Is Quantum Mechanics Self-Consistent?

Question

Quantum mechanics postulates the following (and please correct me if I'm wrong):

• Every physical state of a system is uniquely identified with a ray in a Hilbert space $|\Psi\rangle \in \mathcal{H}$ and viceversa

• Every physical quantity corresponds to an Hermitian operator $\hat{A}$ with spectral projectors $\hat{P_{k}}$, such that $\langle\Psi|\hat{P_{k}}|\Psi\rangle$ is the probability that a measurement of $\hat{A}$ on the state $|\Psi\rangle$ has outcome $k$

• If $|\Psi\rangle_{t}$ is the state of an isolated system, then the state at a later time is given by $\hat{U}|\Psi\rangle_{t}$ where $\hat{U}$ is a unitary operator

• If a measurement of the quantity $\hat{A}$ has outcome $k$ on a state $|\Psi\rangle$, the state after the measurement is given by $\frac{\hat{P_{k}}|\Psi\rangle}{\sqrt{\langle\Psi|\hat{P_{k}}|\Psi\rangle}}$

• Let $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ be the Hilbert state spaces of two systems, then there is a state in $\mathcal{H}_{A} \otimes \mathcal{H}_{B}$ which uniquely describes the global system and viceversa

Suppose now I take a system $|\Psi\rangle$, which at time $t = 0$ consists of a perfectly isolated box containing a particle and a camera, by axioms 1 and 5 we can say that there are unique states for which the isolated box is described by: \begin{equation} |\Psi\rangle_{0} = |p\rangle|C\rangle \end{equation}

At time $t$, the camera takes a picture of the space inside the box, and finds the particle at some unknown location $x$, by applying axiom 4 we can say that at this time, the state of the box must be:

\begin{equation} |\Psi\rangle_{t} = |x\rangle|C_{x}\rangle \end{equation}

Where $x$ is the detected location of the particle, which is unknown to us. On the other hand, by axiom 3 the state $|\Psi\rangle$ at time $t$ has to be the following:

\begin{equation} |\Psi\rangle_{t} = \hat{U}|\Psi\rangle_{0} \end{equation}

The question is then "can we find, for arbitrary initial states $|p\rangle$ and $|C\rangle$, a unitary map $\hat{U}$ for which the two states of $|\Psi\rangle_{t}$ are equal?", we have to impose equality because of axiom 1, for which there must be a unique state describing the isolated box at time $t$:

\begin{equation} |x\rangle|C_{x}\rangle = \hat{U}|p\rangle|C\rangle \end{equation}

Or equivalently, using unitarity and the fact that all states are rays: \begin{equation} \langle p|\langle C|\hat{U^{\dagger}}|x\rangle|C_{x}\rangle = 1 \end{equation}

It's clear now that this condition can't be satisfied in general, it would then seem that the axioms we just stated contradicted each other. That is, applying consistently axioms 3 and 4 leads to a violation of axiom 1, because at time $t$ the state of the box is no longer unique and cannot be made so.

Can somebody please point out to a resolution of this problem, perhaps pointing out where the mistake is, if there is such a mistake?

Answer 1

As hinted by @Connor Behan's answer, the resolution to this "problem" depends on one's interpretation of quantum mechanics (QM), and there is a very long list of such interpretations. But to address the question in the title, quantum mechanics is consistent, because it provides consistent predictions regardless of your interpretation.

To see this, consider a slightly modified version of your setup in which the particle is a spin-1/2 particle and the camera is a device that measures its spin. Suppose also that the 'camera' measures the particle's spin twice. QM tells you that there are two possible outcomes. Either the camera will see the particle in a spin up state both times (call this $|++\rangle$) or spin down both times ($|--\rangle$). (This is of course assuming that there is nothing inside the box that changes the particle's spin state between the measurements.)

Now you can say that the state of the camera is the superposition

$$ \alpha|--\rangle + \beta |++\rangle$$

or that the measurement has 'collapsed' the state to one of the two possible outcomes, but in either case, the conditional probability of observing +/- in the second measurement given the outcome of the first measurement is 100%.

When we are using QM the make predictions, we are always answering questions of the form "given what we know about the state of a system, what is the probability to observe outcome $X$", so those are conditional probabilities. You can think in this view of axiom 4 as merely a practical way of conditioning the probabilities you are calculating on your previous observed outcomes. (Or not, since again that's a matter of your preferred interpretation, but the important thing is that the experimental predictions will always be the same.)

You are correct of course that a projection (axiom 4) is not a unitary transformation. If you assume that the initial state of the particle is some superposition of localized states, you get after applying the unitary time evolution something like:

$$|\Psi\rangle_t = U(t)|\Psi\rangle_0 = \Sigma_i \alpha_i|x_i\rangle|C_{x_i}\rangle$$

which tells you that $|\alpha_i|^2$ is the probability that the camera will record the particle at position $x_i$. If you know that the outcome was some particular value $x_0$ and you want to find the probability that a subsequent measurement at time $t_2$ will record a value $x_j$, you can proceed in two ways:

1. Apply axiom 4, meaning that your state now becomes just $|x_0\rangle|C_{x_0}\rangle$, then apply $U(t_2)$ to this state and find the coefficient of $|x_j\rangle|C_{x_j}\rangle$.
2. Apply $U(t_2)$ directly to the original state $|\Psi\rangle_t$, this will give you the probabilities of all possible pairs of outcomes $(x_i,x_j)$, i.e. something like

$$|\Psi\rangle_{t_2} = U(t_2)|\Psi\rangle_t = \Sigma_{ij} \beta_{ij}|x_j\rangle|C_{x_i,x_j}\rangle$$

but since you are interested in the conditional probability $P(x_j|x_0)$, then using elementary probability

$$ P(x_j|x_0) = \frac{P(x_j,x_0)}{P(x_0)} = \frac{|\beta_{j0}|^2}{\sum_j |\beta_{j0}|^2}. $$

It is a straightforward calculation, due to the linearity of $U(t)$, to see that both of those methods are entirely equivalent.

A final remark: as pointed out in the comments, a hidden assumption here is that the measuring device (camera) is macroscopic enough such that we can safely ignore interference between its states. The problem with the axioms as you stated them is not that they are inconsistent, but that the word "measurement" that is used in axiom 4 is not given a precise definition. However, as demonstrated here, this doesn't matter in practice as long as we can make the above assumption. The only place where you might get conflicting predictions is when interference between macroscopic states can be observed. Indeed it can be argued that such a hypothetical experiment can test, for example, the many-worlds interpretation against the collapse interpretation (see for example this answer). Such experiments however are, and most likely never will be, possible to carry out in practice, meaning that this issue will probably remain a matter of interpretation.

Answer 2

The "resolution" you're asking for is basically a resolution to the measurement problem. This is controversial of course. But the one which can be stated most easily is surprisingly robust against attempts to refute it.

It states that you just remove axiom 4. The universe is an isolated system, all measurements take place within the universe and therefore there is nothing which makes them different from the rest of quantum mechanics. In this point of view, unitary evolution really is unitary no matter what and we have to take seriously the idea that a wavefunction of the universe right now has non-zero support on a number of other unobserved "worlds" which have nevertheless always existed.

Answer 3

This appears to contradict the axioms we just stated, because while the box evolved unitarily in time (axiom 3), the state of the particle did not (axiom 4)

If by a quantum state we mean a ray in a Hilbert space, or a ket $|\Psi\rangle$ in a Hilbert space (understanding the phase factor is immaterial), then if the particle is part of a bigger system, and that bigger system has a quantum state $|\Psi\rangle$, then there is no other quantum state of this kind to the particle. "State of the particle" in this sense is undefined in orthodox QT (QT=quantum theory), as there is no way a ket can be assigned to the particle based on $|\psi\rangle$. In orthodox QT, quantum state of the system $|\psi\rangle$ carries all there is to know about state of that system.

Quantum theory with such quantum states isn't like classical mechanics with coordinates and momenta, where systems can be divided into subsystems and any subsystem has its own coordinates and momenta. Instead, in orthodox QT, only the whole system has a quantum state of this type. Aand all we can do with it is either evolve it in time, or calculate probabilities of the possible values of a quantity that is measured on the system. We can't use it to find which quantum state some subsystem has, or will have.

Of course, this "indivisibility" (implying the well-known entanglement of different subsystems and non-local features) makes quantum theory very strange. One problem with it is that if we apply this kind of description to a sequence of bigger and bigger systems, each one containing the previous one as a part, and make predictions on each level, it is not immediately clear whether all these predictions are mutually consistent. It seems that depends on which quantum states are used to describe the systems in the sequence, but we have no rule to determine the appropriate quantum states from the state of the last big system. We usually apply QT only at one level, where we either assume some plausible quantum state (e.g. a ground state), or have some experimental data to select particular quantum state (e.g. spin measurements).

There is another notion of "state", the so-called density operator $\rho$ (it is similar to probability density in statistical physics; it is meant to give only probabilities of results for any measurement, and not the actual quantum state of single thing). This "statistical" state has the advantage that one can be assigned to the system and one to any of its subsystems, and there is a rule for how to get the latter from the former. However, this has a price; density operator doesn't always correspond to a single quantum state (ket). This is because there are cases where it corresponds to an ensemble of different quantum states (kets). When measurements are done on many systems all of which are described by the same density operator, we get the same statistics of results as for some ensemble of different quantum states, with some fixed statistics. E.g. an ensemble can have 40% systems in $|\Psi_1\rangle$ and 60% of systems in $|\Psi_2\rangle$; such ensemble cannot be described by a single ket.

Answer 4

Your question seems to be related to the so-called measurement problem. The measurement, the camera determining the location of the particle, interrupts the unitary time development of the wavefunction. This measurement and concurrent updating of the wavefunction cannot be described by the unitary time dependent solution of the Schrödinger equation. This is a longstanding problem of Quantum Mechanics which, in spite of a lot of efforts, has not yet been resolved satisfactorily. See, e.g., Wikipedia .

Answer 5

If one prepares a quantum system consisting of a box, camera and particle, then if that system is closed then there is no way to know if the camera has taken a picture after some length of time. Rather the system is in a state of superposition that evolves unitarily. The relevant states are: the system where the camera has made a measurement and the system where the camera has not made a measurement.

If we were capable of knowing that fact then one would have had to measure the system to find out and it would then in fact not be a closed system as assumed, i.e. the system collapsed to a state where the camera had made its measurement.

The apparent paradox results from making a claim about a closed system that is by definition not closed.

The observer matters in QM, if you define it to be part of the system, then the observer evolves unitarily as well, it is just as well as if there were no internal observer because you know nothing more than $|\mathcal U\Psi\rangle$ . On the other hand, if you know something more than that, then there is an external observer somewhere who is not part of the system and $|\Psi\rangle$ collapses. Any attempt to get around this problem is to promote a hidden variable theory, and you will need a non-local one to make a successful attempt.

Answer 6

In classical physics the evolution of a measurable quantity, such as the $x$ location of a particle is described by a function $x(t)$ whose value is the result you would get if you measured $x$ at time $t$.

In quantum theory the equations of motion of a measurable quantity of a system are written in terms of a Hermitian operator called an observable whose eigenvalues are the possible results of measuring that quantity. Quantum theory predicts the probabilities of measuring each of the possible values using the observable and the state of the system being measured.

In general the probabilities of the different measurement results depend on what happens to all of the possible values of the relevant observable. This is called quantum interference, see Section 2 of this paper for an example:

https://arxiv.org/abs/math/9911150

How to get predictions from quantum theory is uncontroversial, but what is happening in reality to produce those outcomes is very controversial. The different explanations of what is happening are usually called interpretations of quantum theory. As you have noted some of the postulates of quantum theory as they are usually stated in textbooks appear to be inconsistent with one another. The different interpretations say different things about how to resolve this problem.

Some of the interpretations, such as spontaneous collapse and pilot wave theories, involve modifying the equations of motion of quantum theory to remove all but one of the possible results of a measurement. These theories also make different predictions than quantum theory

https://arxiv.org/abs/2411.10782

https://arxiv.org/abs/2310.14969

This makes calling them interpretations of quantum theory a little odd. They also don't currently reproduce most of the predictions of quantum theory

https://arxiv.org/abs/2205.00568

In textbooks it is common to see assertions that there is a single outcome to a measurement of an observable with no explanation of how this works: the Copenhagen and statistical interpretations. Without an account of what is happening in reality, there is no such thing as whether an experiment has been set up correctly to test a theory. And with no account of how the single outcome is produced, there is some deviation from the equations of motion of quantum theory somewhere that may lead to different predictions but it is difficult say where since these theories are vague and their advocates haven't clarified what these theories say about what is actually happening.

Another option is to take the equations of motion of quantum theory seriously as a description of reality. When you do that you find that when information is copied out of a quantum system interference is suppressed: this is called decoherence

https://arxiv.org/abs/1911.06282

Any object in everyday life large enough for you to see, e.g. - anything larger than a grain of dust, has information copied out of it by light, collisions with air molecules etc on scales of space and time much smaller than those over which they change significantly and so on those scales interference is negligible. For smaller objects such as electrons in the atoms making up those objects interference is a relatively strong effect. Decoherence doesn't eliminate the multiple versions of a large object but since they don't interfere with one another they evolve autonously to a good approximation. So on the scales of everyday life reality as described by quantum theory without collapse looks a bit like a collection of universes each of which obeys classical physics to a good approximation. This is often called the many worlds interpretation

https://arxiv.org/abs/1111.2189

https://arxiv.org/abs/quant-ph/0104033

There are multiple versions of each measurement result and those different versions all evolve according to the equations of motion of quantum theory. Quantum theory without collapse explains the born rule as a result of symmetry properties of quantum states rather than as a result of one outcome being picked at random:

https://arxiv.org/abs/quant-ph/0405161

https://arxiv.org/abs/0906.2718

This has the advantage of not requiring modifications of the equations of motion of quantum theory and so it doesn't require a new derivation of the experimental predictions of quantum theory. It also has the advantage of being consistent with a broader class of types of measurement such as continuous measurements:

https://arxiv.org/abs/1604.05973