The-Quantum-Foundations-Project

There is no general consensus as to what its fundamental principles are, how it should be taught, or what it really “means.” Every competent physicist can “do” quantum mechanics, but the stories we tell ourselves about what we are doing are as various as the tales of Scheherazade, and almost as implausible (Griffiths (2017)).

Analogy with Classical Mechanics¶

Newton’s Second Law¶

A law of physics can be understood as a rule that describes—usually by means of a mathematical formulation—how certain physical systems behave(Siddiqui (2014)). Newton’s second law, for example, can be expressed by the equation

where $x\left(t\right)$ represents the position of a particle of mass $m$ relative to the origin at the instant $t$. When we solve this differential equation under certain initial conditions, we obtain as a solution a function $x\left(t\right)$. Thus, once the forces acting on the particle and its initial conditions are specified, Newton’s second law determines its temporal evolution: solving the equation provides us with the equation of motion of the particle.

For example, for a free particle($F=0$), we can obtain its equation of motion from Newton’s second law:

Defining the velocity as $v\left(t\right)\equiv\frac{dx}{dt}$, we have:

Therefore, $v\left(t\right)=v\left(t_{0}\right)$, that is, the particle’s velocity is constant. Denoting this constant velocity by $v$, we have:

If $x\left(t_{0}\right)=x_{0}$ and $t_{0}=0$, we obtain:

In this case, $x\left(t\right)$ is easily interpretable as the position of the particle at the instant $t$.

Schrödinger Equation¶

The quantum mechanical analogue of Newton’s second law is the Schrödinger equation. In one dimension and in the non-relativistic case, the Schrödinger equation is:

The corresponding equation for a free particle ($V\left(x,t\right)=0$) is:

Using the method of separation of variables, that is, assuming that the solution can be separated into a spatial part $\Phi\left(x\right)$ and a temporal part $\Pi\left(t\right)$, that is, $\Psi\left(x,t\right)=\Phi\left(x\right)\Pi\left(t\right)$. Then:

Since one side depends only on the spatial dimension ($x$) and the other only on the temporal dimension ($t$), the only way this can be true is if both sides are equal to a constant: $E$. Solving then for the temporal part:

Since the derivative of the function $\Pi\left(t\right)$ is equal to the function itself multiplied by a constant, we have an exponential solution $\Pi\left(t\right)=A\exp\left(-\frac{iE}{\hbar}t\right)$. For the spatial part:

Since this is a second-order linear homogeneous differential equation with constant coefficients, we can look for solutions of the form $\Phi\left(x\right)=e^{\pm ikx}$. In this case, $e^{ikx}$ and $e^{-ikx}$ are two linearly independent solutions to $\frac{d^{2}\Phi}{dx^{2}}=-k^{2}\Phi$. By the linearity of the equation, its general solution is a linear combination of these solutions:

Or, writing $k^{2}=2mE/\hbar^{2}$ and $\omega=\frac{E}{\hbar}$:

where $K_{1}$ and $K_{2}$ are constants determined by the conditions imposed on the solution. Here, some questions already arise. When we solved Newton’s second law, we obtained $x\left(t\right)$, which described the position of the particle. When we solve the Schrödinger equation, we obtain $\Psi\left(x,t\right)$, but what is $\Psi\left(x,t\right)$? What does $\Psi\left(x,t\right)$ describe?

A somewhat more in-depth discussion of this can be found here.

Postulates of quantum mechanics¶

Notation¶

To move forward, we need to know the postulates^[1] of quantum mechanics. The first point to note is that they can be formulated in different ways from one book to another. We will use Professor Jonas Maziero’s material (Maziero (2026)). We need to begin, then, with some comments on Dirac notation.

In this notation, we write a vector as $\left|v\right\rangle$, called a ket. Taking the conjugate transpose of this vector (or adjoint), we obtain $\left\langle v\right|=\left|v\right\rangle ^{\dagger}=\left(\left|v\right\rangle ^{T}\right)^{*}$, which is called a bra. If we represent the vector $\left|v\right\rangle$ in an orthonormal basis $\left\{ \left|b_{j}\right\rangle \right\}$, we can write it as a linear combination of the basis vectors: $\left|v\right\rangle =\sum_{i}\left\langle b_{i}|v\right\rangle \left|b_{i}\right\rangle$. In matrix representation, we have

The product of a bra and a ket is called the inner product and results in a scalar:

On the other hand, the product of a ket and a bra is called the outer product and results in an operator, whose matrix representation is:

With this in mind, we can now move on to the postulates of quantum mechanics.

State Postulate¶

In every physical theory, we seek to describe the possible states of the system of interest, the changes in these states (temporal evolution), and the observable quantities and their measurements. For example, in Newtonian Mechanics (NM), the state of a system at a certain instant of time is specified by the values of position and velocity relative to a certain inertial reference frame, and its temporal evolution is given by Newton’s second law. The observables are essentially the positions, from which we obtain the velocities using the time parameter. It is implicit in NM that the values of the observables are determined even before they are measured, and that all these observables can have well-defined values at a given instant of time. Next, we will see the postulates of QM for describing states, changes in these states, observables, and their measurements. We will see that the measurement part differs greatly from that of NM, mainly because of the existence of observables that cannot have well-defined values at the same time (Maziero (2026)).

The state of a quantum system is described by a normalized vector in a Hilbert space $\mathcal{H}$. For simplicity, we will consider $\mathcal{H}$ as a vector space equipped with an inner product function $\left\langle \cdot|\cdot\right\rangle :\mathcal{H}\rightarrow\mathbb{C}$.

Furthermore, a Hilbert space is a complete complex vector space. To understand what it means to be complete, we first need to understand the concept of a Cauchy sequence. A Cauchy sequence is a sequence whose elements become arbitrarily close to one another as we advance through the sequence. A complete metric space^[2] is one in which every Cauchy sequence converges to a limit that also belongs to the space itself.

By system, we refer to a degree of freedom^[3]. For example, the spin of an electron is a system, another system would be its position, and so on. When we consider two or more degrees of freedom as a single entity, we call it a composite system, and if we divide the composite system into subsystems, its state is described by a normalized vector in a Hilbert space obtained by taking the tensor product of the individual spaces:

Measurement Postulate¶

Observable (measurable) quantities are represented by linear Hermitian operators, $O=O^{\dagger}$, defined on $\mathcal{H}$. When we measure an observable, we will obtain one of its possible values, which are given by the eigenvalues of the Hermitian operator.

Unlike classical mechanics (CM), quantum mechanics (QM), in general, does not allow us to predict with certainty the outcome of a measurement. What QM provides us with are the probabilities of obtaining the different outcomes, calculated by means of the Born rule:

This is the probability of obtaining the value $o$ when measuring the observable $O$ of a system that is in the state $\left|\psi\right\rangle$ immediately before the measurement, and $\left|o\right\rangle$ is the eigenvector of $O$ corresponding to the eigenvalue $o$. $O=O^{\dagger}$ are defined on $\mathcal{H}$. When we measure an observable, we obtain one of its possible values, which are represented by the eigenvalues of the Hermitian operator.

Before moving forward, we can stop to reflect for a moment. When we were working with classical mechanics and Newton’s law, one of the observables of a particle was its position, and Newton’s law directly described how this quantity varied in time.

However, when we approach quantum mechanics, the Schrödinger equation describes how the wave function varies in time (and space). With Dirac notation, as we will see when studying the dynamics postulate, we have a mathematical description of how the state vector of the system varies in time. However, the state vector is not itself an observable.

Even temporarily ignoring the probabilistic aspect, we use the state vector to calculate quantities associated with observables. This already constitutes a fundamental difference between the classical and quantum descriptions of systems. But if we calculate scalar quantities associated with observables from the state vector (or the wave function) of the system, what is the physical meaning of the state vector itself? Does it even have a direct physical meaning?

This is one of the points on which different interpretations of quantum mechanics diverge. There is still no consensus on which interpretation provides the correct description of physical reality, although some interpretations may be ruled out or considered problematic in light of certain theoretical arguments and experimental results.

Likewise, different interpretations diverge on the question of whether all relevant physical information about a system is necessarily contained in its state vector. We will explore different interpretations later on. For now, it is enough to emphasize that the physical meaning of the state vector and the status of the information it contains are profound interpretational questions on which there is no consensus.

Repeatability of measurements¶

However, there is one more detail that, although it is not explicitly contained in the postulates we have presented so far, is significantly different when we are in the quantum domain: performing a measurement affects the system.

When we observe the position of a soccer ball that moves according to classical mechanics, we can, in principle, make this observation without significantly altering the state of the ball. We can therefore consider that the observation was performed approximately “externally” to the system itself, leaving it practically unchanged.

In the quantum domain, however, the situation is different. When we perform a measurement, the system interacts with the measurement apparatus and, in general, its state is significantly altered. One consequence of this is that, if we measure the same observable twice immediately in succession, the result of the second measurement will necessarily be equal to that obtained in the first.

Indeed, suppose that the first measurement of the observable ($O$) produced the result ($o$). After this measurement, the state of the system becomes a state associated with the result ($o$). In the case where the eigenvalue ($o$) is non-degenerate^[4], we can write

A second immediately subsequent measurement of the same observable will therefore once again produce the result ($o$).

Note that, theoretically, before the measurement is performed, quantum mechanics provides us only with the probabilities associated with the possible outcomes. In the laboratory, however, an individual realization of the measurement necessarily produces a definite outcome. This leads us to another fundamental question: the probabilistic nature of quantum mechanics. Its probabilistic predictions concern the frequencies and distributions of outcomes obtained from ensembles of systems prepared in equivalent ways and subjected to the same experimental procedure, and they do not, in general, determine which specific outcome will be obtained in an individual realization.

Before the measurement is performed, we have only the probabilities associated with the different values we may obtain. In this situation, in what state is the system? When dealing with classical mechanics, deterministic dynamics allows us, in principle, to determine the state of a system at an earlier instant $t'<t$ by knowing its state at an instant $t$. Can we extend this line of reasoning to quantum mechanics? Is the probabilistic character of the theory a fundamental feature of nature, or merely an expression of our ignorance about the state of the system? Again, we have no conclusive answers to this question, and different interpretations of quantum mechanics offer different answers.

And finally, still related to the probabilistic nature of measurements, misguided interpretations often support the idea that quantum mechanics is a scientific formulation that naturally incorporates the possibility of free will. However, the probabilistic character of the theory, by itself, provides no support for this conclusion and does not make quantum mechanics compatible with free will. We can discuss this misconception in greater detail when we address the dynamics postulate.

Quantum incompatibility¶

Let $A$ and $B$ be two Hermitian operators defined on a finite-dimensional Hilbert space of dimension $d$. By the spectral theorem, we can write them in their respective spectral decompositions as

The numbers $a_{j}$ and $b_{j}$ are the respective eigenvalues, and the sets$\left\{ \left|a_{j}\right\rangle \right\} ^{d}_{j=1}$ and $\left\{ \left|b_{j}\right\rangle \right\} ^{d}_{j=1}$ form orthonormal bases of eigenvectors of $A$ and $B$, respectively, that is, they diagonalize these operators, so that:

Assuming that the two operators share the same basis $\left|a_{j}\right\rangle =\left|b_{j}\right\rangle \coloneqq\left|j\right\rangle$, then, with the commutator defined as $\left[A,B\right]=AB-BA$, we have:

Then:

Thus, we have that if the operators share the same set of eigenvectors, they commute. And to prove that if they commute, they share the same basis of eigenvectors:

That is, in this case, we must have that $B\left|a_{j}\right\rangle$ and $\left|a_{j}\right\rangle$ belong to the eigenspace of $A$ corresponding to the eigenvalue $a_{j}$. If $a_{j}$ is non-degenerate, its eigenspace has dimension 1. Therefore, any eigenvector of $A$ associated with $a_{j}$ is linearly dependent on $\left|a_{j}\right\rangle$. Since $B\left|a_{j}\right\rangle$ also belongs to this eigenspace, we necessarily have $B\left|a_{j}\right\rangle =b_{j}\left|a_{j}\right\rangle$.

Now, if $a_{j}$ is degenerate, that is, if there are two or more linearly independent eigenvectors corresponding to the same eigenvalue $a_{j}$, then there exists a subspace of $\mathcal{H}$ generated by the eigenvectors of $A$ corresponding to $a_{j}$. It is a little more complex, but it is the same idea applied to a subspace of dimension greater than 1. To understand this, we will base our discussion on another source (Massachusetts Institute of Technology (2013)).

For the degenerate case, we then have more vectors associated with $a_{j}$. Let us call this subspace $\mathcal{H}_{j}$ and, with $\left\{ \left|a^{k}_{j}\right\rangle \right\} ^{d_{j}}_{k=1}$ being any basis of $\mathcal{H}_{j}$, then $A\left|a^{k}_{j}\right\rangle =a_{j}\left|a^{k}_{j}\right\rangle$. Now, since $A$ and $B$ commute, we have $B\left|a_{j}\right\rangle \in\mathcal{H}_{j}$, that is, $B$ transforms vectors from $\mathcal{H}_{j}$ into vectors that also belong to $\mathcal{H}_{j}$. In particular, $B$ transforms an eigenvector of $A$ associated with $a_{j}$ into a vector that remains in the same eigenspace of $A$. If $B\left|a^{k}_{j}\right\rangle \neq0$, this vector is also an eigenvector of $A$ associated with $a_{j}$.

Therefore, each $B\left|a^{k}_{j}\right\rangle$ can be written as a linear combination of the basis vectors of $\mathcal{H}_{j}$. Thus, we can consider the action of $B$ “internally” within the subspace $\mathcal{H}_{j}$. Since $B$ is Hermitian, its restriction to $\mathcal{H}_{j}$ is also Hermitian. And since $\mathcal{H}_{j}$ has finite dimension (and is Hermitian), we can diagonalize this restriction of $B$ in an orthonormal basis of $\mathcal{H}_{j}$. Therefore, there exists a basis of $\mathcal{H}_{j}$, which we can denote by $\left\{ \left|a^{k}_{j}\right\rangle \right\}$, such that $B\left|a^{k}_{j}\right\rangle =b_{j}\left|a^{k}_{j}\right\rangle$. Since this new basis remains a basis of $\mathcal{H}_{j}$, its vectors continue to be eigenvectors of $A$ associated with $a_{j}$.

Thus, the eigenvectors are simultaneously eigenvectors of $A$ and $B$. Repeating this procedure for each eigenspace of $A$, we obtain an orthonormal basis of $\mathcal{H}$ consisting of common eigenvectors of the two operators.

In other words, if the Hermitian operators $A$ and $B$ commute, then they share the same orthonormal basis of eigenvectors. To better understand this proof, we can represent the operator $B$ as a matrix in the basis of eigenvectors of $A$. Using the completeness relation, $\mathbb{I}=\sum_{j}\left|a_{j}\right\rangle \left\langle a_{j}\right|$, we have $B=\mathbb{I}B\mathbb{I}=\sum_{j,k}\left(\left\langle a_{j}\left|B\right|a_{k}\right\rangle \right)\left(\left|a_{j}\right\rangle \left\langle a_{k}\right|\right)$.

Thus, the elements of the matrix representing $B$ in this basis are $B_{a_{j},a_{k}}=\left\langle a_{j}\left|B\right|a_{k}\right\rangle$, or in matrix form:

Then we can think of $B$ as a block-diagonal matrix:

where each $B_{j}$ is a submatrix representing the internal action of $B$ acting on the subspace $\mathcal{H}_{j}$. If the eigenvalue $a_{j}$ is non-degenerate, then the block $B_{j}$ is just a $1\times1$ matrix. If $a_{j}$ is degenerate, then the block $B_{j}$ has dimension $d_{j}\times d_{j}$.

Since $B$ is Hermitian, each block $B_{j}$ is also a Hermitian matrix and, therefore, can be diagonalized. Thus, we can diagonalize $B$ within each subspace $\mathcal{H}_{j}$, obtaining a basis of eigenvectors of $B$ that are also eigenvectors of $A$. Thus, we prove that if and only if two operators $A$ and $B$ commute, then they share a common basis of eigenvectors.

Example 1

In summary, when we have a non-degenerate eigenvalue $a_{j}$, the situation is very simple, since in this case, the only eigenvector of $A$, up to a multiplicative factor, is automatically also an eigenvector of $B$.

When $a_{j}$ is degenerate, its eigenvectors form a subspace $\mathcal{H}_{j}$, corresponding to the eigenvalue $a_{j}$, in which every nonzero vector is also an eigenvector of $A$ associated with $a_{j}$. Since $A$ and $B$ commute, applying $B$ to any vector in $\mathcal{H}_{j}$ results in another vector belonging to $\mathcal{H}_{j}$. Thus, it is sufficient to find, within this subspace, the vectors that are also eigenvectors of $B$.

This possibility is guaranteed by the fact that $B$ is Hermitian: its action restricted to $\mathcal{H}_{j}$ can be represented by a Hermitian submatrix, which can be diagonalized by an orthonormal basis of eigenvectors^[5]. Therefore, we can choose an orthonormal basis of $\mathcal{H}_{j}$ formed by eigenvectors of $B$. Since all vectors in $\mathcal{H}_{j}$ are also eigenvectors of $A$ associated with $a_{j}$, these vectors are common eigenvectors of $A$ and $B$.

Repeating this procedure for each eigenspace of $A$, we obtain an orthonormal basis of $\mathcal{H}$ formed by common eigenvectors of $A$ and $B$.

With the proof of this theorem, we know that if two observables $A$ and $B$ of a system do not commute, they cannot share all eigenvectors. If, by measuring the observable $A$, we prepare the system in a state that is not an eigenvector of the observable $B$, this vector must necessarily be a linear combination of the eigenvectors of with two or more nonzero probability amplitudes.

Example 2

Uncertainty Relation¶

Here we will use the quantum kinematic formalism to obtain general mathematical relations that provide limitations on the product or sum of the “uncertainties” (standard deviations) of two observables.

Heisenberg-Robertson-Schrödinger Uncertainty Relation (HRSUR)

The HRSUR :

with the variance defined by $\mathrm{Var}(A)=\langle(A-\langle A\rangle)^{2}\rangle.$ The quantities on the right-hand side of the inequality will be defined below.

To prove the HRSUR, we will use the Cauchy-Schwarz inequality (CSI) as one of the main tools:

with equality obtained if and only if $|\psi\rangle$ and $|\phi\rangle$ are collinear ($|\psi\rangle\propto|\phi\rangle$). This result is valid for any pair of vectors $|\psi\rangle,|\phi\rangle$ in a Hilbert space $\mathcal{H}.$ Consider $A$ and $B$ to be two observables of a physical system prepared in the state $|\xi\rangle$. We will use

to denote the expectation value of any operator $X$ and $I$ to denote the identity operator in $\mathcal{H}$. Before proceeding, I would like to explore this result further. Let us consider a system in the state $\left|\psi\right\rangle =\sum_{n}c_{n}\left|\psi\right\rangle$, and an operator $X$ with eigenvalues $x_{n}$ associated with the eigenvectors $\left|x_{n}\right\rangle$. By the definition of a weighted average, the mean value of the results is $\left\langle O\right\rangle =\sum_{n}o_{n}P\left(o_{n}\right)$, and since $P\left(o_{n}\right)=\left|c_{n}\right|^{2}$, we can write $\left\langle O\right\rangle =\sum_{n}o_{n}\left|c_{n}\right|^{2}$. Since $\left\langle o_{n}|\psi\right\rangle =c_{n}$, then $\left\langle O\right\rangle =\sum_{n}o_{n}\left\langle \psi|o_{n}\right\rangle \left\langle o_{n}|\psi\right\rangle$, since $\left|\left\langle \psi|o_{n}\right\rangle \right|=\sqrt{\left\langle o_{n}|\psi\right\rangle \left\langle \psi|o_{n}\right\rangle }$. Then, by the spectral decomposition $O=\sum_{n}o_{n}\left|o_{n}\right\rangle \left\langle o_{n}\right|$, we have:

We will also define the following vectors:

which we will replace at CSI. Let us note that

with $\mathrm{Var}(X)=\mathrm{Var}(X)_{\xi}=\langle\xi|(X-\langle X\rangle I)^{2}|\xi\rangle=\langle(X-\langle X\rangle I)^{2}\rangle$ being the variance of $X$. We also use $(A-\langle A\rangle I)^{\dagger}(A-\langle A\rangle I)=(A-\langle A\rangle I)^{2}$ since $A=A^{\dagger}$ and $\left\langle A\right\rangle =\left\langle A\right\rangle ^{*}$. We can see that this follows from the fact that $A$ (and $I$) are Hermitian, and:

Furthermore, we have to

And since:

are, respectively, the commutator and anticommutator of $A$ and $B$. Summing them:

Then:

So far, then, we have:

Since $\{A,B\}$ and $[A,B]$ are, respectively, Hermitian ($\{A,B\}^{\dagger}=\{A,B\}$) and anti-Hermitian ($[A,B]^{\dagger}=-[A,B]$) operators, their expectation values are, respectively, purely real and purely imaginary numbers. We have already verified that the expectation value of a Hermitian operator is a real number; now, for an anti-Hermitian operator:

Now, if $\left\langle A\right\rangle =a+ib$, its conjugate $\left\langle A\right\rangle ^{*}=a-ib$, and if $\left\langle A\right\rangle ^{*}=-\left\langle A\right\rangle$, then $a+ib=-\left(a-ib\right)$, that is, $a+ib=-a+ib$. For this to be equal, we must have $a=0$. Thus, we are left with $\left\langle A\right\rangle =ib$. Since $\left\langle A\right\rangle$ and $\left\langle B\right\rangle$ are also real numbers, we can verify this for $\left[A,B\right]$ and $\left\{ A,B\right\}$.

That is, if $\left[A,B\right]=AB-BA$ then $\left[A,B\right]^{\dagger}=-\left[A,B\right]$. And if $\left\{ A,B\right\} =AB+BA$ then $\left\{ A,B\right\} ^{\dagger}=\left\{ A,B\right\}$. Thus, we can write:

We then have:

Note: Since $2^{-1}\langle[A,B]\rangle=ib$ is a purely imaginary number, multiplying by $-i$: $-i\cdot\left(ib\right)=b$, gives us its real coefficient. Thus, considering that

We then have:

Now let’s open it:

Furthermore, since it is purely imaginary, then:

We are then left with $\Im\left(\left\langle \psi|\phi\right\rangle \right)^{2}=2^{-2}\left|\left\langle \left[A,B\right]\right\rangle \right|^{2}$. And our inequality:

It now remains for us to demonstrate that:

Where the covariance is $\mathrm{Cov}(X,Y)=\langle XY\rangle-\langle X\rangle\langle Y\rangle$ between the observables $X$ and $Y$, therefore:

And this $\Re\left(\left\langle \psi|\phi\right\rangle \right)=2^{-1}\left\langle \left\{ A,B\right\} \right\rangle -\left\langle A\right\rangle \left\langle B\right\rangle$ is the result we had obtained. We can then write $\Re\left(\left\langle \psi|\phi\right\rangle \right)^{2}=\text{CovQ}\left(A,B\right)^{2}$ and:

It is worth noting that if $A$ and $B$ are compatible observables, i.e., if $[A,B]=0$, then we will have $\mathrm{CovQ}(A,B)=\mathrm{Cov}(A,B)$.

Let us now analyze the problem of the possible triviality of the HRSUR, whose general objective is to determine how small the product of the variances of a pair of observables can be. For this, let us assume that the system is prepared in a state that is an eigenvector of the observable $A$, i.e.,

with $A|a\rangle=a|a\rangle$ and $a\in\mathbb{R}$. In this case, it can be verified that

And with that, HRSUR gets back to us:

Therefore, in this case, the HRSUR provides no information whatsoever about the possible incompatibility of $A$ and $B$. The same type of limitation applies if $|\xi\rangle=|b\rangle$ for $B|b\rangle=b|b\rangle.$ In any case, let us verify. For the commutator: