Intuition for why linear algebra is needed in quantum physics

[Shirish]

I'm watching a nice video that tries to explain how linear algebra enters the picture in quantum physics. A quick summary:

Classical physics requires that physical quantities are single-valued and vary smoothly as they evolve in time. So a natural way to model classical physical quantities is as continuous functions. But if we try to measure energies emitted by an electron in a hydrogen atom, we see that:

1. They take only a specific set of values
2. We can't say that the energy is single-valued since prior to observation. It could be anything out of the permissible set of values
3. Each of the possible values comes up with specific probabilities

So we need a mathematical structure that accounts for all 3 of the above. The video explains it as follows:

Let's say that we know for sure that the particle has energy $A$, and let's say that the particle is represented by the mathematical object $M_A$ (which could be a function/element of a ring/manifold - whatever). So we have a mathematical object $M$ for every possible outcome we could get for a particle.

Somehow the particle is represented as an amalgamation of all these mathematical objects, holding on to each outcome until we make a measurement. We need to put all these objects together somehow, which can be done by some mathematical operation.

The next step is to encode the probabilities corresponding to each outcome - a natural way is to use probabilities as coefficients for these mathematical objects, which gives us $p_AM_A \circ p_BM_B\circ p_CM_C\circ p_DM_D$. Again, a natural thing that springs to mind is linear combination - so we can try to model $M$'s as vectors and $\circ$ as $+$.

The paragraph above makes sense to me and I'm comfortable with the logic of deciding on the appropriate mathematical object and operation. But I'm confused about why we made the jump to a linear combination representing the particle.

From what I understand, the entire motivation was to resolve the weird behavior of physical quantities in the quantum world. So the most natural way to resolve that would've been - treat $A,B,C,D$ as vectors, and their linear combination with probability coefficients would then represent the energy. i.e. $E=p_AA+p_BB+p_CC+p_DD$. Where and why did particles come into the picture?

[1]:
[2]: https://i.stack.imgur.com/Iezp6m.png

 

[PeroK]

In introductory QM, the state of the particle is represented by a vector. Not, strickly speaking, the particle itself. The state specifies the possible measurement outcomes and their probabilities for all observables. Each observable is represented by a linear operator.

 

[Shirish]

In introductory QM, the state of the particle is represented by a vector. Not, strickly speaking, the particle itself. The state specifies the possible measurement outcomes and their probabilities for all observables. Each observable is represented by a linear operator.

So my understanding is, the following approach isn't what we follow: "Let $E$ be an observable with possible values (represented as vectors) $v_1,\ldots,v_n$, then $E$ is also a vector that's a linear combination of the possible values, with probabilities as coefficients."

Rather we encode specific values of all possible observables into some specific vector, and then we say that the state of a particle is a linear combination of all such possible vectors. e.g. if we consider energy and spin as observables, with possible energy values $E_1,\ldots,E_n$ and spin values $s_1,\ldots,s_m$, then the observables' values are encoded into vectors like $(E_1,s_3)$, or $(E_5,s_2)$, etc. And then the state $S$ is a linear combination of all possible $(E_i,s_j)$.

Is the above understanding correct? One caveat is the encoding of observables' values probably isn't as simple as $(E_i,s_j)$, but I only used it as an example.

 

[PeroK]

You start with an appropriate Hilbert Space of all possible states for the system you are studying. Each observable is represented by a Hermitian linear operator on that Hilbert space.

The possible measurement outcomes are the eigenvalues of that operator. Eigenvalues are real numbers. Not vectors. To each eigenvalue there corresponds a subspace of eigenstates (or eigenvectors).

Each state can be represented as a linear combination (also called a superposition) of eigenstates of any observable (operator). The coefficients of this linear combination are the probability amplitudes for the eigenstates. The probability of each measurement outcome is the modulus squared of these amplitudes.

The simplest system is where we consider only the spin on an electron. The Hilbert space is two-dimensional complex vectors. Measurements of spin are represented by the Pauli matrices. The eigenvalues are $\pm \dfrac \hbar 2$. And the corresponding eigenstates are called spin-up and spin-down.

 

[Shirish]

You start with an appropriate Hilbert Space of all possible states for the system you are studying. Each observable is represented by a Hermitian linear operator on that Hilbert space.

The possible measurement outcomes are the eigenvalues of that operator. Eigenvalues are real numbers. Not vectors. To each eigenvalue there corresponds a subspace of eigenstates (or eigenvectors).

Each state can be represented as a linear combination (also called a superposition) of eigenstates of any observable (operator). The coefficients of this linear combination are the probability amplitudes for the eigenstates. The probability of each measurement outcome is the modulus squared of these amplitudes.

The simplest system is where we consider only the spin on an electron. The Hilbert space is two-dimensional complex vectors. Measurements of spin are represented by the Pauli matrices. The eigenvalues are $\pm \dfrac \hbar 2$. And the corresponding eigenstates are called spin-up and spin-down.

Thanks! From your explanation the case of 1 observable is very clear now. Just one doubt though for the case with multiple observables (let's say 2 for now) - will there be separate Hilbert spaces corresponding to separate observables?

Or is it as follows: there is only one Hilbert space, wherein we'll have 2 operators - say $V,W$, with eigenstates $v_1,\ldots,v_n$ and $w_1,\ldots,w_m$ respectively. And then any point in that Hilbert space is a linear combination of $v_1,\ldots,v_n,w_1,\ldots,w_m$?

 

[martinbn]

So my understanding is, the following approach isn't what we follow: "Let $E$ be an observable with possible values (represented as vectors) $v_1,\ldots,v_n$, then $E$ is also a vector that's a linear combination of the possible values, with probabilities as coefficients."

No, the possible values are numbers say $a_1,a_2,\dots a_n$. The observable is modeled by an operator whose eigenvalues are those numbers. Each of those numbers correspond to an eigenvector. Let's say they are $e_1,e_2,\dots e_n$. So if the value of the observable $E$ is $a_1$, then the particle's state is the vector $e_1$ and so on. The state of the particle can be any combination of those vectors. Then the observable $E$ doesn't have a specific value. The coeficients in the combination are related to the probability to get the corresponding value if you measure $E$.

 

[PeroK]

Thanks! From your explanation the case of 1 observable is very clear now. Just one doubt though for the case with multiple observables (let's say 2 for now) - will there be separate Hilbert spaces corresponding to separate observables?

Or is it as follows: there is only one Hilbert space, wherein we'll have 2 operators - say $V,W$, with eigenstates $v_1,\ldots,v_n$ and $w_1,\ldots,w_m$ respectively. And then any point in that Hilbert space is a linear combination of $v_1,\ldots,v_n,w_1,\ldots,w_m$?

Both sets of eigenstates span the Hilbert space. That said, there are complications around degenerate eigenvalues. And, if you consider more than the spin on the electron, you need a larger Hilbert space that involves, spatial wave functions combined with spin states.

 

[LittleSchwinger]

The state spaces of probability theories are generally embeddable in a linear space.

A very simple example, consider the classical probability case of a dice. Here our states are lists $p_{i}, i=1\ldots 6$ obeying $\sum_{i} p_{i} = 1$. Strictly speaking this isn't a linear space, but it can be immediately represented as a subspace of $\mathbb{R}^{6}$ which is linear.

The same holds true for generalisations to probability theory like Quantum Theory. The state space being linear immediately calls for linear algebra.