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Yes, except for the specific case in which the state is in an eigenstate of the measurement operator. Measurements with deterministic outcomes More specifically, for any quantum (pure) state $|\psi\rangle$, there is just one class of measurements that can be performed without changing the state, and these are the measurements which ask questions of the ...

-1

In quantum mechanics in fact there is no isolated systems. Interactions with environment are impossible to avoid partially because of real influence of small perturbations, partially because of quantum entanglement. Saying that it is clear that measurements process, based on interaction between quantum system and macroscopic measurement aparatus is ...

4

Occam's razor favors simplicity, but simplicity is subjective. I've heard people use Occam's razor for exactly the opposite purpose, to argue that it's extravagant to talk about many worlds, so we should favor the Copenhagen interpretation (CI) over the many-worlds interpretation (MWI). I think wrangling over CI versus MWI is pretty pointless, because we ...

4

Consider the Many-Worlds approach. You have a wavefunction (an immensely complicated one, of course). Your amplitude for having heard a click steadily grows in magnitude. No paradox if you look at it like this.

4

Your statements treat the quantum mechanical distribution as physical, whereas it is a mathematical function fitting the boundary condition of your experiment, i.e. it is the mathematical function describing a particle's probability of decay. Probabilities are the same in classical mechanics, in economics in gambling, in population interactions. Take the ...

2

I think that “listening” even in the case of silence is already the measurement. You can only hope to hear something when there is a medium (air) that will carry the sound waves. This medium causes a continuous interaction between you and the Geiger counter. Only without the medium there is no interaction but then you can also not tell that the Geiger ...

9

My take on this is that in the original thought experiment, you don't get to monitor the detector. When the detector detects, it kills the cat. But it doesn't tell you then. You only find out when you open the box. If it tells you immediately, then you know immediately. And then there's the question whether the detector detects 100%. If the Geiger counter ...

16

Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of measurement, and a variable with a corresponding hermitian operator $x, p, H$, etc is measured. However there are questions which can be asked, like this one, which ...

9

No, the detector is not always collapsing the state. When the particle is in an undecayed state its wave function is physically localised with a vanishingly small amplitude in the region of the detector, so the detector doesn't interact with it and isn't 'always' measuring it. It is only when the particle's state evolves to the point at which it has a ...

1

The idea of the collapse of the state is not a fundamental part of quantum mechanics. It's a feature of the Copenhagen interpretation (CI). The CI is not the only way to think about quantum mechanics. Even within CI, it's not necessarily true that measurement must disturb the system, depending on what you mean by "measurement" and "disturb." In the Stern-...

1

Answering the question in the title: a measurement process is intrinsically non-unitary. One way to see this is to realise that the unitarity of a process is equivalent to its being reversible. A measurement process is intrinsically non-reversible, as some information gets lost. For example, measuring $(|0\rangle+|1\rangle)/\sqrt2$ in the computational ...

0

As the other answers point out, your question is very confusing and it's not entire clear what you're asking. I think what you're asking is what happens if you measure whether the system is in any of the states $|\psi_i\rangle$ with $i \geq 3$, i.e. you measure the value of the observable that is the projection operator \hat{P} = \sum_{i \geq 3} |\psi_i\...

2

In QM, a measurement always amounts to a choice of a basis (or more generally, a set of projectors summing to the identity) with respect to which the wavefunction collapses. In other words, any measurement of a state $|\Psi\rangle$ can be described via a set of orthogonal projectors $P_k$ such that $\sum_k P_k=I$, by writing the state as \$|\Psi\rangle=\sum_k ...

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