Intuition for why linear algebra is needed in quantum physics

In summary, the video explains how linear algebra is used in quantum physics to model the behavior of physical quantities. Classical physics uses continuous functions to represent physical quantities, but in quantum physics, these quantities can only take specific values with probabilities. To account for this, a mathematical structure is needed, where each possible outcome is represented by a mathematical object. These objects are combined with probability coefficients to represent the state of a particle. The state is then a linear combination of all possible outcomes, with the coefficients representing the probability amplitudes. This approach is different from treating the values of observables as vectors, as each observable is represented by a linear operator on a Hilbert space, and the state of a particle is a superposition of eigenstates for
  • #1
Shirish
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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.
1672824740793.png

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
 
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  • #2
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.
 
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  • #3
PeroK said:
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.
 
  • #4
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.
 
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  • #5
PeroK said:
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##?
 
  • #6
Shirish said:
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##.
 
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  • #7
Shirish said:
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.
 
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  • #8
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.
 
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1. What is linear algebra and how is it related to quantum physics?

Linear algebra is a branch of mathematics that deals with vector spaces and linear transformations. It is used in quantum physics to describe the state of a quantum system and the evolution of that state over time. Quantum systems are represented by vectors in a complex vector space, and linear algebra provides the tools to manipulate and analyze these vectors.

2. Why is linear algebra necessary for understanding quantum mechanics?

Quantum mechanics is a mathematical framework that describes the behavior of particles at the atomic and subatomic level. Linear algebra is necessary because it provides a rigorous mathematical structure for representing and analyzing the complex and abstract concepts of quantum mechanics. Without linear algebra, it would be difficult to make precise predictions and calculations in the quantum world.

3. How does linear algebra help in solving problems in quantum physics?

Linear algebra provides a powerful set of tools for solving problems in quantum physics. These tools include vector spaces, matrices, eigenvalues and eigenvectors, and linear transformations. These concepts are essential for understanding the behavior of quantum systems and making predictions about their properties and interactions.

4. Can quantum physics be understood without knowledge of linear algebra?

It is possible to have a basic understanding of some concepts in quantum physics without knowledge of linear algebra. However, to fully grasp the mathematical principles and make precise calculations, a solid understanding of linear algebra is necessary. Without it, many of the fundamental concepts and equations in quantum physics would be difficult to comprehend.

5. How can learning linear algebra improve my understanding of quantum physics?

Learning linear algebra can greatly enhance your understanding of quantum physics. It provides a rigorous mathematical foundation for understanding the complex concepts of quantum mechanics and allows for more accurate predictions and calculations. Additionally, many advanced topics in quantum physics, such as quantum entanglement and quantum computing, rely heavily on linear algebra, so a strong grasp of this subject can open up new areas of study and research in the field of quantum physics.

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