rangle \Big) \Big(\sum_k \langle p_j | E_k \rangle -1 \Big)$$

Obviously:

$$1 \neq \sum_k \langle p_j | E_k \rangle$$

Hence,

$$0 = \sum_{\lambda '} \sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_2|E'_{\lambda'} \rangle \langle \psi^{(1)}_{\lambda '}|\psi^{(0)}_{n} \rangle$$

Re-substituting this in the $$\tilde t^-$$ equation (without time dependency) we get:

$$0=\sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_{\text{non-int}} |\psi^{(1)}_{n} \rangle$$

Let the above equations be known as the "first-order strong summation condition".

## Detour about $$\epsilon$$ and $$\tilde \epsilon_\pm$$

I would like to supplement why the perturbation approximation is a good one. Let's say we are in the Heisenberg picture and we want to know the time evolution an operator in system $$1$$. A natural question arises which time evolution to use $$H_{net}$$ or $$H_{1}$$?The answer is the time evolution is the same (approximately):

$$\langle m| \hat O_1(t') | n \rangle = \langle m|e^{\frac{i H_1 t' }{\hbar}} \hat O_1 e^{\frac{-i H_1 t '}{\hbar}}| n \rangle = \langle m|e^{\frac{i (H_1 + H_2 + H'_{12}) t' }{\hbar}} \hat O_1 e^{\frac{-i (H_1 + H_2 + H'_{12}) t' }{\hbar}}| n \rangle$$

The above makes sense iff,

$$\langle k |e^{\frac{-i H'_{12} t' }{\hbar}}| l \rangle \to 1$$

with $$t' \neq t$$ (the time of the measurement $$\tilde \epsilon_\pm \neq 0$$) where $$|k \rangle$$ and $$| l \rangle$$ can be any basis element.

• I just glanced through the details but your idea seems related to weak measurements, where the ancilla would be your system 1. If that is the case, the theory is pretty much well established already - so what would be your question? – Stéphane Rollandin Jun 18 at 16:06
• thank you so much for looking at a long post! I'm not sure if this part is also in the literature but I obtain a condition: $0 = \sum_{\lambda '} \sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_2|E'_{\lambda'} \rangle \langle \psi^{(1)}_{\lambda '}|\psi^{(0)}_{n} \rangle =\sum_{n} \langle \psi^{(0)}_{n}| \ \psi_{net} \rangle \langle \psi_{\text{non-int}} |\psi^{(1)}_{n} \rangle$ ... My question is since constrains the net system's wavefunction ... How does that go with the usual measurment postulates ... – More Anonymous Jun 18 at 16:15
• Also my line of reasoning is given a measurment happens this equation is true ... Is the opposite true? Given the equation is a measurement is then guaranteed ... – More Anonymous Jun 18 at 16:16
• @StéphaneRollandin Also you say "idea seems related to weak measurements" ... I'm not sure if it has to be a weak measurement ... I think this works for any system which undergoes unitary evolution after the measurment ... See detour (last section) ... – More Anonymous Jun 18 at 16:23
• I am sorry but I do not have the skills to seriously review your analysis. Now for the sake of clarity I think it would be worth elaborating a little about what you are exactly asking on the conceptual level, before launching the fireworks of formulas :) For example, is there an actual measurement assumed somewhere, or is this supposed to be a model for the measurement process? Note that the justification you give for introducting your system 2 is problematic, because it relies on your previous question having received a positive answer, which is not the case as far as I can tell. – Stéphane Rollandin Jun 18 at 16:39