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 (˜t−)and after (˜t+)
Introduction
There are 2 systems 1 and 2. Let the Hamiltonian of system 1 be H1 and let it be in an energy eigenstate:
ˆH1|Em⟩=Em|Em⟩
Now, a measurement is done (forcing the system to a momentum eigenstate):
ˆp|pj⟩=pj|pj⟩
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 H2. Let us express the net Hamiltonian as:
ˆHnet=ˆH1+ˆH2+ˆH′int(1,2)
Where, Hint(1,2) is the interaction Hamiltonian between the systems. But let us consider another system before we proceed:
ˆHnon-int=ˆH1⊗ˆ1+ˆ1⊗ˆH2
where ˆ1 is the identity matrix. In the non-interacting case we have a separable wave function:
|ψnon-int⟩=|ψ1⟩⊗|ψ2⟩=|ψ1,ψ2⟩
where |ψ1⟩ and |ψ2⟩ 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=∑λ′∑n⟨ψ(0)n| ψnet⟩⟨ψ2|E′λ′⟩⟨ψ(1)λ′|ψ(0)n⟩=∑n⟨ψ(0)n| ψnet⟩⟨ψnon-int|ψ(1)n⟩
Questions
Now let's say I want to perform a measurement on |ψnet⟩ does the above equation add an additional constraint? If so does it interfere with the usual measurement postulates (and how badly) |ψnet⟩→|eigenstate⟩ and the probability of the eigenstate is |⟨ψnet|eigenstate⟩|2? If it invalidates the model I'm curious to know what was the false assumption?
Also, the current order of logic is: Measurement in system 1⟹first-order strong summation condition. Is the inverse true? first-order strong summation condition⟹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):
ˆHnet=ˆHnon-int+ϵˆHint(1,2)
Using perturbation theory (upto first order correction) in the energy eigenstates:
|ψnet−n⟩=|ψ(0)non-int⟩+ϵ|ψ(1)non-int⟩=|ψ(0)n⟩+ϵ|ψ(1)n⟩
Where:
|ψ(1)n⟩=∑k≠n⟨ψ(0)k|ˆHint(1,2)|ψ(0)n⟩E(0)n−E(0)k|ψ(0)k⟩
Case ˜t−:
(When not mentioned the kets are at time ˜t− where ˜t−=t−˜ϵ− and ˜t+=t+˜ϵ+)
Let ψnet be in some superposition of energy eigenstates:
|ψnet⟩=∑ncn|ψnet−n⟩
Let, us assume the measurement was done at a time t. Hence,
|ψ1(˜t−)⟩=|Em⟩
On the other hand, let system 2 be in some superposition of energy eigenstates:
|ψ2⟩=∑λ′c′λ′|E′λ′⟩
Putting things together:
|ψnet(˜t−)⟩=∑ncn|ψnet−n(˜t−)⟩=∑ncn(|ψ(0)n(˜t−)⟩+ϵ|ψ(1)n(˜t−)⟩)
However, we know,
|ψ(0)n(˜t−)⟩=|Em,E′λ′⟩n(m,λ)
where n(m,λ) is a function which puts ˜En=Em+E′λ′ (˜En is the energy of the non-interacting Hamiltonian) in ascending order (ignoring degeneracy). Also, to relate coefficients by the below procedure:
|ψnon-int(˜t−)⟩=|ψ1,ψ2⟩=∑λ′c′λ′|Em,E′λ′⟩
But we can also throw light on: |ψnet−n⟩=|ψ(0)n⟩+ϵ|ψ(1)n⟩⟹∑λ′c′λ′|ψnet−λ⟩=|ψnon-int⟩+ϵ∑λ′c′λ′|ψ(1)λ′⟩
Taking the inner product with |ψnet⟩=∑ncn|ψnet−n⟩=∑ncn(|ψ(0)n⟩+ϵ|ψ(1)n⟩) and the right side of the above equation:
0=ϵ(∑ncn⟨ψnon-int|ψ(1)n⟩+(∑λ′ˉc′λ′⟨ψ(1)λ′|)(∑ncn|ψ(0)n⟩))+O(ϵ2)
Ignoring ϵ2:
0=∑ncn⟨ψnon-int|ψ(1)n⟩+(∑λ′ˉc′λ′⟨ψ(1)λ′|)(∑ncn|ψ(0)n⟩)
Let us write the above more generally:
0=∑n⟨ψ(0)n| ψnet⟩⟨ψnon-int|ψ(1)n⟩+∑λ′∑n⟨ψ(0)n| ψnet⟩⟨ψnon-int|Em,E′λ′⟩⟨ψ(1)λ′|ψ(0)n⟩
Let, the above equation be called the ˜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.
