Are all (pure) states physically realizable?

[einheit]

TL;DR Summary

For example, given a spin-1/2 particle, a pure state orthogonal to its two-dimensional spin subspace would have probability zero for both spin-up and spin-down, which doesn't seem possible.

To elaborate that summary a bit, suppose $\mathcal H$ is the Hilbert space of the particle, with $\mathcal{H}_2\subseteq\mathcal{H}$ its two-dimensional spin subspace. Now consider any $|x\rangle\in\mathcal{H}$ such that $|x\rangle\perp\mathcal{H}_2$, i.e., $\forall ~ |s\rangle\in\mathcal{H}_2~:~\langle x|s\rangle=0$. Then, as per the Summary, the spin-up and spin-down probabilities are both zero, no matter what basis you choose. So that seems unphysical since the particle must possesses some spin value when measured. But then our $|x\rangle$ doesn't represent a physical state, whereby the complete (containing all limits) $\mathcal H$ isn't precisely the particle's state space. On the other hand, if that were really correct, I'm sure I'd have learned/read/whatever about it. So what am I getting wrong?

 

[vanhees71]

A spin-1/2 particle has only a two-dimensional spin space. There's simply nothing else. Also if you measure a spin component you get either $\hbar/2$ or $-\hbar/2$ and for any state the probabilities for the occurance of these measurement outcomes of course must add to 1.

To answer your question. In principle you have a Hilbert space for a given quantum system. In principle with two vectors (representing pure states) also any of their superpositions also represents a pure state, but there are indeed exceptions, dictated by symmetries. E.g., it's impossible to have superpositions of angular momentum states of integer and half-integer representations or superpositions of states referring to states with different electric charge. This is known as superselection rules.

 

[PeroK]

Summary:: For example, given a spin-1/2 particle, a pure state orthogonal to its two-dimensional spin subspace would have probability zero for both spin-up and spin-down, which doesn't seem possible.

To elaborate that summary a bit, suppose $\mathcal H$ is the Hilbert space of the particle, with $\mathcal{H}_2\subseteq\mathcal{H}$ its two-dimensional spin subspace. Now consider any $|x\rangle\in\mathcal{H}$ such that $|x\rangle\perp\mathcal{H}_2$, i.e., $\forall ~ |s\rangle\in\mathcal{H}_2~:~\langle x|s\rangle=0$. Then, as per the Summary, the spin-up and spin-down probabilities are both zero, no matter what basis you choose. So that seems unphysical since the particle must possesses some spin value when measured. But then our $|x\rangle$ doesn't represent a physical state, whereby the complete (containing all limits) $\mathcal H$ isn't precisely the particle's state space. On the other hand, if that were really correct, I'm sure I'd have learned/read/whatever about it. So what am I getting wrong?

What sort of state is $|x\rangle$ supposed to be? Are you thinking of combining the spin state with the state that describes the other dynamics properties of the particle, such as position and momentum?

 

[einheit]

A spin-1/2 particle has only a two-dimensional spin space. There's simply nothing else.

That alone would answer my question. But excuse me for disagreeing. For example, an electron is spin-1/2 but also has energy. Indeed, just about all systems have Hamiltonians, whereby there's pretty much always "something else." Usually lots of other things, i.e., other observables.

And it's all those other observables I was thinking about. A system's Hilbert space $\mathcal{H}$ would then be the tensor product of subspaces, each spanning the eigenvectors of one observable, with spin being just one of them. And I guess there's the way more formal/rigorous GNS construction, if we want to get into the C*-algebra of observables characterizing a given system, but that's hopefully not necessary for the purposes of this discussion.

And now we could choose any observable, and prepare a pure state $|x\rangle$ orthogonal to that observable's subspace, whereby there'd now be zero probability for subsequently observing every one of that observable's eigenvalues. Spin just seemed like the easiest illustration. And again, by the way, this argument's pretty obviously incorrect, somehow, but I'm not seeing how.

 

[PeroK]

That alone would answer my question. But excuse me for disagreeing. For example, and electron is spin-1/2 but also has energy. Indeed, just about all systems have Hamiltonians, whereby there's pretty much always "something else." Usually lots of other things, i.e., other observables.

And it's all those other observables I was thinking about. A system's Hilbert space $\mathcal H$ would then be the tensor product of subspaces, each spanning the eigenvectors of one observable, with spin being just one of them. And I guess there's the way more formal/rigorous GNS construction, if we want to get into the C*-algebra of observables characterizing a given system, but that's hopefully not necessary for the purposes of this discussion.

And now we could choose any observable, and prepare a pure state $|x\rangle$ orthogonal to that observable's subspace, whereby there'd now be zero probability for subsequently observing every one of that observable's eigenvalues. Spin just seemed like the easiest illustration. And again, by the way, this argument's pretty obviously incorrect, somehow, but I'm not seeing how.

If you describe the total state of an electron as $|x\rangle \otimes |s\rangle$, where $|s\rangle$ is the spin state, then $|s\rangle = 0$ is not a valid spin state, hence not a valid state for a spin 1/2 particle.

 

[einheit]

What sort of state is $|x\rangle$ supposed to be? Are you thinking of combining the spin state with the state that describes the other dynamics properties of the particle, such as position and momentum?

Yeah, exactly. Please see reply to vanhees71, where I elaborated all that, partly in reply to your post. But I meant that I'm "combining" all the system's observables to construct the Hilbert space representing its space of physical states. And then, presumably, any pure state $|x\rangle\in\mathcal{H}$ can in principle be prepared by some preparation procedure. So then I'm saying: choose an observable (e.g., spin) and then prepare an $|x\rangle$ orthogonal to that observable's subspace. Exactly what "$|x\rangle$ is supposed to be" (to quote you), I don't know. Any/all $|x\rangle\in\mathcal{H}$ should be preparable, so just go prepare it. And if you're saying it can't be prepared, then you just proved my conjecture -- there exist (normalized) vectors $|x\rangle\in\mathcal H$ that don't represent physical states of the system. So you can't say that, I don't think. Nevertheless, like I've said several times, this argument's got to be wrong, somehow, but I'm just not seeing the flaw(s).

 

[PeroK]

Nevertheless, like I've said several times, this argument's got to be wrong, somehow, but I'm just not seeing the flaw(s).

The flaw is having a spin state of zero for a spin 1/2 particle. Note that you are actualy talking about using the zero vector, rather than a vector that represents a spin measurement of $0$. States must be normalisable for one thing and any state of the form $|x \rangle \otimes |0 \rangle$ has zero norm.

 

[einheit]

If you describe the total state of an electron as $|x\rangle \otimes |s\rangle$, where $|s\rangle$ is the spin state, then $|s\rangle = 0$ is not a valid spin state, hence not a valid state for a spin 1/2 particle.

Thanks, PeroK, I guess that's my problem.

 

[einheit]

The flaw is having a spin state of zero for a spin 1/2 particle. Note that you are actualy talking about using the sero vector, rather than a vector that reprsents a spin measurement of $0$. States must be normalisable for one thing and any state of the form $|x \rangle \otimes |0 \rangle$ has zero norm.

Thanks, again, that makes the problem even clearer.

 

[Gaussian97]

I think there are 3 key points here.
1) Are all (pure) states physically realizable?
The answer is no, not all states are physically realizable. For example, you cannot mix integer and half-integer spin states. And in a theory with Galilean invariance, you cannot mix different mass states.

2) The second point of discussion is your particular argument, which I think has been the main discussion until now and I won't add more arguments.

3) Another important thing is that even if your argument was correct, and you could find such state $\left| x \right>$, it doesn't imply that it is not physical.

 

[vanhees71]

That alone would answer my question. But excuse me for disagreeing. For example, an electron is spin-1/2 but also has energy. Indeed, just about all systems have Hamiltonians, whereby there's pretty much always "something else." Usually lots of other things, i.e., other observables.

And it's all those other observables I was thinking about. A system's Hilbert space $\mathcal{H}$ would then be the tensor product of subspaces, each spanning the eigenvectors of one observable, with spin being just one of them. And I guess there's the way more formal/rigorous GNS construction, if we want to get into the C*-algebra of observables characterizing a given system, but that's hopefully not necessary for the purposes of this discussion.

And now we could choose any observable, and prepare a pure state $|x\rangle$ orthogonal to that observable's subspace, whereby there'd now be zero probability for subsequently observing every one of that observable's eigenvalues. Spin just seemed like the easiest illustration. And again, by the way, this argument's pretty obviously incorrect, somehow, but I'm not seeing how.

Then you need to take into account the complete description. In non-relativistic QM you need to choose a complete set of compatible (independent) observables to define a complete orthonormal basis. Common choices are the position-spin or the momentum spin basis. The complete sets here are $\hat{\vec{x}},\hat{s}_3$ or $\hat{\vec{p}},\hat{s}_3$. Formally you have of course for the eigenvectors
$$|\vec{x},\sigma \rangle = |x_1 \rangle \otimes |x_2 \rangle \otimes x_3 \rangle \otimes |\sigma \rangle.$$

I don't understand what you mean with to choose a pure state "orthogonal to that observable's subspace". Can you give a concrete example for what you mean by such a pure state?

 

[mattt]

@vanhees71 , PeroK solved einheit's doubt in post n° 7.