Consider general state of a system with spin-$1/2$

$$ \psi = \frac{1}{\sqrt{2}}\left[\phi_{+1/2}(x) \left( \begin{array}{cc} 1 \\ 0 \\ \end{array} \right) + \phi_{-1/2}(x) \left( \begin{array}{cc} 0 \\ 1 \\ \end{array} \right)\right]. $$

If I created an apparatus which measures only spin would that imply that the $x$-dependence of $\psi$ remains uncertain after measurement? E.g. if the system appeared to be in $+1/2$ state then the coordinate part of wave function (w.f.) didn't undergo collapse

$$ \psi' = \phi_{+1/2}(x) \left( \begin{array}{cc} 1 \\ 0 \\ \end{array} \right) .$$

Maybe such an apparatus cannot be created (why?) or if one "part" of w.f. is measured then all other "parts" necessarily collapse too, but their values go by unnoticed?


In a more general way: I have a set of "uncoupled" observables $\alpha$, $\beta$ (say spin and isospin). If I measure $\alpha$ does that mean that $\beta$ will also take a definite value (will be measured indirectly).


1 Answer 1


In practice, the apparatus measuring the spin should be localized somewhere in space (it cannot fill the whole universe!) and this fact implies that you always make a measurement of position (actually very rough in general), even if you are measuring the spin. Suppose that $\Omega \subset R^3$ is the bounded region in $R^3$ where the apparatus is localized. The simplest (naive) mathematical model of the apparatus I could imagine is the following.

The YES-NO observable associated with the apparatus measuring, say, if the spin is directed along z+, has the form of the orthogonal projector: $$P_{\Omega} \otimes P_{z+}$$ Here $P_{z^+} = |z+\rangle \langle z+|$ is the obvious projector in $C^2$ along the states with spin $z+$-directed , whereas $P_\Omega$ is the operator (orthogonal projector in $L^2(R^3)$) $$(P_\Omega \psi)(x) = \chi_\Omega (x) \psi(x)\:.$$ This observable admits two values (its eigenvalues) $0=$ NO and $1=$YES. YES means that the particle is found in $\Omega$ AND the spin is found to be directed along $z+$. NO means that the the particle is not found in $\Omega$ OR the spin is not along $z+$.

There is another elementary YES-NO observable associated with the spin detected along the direction $-z$, with analogous meaning. It is the orthogonal projector:

$$P_{\Omega} \otimes P_{z-}\:.$$

The observable associated with the spin along $z$ --referring to this experiment-- is not the standard operator $S_z = \sigma_z/2$ (I am assuming $\hbar =1$). It is instead constructed, via spectral decomposition, taking the above elementary observables (projectors) into account and combining them with the corresponding values of the spin (which turn out to be the eigenvalues of the overall observable).

$$\Gamma_{z,\Omega} = \frac{1}{2}P_{\Omega} \otimes P_{z+} - \frac{1}{2} P_{\Omega} \otimes P_{z-} = P_\Omega \otimes S_z\:.$$

You see that it includes a rough measurement of the position: it just checks if the position of the particle is in $\Omega$. To measure the spin of the particle the supports of the components $\phi_i$ must have a non-negligible intersection with $\Omega$. In general the measurement procedure of the spin, for instance along $z+$, even affects the surviving component $\phi_{+1/2}$. You see that, only if the support of $\phi_{+1/2}$ is completely included in $\Omega$, the wavefunction is not affected by the measurement of the spin, otherwise, after the procedure (supposing to have found spin $+1/2$), the state, up to a normalization constant, is described by:

$$P_\Omega \phi_{+1/2} \otimes |z+\rangle \:.$$

It is questionable if we have defined an observable $\Gamma_{z,\Omega}$ in that way. The point is that $\Gamma_{z,\Omega}$ admits a third eigenvalue, $0$, associated with the projector $P_{R^3-\Omega} \otimes I$. Actually, there is no real measurement in the region $R^3-\Omega$, since we are not assuming that there are detectors therein. We are simply using the argument: "if the particle is not found in $\Omega$ it must be found outside it". For several reasons I am always a bit suspicious to this sort of formal arguments. A more physically safe interpretation could be that we are performing a conditioned measurement of the spin. $P_\Omega$ is nothing but a filter: only the particles which pass through it are measured.

  • $\begingroup$ Thanks for a detailed answer! I would also like to see your thoughts on a more general problem, please see upd $\endgroup$
    – xaxa
    Commented Dec 8, 2013 at 18:19
  • $\begingroup$ Concerning your upd, take into account that measurements of positions are very particular because instruments devoted to measure any observable are, after all, localized in space. So, in practice, some sort of position measurement is always present in any measurement of any observable. If, instead, you consider other observables like those you mentioned, spin and isospin, I do not know what actually happens in real experiments (I am a mathematical physicist). Certainly, if you consider spin-isospin entangled states, measurements of spin may include indirect measurements of isospin. $\endgroup$ Commented Dec 8, 2013 at 20:41
  • $\begingroup$ Taking away all complications due to measuring devices (so that observing $\alpha$ and $\beta$ does not interfere), what does a standard QM interpretation say about this? Or does it state that such an ideal measuring device is impossible to build $\endgroup$
    – xaxa
    Commented Dec 8, 2013 at 21:01
  • $\begingroup$ Suppose the Hilbert space is $H \otimes K$ ($H$ for spin and $K$ for isospin). For this system the spin observable $S_z$ along $z$ is not $\sigma_z/2$, but is $\sigma_z/2 \otimes I$. The projector on the $+1/2$ eigenspace is $P_+\otimes I$. Let the initially measured state be $\psi := |+>\otimes |+> + |->\otimes |->$ and that, measuring $S_z$ you find $+1/2$. QM says that the state after the measurement is $P_+\otimes I \psi = |+>\otimes |+>$. This is the same result as if measuring the isospin $T_3= I \otimes \sigma_3/2$. So, measuring the spin you also have measured the isospin. $\endgroup$ Commented Dec 8, 2013 at 21:20
  • $\begingroup$ In the case I have discussed it is impossible measuring the spin without measuring the isospin simultaneously. This is a direct consequence of basic assumptions of QM. $\endgroup$ Commented Dec 8, 2013 at 21:24

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