- #1
Peter Morgan
Gold Member
- 274
- 77
One sees "Collapse" language all the time, and yet it seems there's a very simple argument that shows that it's not necessary. Suppose we measure ##\hat A## twice, at an earlier and at a later time, then the joint probability density for the measurement results being ##u## and ##v## respectively, in a state ##\rho(\hat A)=\mathsf{Tr}[\hat A\hat\rho]## is ##\rho\left(\delta(\hat A-u)\delta(\hat A-v)\right)##. If we take the Fourier transform of that and perform a single change of variable, we obtain
\begin{eqnarray*}
\quad\rho\!\left(\delta(\hat A-u)\delta(\hat A-v)\right)
&=&\int\hspace{-0.7em}\int\!\!\rho\!\left(\mathrm{e}^{\mathrm{j}\lambda\hat A}\mathrm{e}^{\mathrm{j}\mu\hat A}\right)
\mathrm{e}^{-\mathrm{j}\lambda u-\mathrm{j}\mu v}\frac{\mathrm{d}\lambda}{2\pi}\frac{\mathrm{d}\mu}{2\pi}\cr
&=&\int\hspace{-0.7em}\int\!\!\rho\!\left(\mathrm{e}^{\mathrm{j}\alpha\hat A}\right)
\mathrm{e}^{-\mathrm{j}\alpha u-\mathrm{j}\mu(v-u)}\frac{\mathrm{d}\alpha}{2\pi}\frac{\mathrm{d}\mu}{2\pi}
\quad\mbox{(substituting $\lambda=\alpha-\mu$)}\cr
&=&\rho\!\left(\delta(\hat A-u)\right)\delta(v-u)
\end{eqnarray*}
[we can think of the integrands above as generating functions for moments ##\rho\left(\hat A^n\right)##.] This asserts that there's a 100% correlation between the results one obtains for the two measurements, because ##v## must be exactly equal to ##u##, and yet there's been no "collapse" of the state ##\rho(\cdot)##, we've just used the one state at different times. If ##\hat A## commutes with the Hamiltonian, the same measurement procedure will give the same result, whereas if ##[\hat A,\hat H]\not=0##, a different measurement procedure will have to be used to give the same measurement result as was obtained at the earlier time, but if we make the same measurement at different times we will, according to this calculation, obtain precisely the same result. Conversely, if we don't obtain the same, perfectly correlated results, we haven't made the same measurement (the proof of how well we understand what measurements we've made is in the measurement results).
This is a very simple calculation, but it seems to say that just using a multi-time formalism correctly (which is very natural for quantum fields such as quantized EM/quantum optics, but trickier, though it can be done, if we use a one-time Hilbert space unitarily evolved to different times) is enough, we don't ever have to collapse the wavefunction. Note that although the wavefunction does not determine what the measurement result for either the first or the second measurement will be, the wave function determines the conditional probability density for both the first and the second measurement, given the measurement result of either one, which does determine the measurement result of the other. Indeed, adopting this conditional probability approach is perhaps less likely to lead our intuition astray, insofar as one might not want to say that the second measurement "collapsed" the result for the first measurement.
Of course it has been argued before that no-collapse interpretations are possible, but this is an entirely mathematical argument that seems to me to have relatively little metaphysical/ontological baggage. I haven't seen the argument reduced to the three line derivation above (short is how I like 'em), but is it out there in a closely comparable form?
\begin{eqnarray*}
\quad\rho\!\left(\delta(\hat A-u)\delta(\hat A-v)\right)
&=&\int\hspace{-0.7em}\int\!\!\rho\!\left(\mathrm{e}^{\mathrm{j}\lambda\hat A}\mathrm{e}^{\mathrm{j}\mu\hat A}\right)
\mathrm{e}^{-\mathrm{j}\lambda u-\mathrm{j}\mu v}\frac{\mathrm{d}\lambda}{2\pi}\frac{\mathrm{d}\mu}{2\pi}\cr
&=&\int\hspace{-0.7em}\int\!\!\rho\!\left(\mathrm{e}^{\mathrm{j}\alpha\hat A}\right)
\mathrm{e}^{-\mathrm{j}\alpha u-\mathrm{j}\mu(v-u)}\frac{\mathrm{d}\alpha}{2\pi}\frac{\mathrm{d}\mu}{2\pi}
\quad\mbox{(substituting $\lambda=\alpha-\mu$)}\cr
&=&\rho\!\left(\delta(\hat A-u)\right)\delta(v-u)
\end{eqnarray*}
[we can think of the integrands above as generating functions for moments ##\rho\left(\hat A^n\right)##.] This asserts that there's a 100% correlation between the results one obtains for the two measurements, because ##v## must be exactly equal to ##u##, and yet there's been no "collapse" of the state ##\rho(\cdot)##, we've just used the one state at different times. If ##\hat A## commutes with the Hamiltonian, the same measurement procedure will give the same result, whereas if ##[\hat A,\hat H]\not=0##, a different measurement procedure will have to be used to give the same measurement result as was obtained at the earlier time, but if we make the same measurement at different times we will, according to this calculation, obtain precisely the same result. Conversely, if we don't obtain the same, perfectly correlated results, we haven't made the same measurement (the proof of how well we understand what measurements we've made is in the measurement results).
This is a very simple calculation, but it seems to say that just using a multi-time formalism correctly (which is very natural for quantum fields such as quantized EM/quantum optics, but trickier, though it can be done, if we use a one-time Hilbert space unitarily evolved to different times) is enough, we don't ever have to collapse the wavefunction. Note that although the wavefunction does not determine what the measurement result for either the first or the second measurement will be, the wave function determines the conditional probability density for both the first and the second measurement, given the measurement result of either one, which does determine the measurement result of the other. Indeed, adopting this conditional probability approach is perhaps less likely to lead our intuition astray, insofar as one might not want to say that the second measurement "collapsed" the result for the first measurement.
Of course it has been argued before that no-collapse interpretations are possible, but this is an entirely mathematical argument that seems to me to have relatively little metaphysical/ontological baggage. I haven't seen the argument reduced to the three line derivation above (short is how I like 'em), but is it out there in a closely comparable form?