Can one construct a non-zero (conditional) probability in this situation?

Question

Background

Inspired by this video I'd like to ask the following question:

I'm not an expert at QED or optics. But this is my $2$ cents on the problem: "Do we see the light or the object?"

Problem statement: Given I "see" the light of an object can I infer the existence of the "source"?

My take: What one "sees" is an eigenvalue linked to that of a photon. In fact, one sees is a stream of photons at different times (but as time is not an observable but a parameter it should not affect the thrust of the argument). With each eigenvalue one can determine there is some "source" (the word source seems to imply it's spatially confined and uniqueness of hamiltonian producing these photons). Now I know of no theoretical limits on having multiple sources to produce the same set of eigenvalues measured (though there would be some non-zero probability of one measurement of being the same in both cases).

Thus I conclude due to existence of these cases one cannot conclude the uniqueness of the source.

I think it becomes a question of conditional probability as what is the probability I deduce a Hamiltonian (assuming uniform probability of each Hamiltonian) given I measure an eigenvalue at $t_1$. Note. This becomes harder when one says what is the probability of it being the same a Hamiltonian at time $t_2 > t_1$. The question then becomes can one construct a non-zero (conditional) probability in this situation .

The (incomplete) math/ My attempt

This is an incomplete answer but let's restrict ourselves to energy eigenvalues and same dimensions Hamiltonian. If one uses a different observable then another layer of proabability would be added. We wont even mention time evolution or multiple measurements. Now:

From the spectral theorem I can reconstruct a particular Hamiltonian, see:

$$ \sum_{i} \lambda_i |\lambda_i \rangle \langle \lambda_i| = H_{\lambda} $$

Where $\lambda_j$ is part of $\lambda_i$ summation. However, given I only have a Hamiltonian it can also be part of a different Hamiltonian:

$$ \sum_{i} \beta_i |\beta_i \rangle \langle \beta_i| = H_{\beta} $$

where $\beta_k$ is part of $\beta_i$ summation. We add, $\beta_k = \lambda_j$. Now, let us use matrices

$$ \begin{pmatrix} |\lambda_1 \rangle \langle \lambda_1| & |\lambda_2 \rangle \langle \lambda_2| & \dots \\ |\beta_1 \rangle \langle \beta_1| & |\beta_2 \rangle \langle \beta_2| & \dots \\ \vdots & \vdots \end{pmatrix} \begin{pmatrix} \lambda_1 & \beta_1 & \dots \\ \lambda_2 & \beta_2 & \dots \\ \vdots & \vdots \end{pmatrix} = \begin{pmatrix} H_{\alpha} & ? & \dots\\ ? & H_{\beta} & \dots \\ \vdots & \vdots \end{pmatrix} $$

We multiply a $S$ and $S^{-1}$ such that: $$ S \begin{pmatrix} |\lambda_1 \rangle \langle \lambda_1| & |\lambda_2 \rangle \langle \lambda_2| & \dots \\ |\beta_1 \rangle \langle \beta_1| & |\beta_2 \rangle \langle \beta_2| & \dots \\ \vdots & \vdots \end{pmatrix} \begin{pmatrix} \lambda_1 & \beta_1 & \dots \\ \lambda_2 & \beta_2 & \dots \\ \vdots & \vdots \end{pmatrix} S^{-1}= \begin{pmatrix} H_{\alpha} & 0 & \dots\\ 0 & H_{\beta} & \dots \\ \vdots & \vdots \end{pmatrix} $$

Now, to quantify probabilities. Let us take trace:

$$ \text{Tr} \begin{pmatrix} |\lambda_1 \rangle \langle \lambda_1| & |\lambda_2 \rangle \langle \lambda_2| & \dots \\ |\beta_1 \rangle \langle \beta_1| & |\beta_2 \rangle \langle \beta_2| & \dots \\ \vdots & \vdots \end{pmatrix} \begin{pmatrix} \lambda_1 & \beta_1 & \dots \\ \lambda_2 & \beta_2 & \dots \\ \vdots & \vdots \end{pmatrix} = H_\alpha + H_{\beta} + \dots $$

Hence,

$$H_\alpha + H_{\beta} + \dots = \dots+\lambda_j (| \lambda_j \rangle \langle \lambda_j | + |\beta_k \rangle \langle \beta_k | + \dots) + \dots$$

Given on measures $\lambda_j$ one can conclude the following $| \lambda_j \rangle \langle \lambda_j | + |\beta_k \rangle \langle \beta_k | + \dots$ becomes relevant. To quantify how relevant one has to use the Born rule.

Question

Can someone construct a toy model to prove or disprove my intuition? I think I must be wrong as I'm up against the founder of QED.

Answer 1

You are mixing up two different frameworks, the quantum mechanical where the photons exist and have a wavefunction which is a probability map, and the classical , where electromagnetic waves that are reflected etc as classical wave on a medium, with energy transfer.through the medium ( water waves, sound waves and light waves)

Hamiltonians and eigenfunctions are an underlying level from which the classical light wave emerges. When you deduce that there is a water source or a water sink in a lake, you do not have to think of the eigen values of all those zillions of water molecules that make up the water wave and its reflections.

Light emerges from the photons in a complicated way as can be shown in QED. It is not additive, and even though the frequency of the light is connected to the energy of the photon the mathematics is not simple of how probabilities end up in energy transfers macroscopically. It is those energy waves composed by zillions of photons that define the objects by the light reflection for us.

addition:

This hypothesis of yours is not correct:

In fact, one sees is a stream of photons at different times

One sees a confluence of photons, as the link I give of how the classical emerges from the quantum. This is tightly tied to quantum mechanical probability space, and the real electromagnetic field emerges from zillions of photons' wavefunctions adding up, complex conjugated to give a probability amplitude for how the energy of all this photons generates the classical electromagnetic energy motion in space.