Suppose we have a spin $\frac{1}{2}$ particle in the spin-up state along the $z$-axis, $\lvert \uparrow \rangle$, and after $t$ seconds of evolution under the Schrodinger equation it is in state $\alpha\lvert \uparrow \rangle + \beta\lvert \downarrow \rangle$. Now, according to the Copenhagen interpretation, if we measure its spin along the $z$-axis, there is a probability $\lvert \alpha \rvert ^{2}$ that the state collapses to $\lvert \uparrow \rangle$ and a probability $\lvert \beta \rvert ^{2}$ that it collapses to $\lvert \downarrow \rangle$. Then, after a further $t$ seconds, the state will be $\alpha\lvert \uparrow \rangle + \beta\lvert \downarrow \rangle$ with probability $\lvert \alpha \rvert ^{2}$ and $\gamma\lvert \uparrow \rangle + \delta\lvert \downarrow \rangle$ with probability $\lvert \beta \rvert ^{2}$ for some $\gamma$ and $\delta$. Therefore, if we measure the particle's spin along the $z$-axis again, the probability that we will find it to be spin-up is $\lvert \alpha \rvert ^{4}+\lvert \beta \rvert ^{2}\lvert \gamma\rvert ^{2}$.

I am trying to understand the Everettian interpretation of quantum mechanics. According to the Everettian view as I understand it, after we make our first "measurement" in the scenario above, our measurement device and the particle would be entangled with a joint state of $\alpha\lvert \uparrow \rangle\lvert\frac{1}{2}\rangle + \beta\lvert \downarrow \rangle\lvert-\frac{1}{2}\rangle$, where $\lvert\frac{1}{2}\rangle$ and $\lvert-\frac{1}{2}\rangle$ are supposed to represent the states of the measuring device corresponding to it having measured spin-up and spin-down respectively. Before our second measurement then, our system would presumably be in the state $\alpha^{2}\lvert \uparrow \rangle \lvert\frac{1}{2}\rangle + \beta\gamma\lvert \uparrow \rangle\lvert-\frac{1}{2}\rangle + \alpha\beta\lvert \downarrow \rangle \lvert\frac{1}{2}\rangle + \beta\delta\lvert \downarrow \rangle\lvert-\frac{1}{2}\rangle$. Now, when we make our second measurement, the system is still supposed to evolve under the Schrodinger equation, which is linear, so to work out what will happen, we can just work out what state we would have ended up with if the state had initially been each of the four parts of our superposition and then take the appropriate linear combination. For example, if the initial state had been $\beta\gamma\lvert \uparrow \rangle\lvert-\frac{1}{2}\rangle$, then, we would presumably have ended up with the state $\beta\gamma\lvert \uparrow \rangle\lvert\frac{1}{2}\rangle$, up to some phase, after we brought our measuring device and our particle back together. By this logic, it would seem that we would end up in the state $\left(\alpha^{2} + e^{i\theta_{1}}\beta\gamma\right)\lvert \uparrow \rangle\lvert\frac{1}{2}\rangle + \left(e^{i\theta_{2}}\alpha\beta + e^{i\theta_{3}}\beta\delta\right)\lvert \downarrow \rangle\lvert-\frac{1}{2}\rangle$ for some $\theta_{1}$, $\theta_{2}$ and $\theta_{3}$. The probability of measuring spin-up on our second measurement would apparently then be $\lvert\alpha^{2} + e^{i\theta_{1}}\beta\gamma\rvert^{2}$.

If so, the predictions of the Copenhagen and Everettian interpretations of quantum mechanics could be in agreement with one another only if the Schrodinger equation made sure that $\theta_{1}$ was such that $\lvert \alpha \rvert ^{4}+\lvert \beta \rvert ^{2}\lvert \gamma\rvert ^{2} = \lvert\alpha^{2} + e^{i\theta_{1}}\beta\gamma\rvert^{2}$, i.e. $\Re\left({\alpha^{*}}^{2}e^{i\theta_{1}}\beta\gamma\right) = 0$. My questions then are:

  1. Is my above argument correct?
  2. If so, does the Schrodinger equation make sure the phases are such that the predictions of the two interpretations agree and how can I see that it does so or does not?

2 Answers 2


The many worlds interpretation (MWI) involves working out the implications of quantum mechanics without modifying it to include collapse.

In the MWI if you make a measurement in general the measurement device becomes correlated not just with the measured system but with many other systems, such as photons reflected from the measurement device, air molecules, grains of dust and so on: these other systems are called the environment. So if you have a system $S$, a measurement device $M$ and an environment $E$ you actually end up with the evolution: $$(\alpha|\uparrow\rangle_S+\beta|\downarrow\rangle_S)|0\rangle_M|0\rangle_E\to \alpha|\uparrow\rangle_S|\uparrow\rangle_M|\uparrow\rangle_E+\beta|\downarrow\rangle_S|\downarrow\rangle_M|\downarrow\rangle_E$$

There is no unitary evolution on $S$ and $M$ that will undo their entanglement with the environment and the environment isn't under our control. There is a large literature on this effect under the name of decoherence:



So doing the kind of experiment you describe requires isolating $S$ and $M$ from the environment In particular it means that there can't be any record at all outside of $S$ and $M$ of the results of the previous measurement.

You ask about the predictions of the Copenhagen interpretation (CI). The CI doesn't specify the mechanism by which the collapse takes place, which makes its predictions vague at best. There are theories that specify mechanisms of collapse, such as spontaneous collapse theories, and those theories in general make different predictions than quantum mechanics:


There are unsolved problems for such theories that don't exist in the MWI:


  • $\begingroup$ Is the idea then that there is no way, even in principle, to have S and M be isolated from the wider environment such that there is no record of prior measurements or is it theoretically possible to construct such an experiment and test the two different predictions? $\endgroup$
    – MBar2269
    Nov 28, 2023 at 17:35
  • $\begingroup$ You can get arbitrarily close to isolating a measurement device and system from the environment in at least some cases. However, the CI doesn't give a precise account of what counts as a measurement so it's not clear that they would count that as a test of the CI. For example, if you're measuring an atom and using another atom as the measurement device and isolate them both, then CI advocates could say that doesn't count as a measurement. $\endgroup$
    – alanf
    Nov 28, 2023 at 18:09

The right way to discuss this calculation is with density matrices. You also seem to have missed the second measurement in your Everettian version. Let us do the Everettian version again us go step by step:

  1. Prepare state $|\uparrow\rangle$
  2. Evolve, obtain $\alpha |\uparrow\rangle+\beta|\downarrow\rangle$
  3. Measure, entangled with the system $\alpha |\uparrow\rangle|\tfrac12\rangle+\beta|\downarrow\rangle|-\tfrac12\rangle$
  4. Evolve, we obtain $\alpha^2 |\uparrow\rangle|\tfrac12\rangle+\alpha\beta |\downarrow\rangle|\tfrac12\rangle+\beta\gamma|\uparrow\rangle|-\tfrac12\rangle+\beta\delta|\downarrow\rangle|-\tfrac12\rangle$
  5. Measure once more and entangle, $|\Psi\rangle=\alpha^2 |\uparrow\rangle|\tfrac12,\tfrac12^*\rangle+\alpha\beta |\downarrow\rangle|\tfrac12,-\tfrac12^*\rangle+\beta\gamma|\uparrow\rangle|-\tfrac12,\tfrac12^*\rangle+\beta\delta|\downarrow\rangle|-\tfrac12,-\tfrac12^*\rangle$ where the second number with the $*$ is the second measurement.
  6. Write density matrix $\rho=|\Psi\rangle\langle\Psi|$
  7. Trace out the measurement device $\tilde\rho=\langle\tfrac12\tfrac12^*|\rho|\tfrac12\tfrac12^*\rangle+\langle\tfrac12,-\tfrac12^*|\rho|\tfrac12,-\tfrac12^*\rangle+\langle-\tfrac12,\tfrac12^*|\rho|-\tfrac12,\tfrac12^*\rangle+\langle-\tfrac12,-\tfrac12^*|\rho|-\tfrac12,-\tfrac12^*\rangle$
  8. We get $\tilde\rho=(|\alpha|^4+|\beta|^2|\gamma|^2)|\uparrow\rangle\langle\uparrow|+|\beta|^2(|\alpha|^2+|\delta|^2)|\downarrow\rangle\langle\downarrow|$ which is diagonal as expected
  9. Read probability to have $|\uparrow\rangle$ directly from the diagonal, that is $|\alpha|^4+|\beta|^2|\gamma|^2$

This result is exactly the same as the one obtained from your Copenhagen calculation.

  • $\begingroup$ An important difference between your calculation and mine seems to be that you assume that the state of the measuring device after the second measurement depends on both measurement outcomes, whereas I assumed it only depends on the most recent measurement so, for instance, $\lvert \frac{1}{2}, {\frac{1}{2}}^{*} \rangle$ and $\lvert \frac{1}{2}, {-\frac{1}{2}}^{*} \rangle$ would be the same state under my assumption. Am I right in thinking that's a key difference? If so, could you justify your assumption? $\endgroup$
    – MBar2269
    Nov 27, 2023 at 23:08
  • $\begingroup$ @MBar2269 It seems indeed that here is where we diverge but I am not sure what are our differences. How is $|\tfrac12,\tfrac12^*\rangle$ the same as $|\tfrac12,-\tfrac12^*\rangle$? in the former you get the state whose probability you are trying to determine. In the latter, you do not because you measure -1/2 or downstate. $\endgroup$
    – Mauricio
    Nov 27, 2023 at 23:31
  • $\begingroup$ I am sorry but I cannot edit my comment. I meant to say $\lvert \frac{1}{2}, {\frac{1}{2}}^{*} \rangle$ and $\lvert -\frac{1}{2}, {\frac{1}{2}}^{*} \rangle$ would be the same. The picture I have in my head for the measuring device is of a needle pointing either to $\frac{1}{2}$ or $-\frac{1}{2}$, depending on what it has most recently measured, leaving no trace of any prior measurements. $\endgroup$
    – MBar2269
    Nov 27, 2023 at 23:44
  • $\begingroup$ I guess I should have traced out the first measurement and the quantum state, in the end what you want is to focus on the last measurement device. The point is that once you measure a second time, there is a distinction between these two states or "worlds" in terms of probabilities. $\endgroup$
    – Mauricio
    Nov 28, 2023 at 9:24
  • $\begingroup$ So $\lvert| \frac{1}{2}, {\frac{1}{2}}^{*}$ and $-\lvert| \frac{1}{2}, {\frac{1}{2}}^{*}$ should be considered separate, orthogonal states because they correspond to different worlds? Don't you then have to add some postulate to specify when a new world is created? $\endgroup$
    – MBar2269
    Nov 28, 2023 at 14:30

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