0
$\begingroup$

Let $A$ an $B$ be two discrete observables (like spins). When exactly and why we have to consider their tensor product when talking about the mutual observation of the corresponding phenomena?

$\endgroup$
  • $\begingroup$ Yes. My question is a duplicate or even a moreplicate. The reason for asking is exactly that I have never been given an answer. Answers like (1) "it is reasonable to... " or (2) "it is an axiom..." do not satisfy me. Do you,physicists, know the answer or do you not? $\endgroup$ – zoli Feb 25 '15 at 20:24
  • $\begingroup$ @zoli what is "mutual observation"? The observables $A$ and $B$ belong to different particles? If they belong to one and the same particle there is no tensor product there. $\endgroup$ – Sofia Feb 25 '15 at 20:24
  • $\begingroup$ @Sofia: Of course it is not about one particle bothered twice. It is about a pair of particles unleashed and then measured independently. When to tensor and when not to tensor. In mathemathics we cannot be so uncertain about our statements. (With all do respect...Sofia.) $\endgroup$ – zoli Feb 25 '15 at 20:29
  • $\begingroup$ @zoli: If you want just mathematical reasons, here they are: physics.stackexchange.com/q/148378 - putting it very differently, you want two observables that are measured on two different subsystems to have independent outcomes, if the states are uncorrelated. To have this, you need the product measure, which descends to a tensor product of Hilbert spaces. But all of this has been said in various threads... $\endgroup$ – Martin Feb 25 '15 at 21:08
0
$\begingroup$

For describing the behavior of a particle we usually write the Schrodinger equation, or Dirac equation, or other. All the possible solutions of the equation form a space of functions with certain properties, and we name such a space of functions Hilbert space. Now, if we have two particles, $1$ and $2$, which don't interact, behave as if none of them is aware if the presence of the other. The behavior of the two particles can be described, by anyone of the functions in the Hilbert space $\mathcal H_1$ for the particle $1$, and anyone of the functions in $\mathcal H_2$ for the particle $2$. We denote this situation by saying that the Hilbert space of the two particles is $\mathcal H = \mathcal H_1 \otimes \mathcal H_2$. This is what means this notation.

So, picking a state $|a\rangle$ from $\mathcal H_1$ and a state $|b\rangle$ from $\mathcal H_2$, the state $|a\rangle \otimes |b\rangle$, which as explained above belongs to $\mathcal H_1 \otimes \mathcal H_2$, can be very well a state describing the behavior of the pair of particles $1$ and $2$.

However we may have a more complicated situation, name entanglement. An entanglement has no classical equivalent, it is a topic of which you have to read separately. In short, without interacting with one another by any classical field, the particle aren't though independent. If you heard of the spin singlet, the state of the two fermions is

$$|S\rangle = \frac {|\uparrow\rangle_1 \otimes |\downarrow\rangle_2 - |\uparrow\rangle_2 \otimes |\downarrow\rangle_1}{\sqrt {2}}.$$

Well, in order to economize the subscripts $1$ and $2$ we use to write first the particle $1$ and second the particle $2$,

$$|S\rangle = \frac {|\uparrow\rangle \otimes |\downarrow\rangle - |\downarrow\rangle \otimes |\uparrow\rangle}{\sqrt {2}}.$$

You see the state $|S\rangle$ still belongs to the space $\mathcal H_1 \otimes \mathcal H_2$, but it is no more the simple tensor product of one element from $\mathcal H_1$ with one element from $\mathcal H_2$, but a superposition of two such products.

Well, I hope that this time the situation is more clear.

$\endgroup$
  • $\begingroup$ "If one understands a physical phenomenon, one is able to explain it without formulas." I've read that statement recently... $\endgroup$ – zoli Feb 25 '15 at 21:55
  • $\begingroup$ @zoli I understand that my answer doesn't satisfy you? Then tell me what exactly. $\endgroup$ – Sofia Feb 25 '15 at 22:03
  • $\begingroup$ It does not. But this is not your fault. So I accept your answer with one condition: you, please, write to me an email. And we could continue this unfortunate dispute. $\endgroup$ – zoli Feb 25 '15 at 23:31
  • $\begingroup$ @zoli you don't need to put me conditions. I understand that there is a problem if you say "unfortunate dispute". We help people gladly. NOW, there is a chat room. Do you want that we move there? $\endgroup$ – Sofia Feb 25 '15 at 23:52
  • $\begingroup$ @zoli if you insist on email, so be it. But, give me an email-address. $\endgroup$ – Sofia Feb 25 '15 at 23:54

Not the answer you're looking for? Browse other questions tagged or ask your own question.