What is the actual use of Hilbert spaces in quantum mechanics?

Question

I'm slowly learning the quirks of quantum mechanics. One thing tripping me up is... while (I think) I grasp the concept, most texts and sources speak of how Hilbert spaces/linear algebra are so useful in quantum calculations, how it's the fundamental language, how it supposedly simplifies the calculations immensely, when in essentially every calculation I've seen (e.g., particle in a box, harmonic oscillator, hydrogen atom, etc) Hilbert spaces are pretty much never mentioned. It's just solving the Schrodinger equation for the Wavefunction, then determining energy levels and expectation values etc. I understand the premise of state vectors and what not, just don't quite see the use.

So, how does the language of linear algebra (which I have a basic grasp of) actually play a part in calculation beyond seemingly redundant formalism? Could someone point me to problems in QM where the language of linear algebra is actually used to do calculations and solve problems? Perhaps someone could show me how one of the aforementioned problems connect?

Answer 1

I think there are two distinct ways that one could interpret this question, so I'll try my hand at answering both.

Interpretation 1: I am learning the standard formulation of quantum mechanics and solving problems like the particle in a box. I am comfortable performing all of these calculations, but I do not understand why I have to know about Hilbert spaces or linear algebra.

This one is pretty straightforward. If you can add things together, multiply them by constants, and take inner products, then you're essentially$^\dagger$ working with a Hilbert space. The wavefunctions which you are comfortable solving for are elements of such a space, and the self-adjoint operators which represent observables are linear maps from one element of the space to another.

Linear algebra is just the study of vector spaces and linear maps between them, so it is in particular the backdrop of all of the calculations you are performing. When you solve an eigenvalue equation like the Schrodinger equation, you're doing linear algebra. When you expand a generic state as a superposition of eigenstates of some observable, you're doing linear algebra. When you are confident that such a set of eigenfunctions even exists and that the corresponding eigenvalues are real, it is because you have learned the spectral theorem for self-adjoint operators, which is a central result in (you guessed it) linear algebra (or functional analysis, which is essentially linear algebra in infinite dimensional spaces).

In that sense, it's not so much that linear algebra is useful in the standard formulation of quantum mechanics; its that the standard formulation of quantum mechanics is linear algebra, whether you'd like to call it that or not. To be sure, particular techniques, theorems, and general mindsets from linear algebra can be extremely useful for performing calculations, making up models, etc - but the fact remains that no matter how you slice it, what you are doing is linear algebra on a Hilbert space.

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$^\dagger$Actually this describes what's called a pre-Hilbert space. To be a full Hilbert space, there's an additional technical requirement called completeness. Loosely, this means that sequences which "look like" they should converge actually do. This is important whenever you use a limit, which appears when you differentiate (e.g. the $\frac{d}{dt}$ in the Schrodinger equation) and whenever you expand a wavefunction as an infinite series of eigenvectors of some observable.

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Interpretation 2: I understand that the formulation of quantum mechanics that I'm currently learning is based around the Hilbert space (i.e. a vector space with an inner product) as a central concept, but I don't understand why such a construct provides the correct description of nature.

This is a much deeper question. At the deepest level, a physical theory is nothing more or less than a mechanism for assigning probabilities to the possible outcomes of measurements. The standard formulation of quantum mechanics accomplishes this in a somewhat peculiar way, wherein we devise a correspondence between measurable properties of the system and linear maps on some Hilbert space (and then proceed as you have learned).

This approach works, as has been shown by many thousands of experiments over the past hundred years, but it's far from obvious why this is the right path to take. Some insight may be obtained from the algebraic formulation of quantum mechanics, wherein the central object under consideration is the so-called algebra of observables, whose elements represent the various measurable properties of a given system.

On one hand, this is very nice - we're working and referring directly to things we intend to measure, and in many cases it's even possible to obtain this quantum algebra of observables by suitably fiddling with a corresponding classical algebra of observables (though I should say, the latter is a different variety of algebra). The downside is that this formulation of quantum mechanics is very abstract and very sophisticated - so much so that I would be willing to bet that the substantial majority of working physicists are at best only tangentially aware of its existence.

Luckily, for those of us who are not interested in studying $C^*$-algebras until our eyes bleed, there is an alternative to this heavy mathematical abstraction. According to the Gelfand-Naimark theorem, any such algebra of observables can be concretely realized as operators on some Hilbert space. In that way we are led back to the standard formulation of quantum mechanics, but with a fresh perspective: the seemingly arbitrary choice to model a quantum system around a Hilbert space is a choice born not of necessity but rather of convenience, because it provides a concrete realization of what would otherwise be a terribly abstract description of nature.

Answer 2

The Schrodinger representation privileges the position basis for representing the state of a system. This is not necessary, since quantum mechanics works in any basis (such as the momentum basis, or the energy basis), so it's worthwhile to learn a more general formalism that treats all bases on equal footing (eg the Dirac bra-ket notation, which I'm going to refer to as "the operator language").

There are several practical examples of ways in which using the operator language gives you advantages over just solving the Schrodinger equation

• The ladder operator method applied to the quantum harmonic oscillator would be my "starter example" of a way that linear algebra, Hilbert spaces, and operator methods are actually used to solve problems and give you more insight than just the Schrodinger equation.
• Another example from intro QM is the derivation that angular momentum is quantized, which also uses ladder operators constructed from the commutation relationships of the angular momentum operators.
• As a more advanced example, you can solve the hydrogen atom using ladder operators and a hidden $SO(4)$ symmetry.
• Moving away from exact solutions into messy real world applications, in degenerate perturbation theory (for instance applied to computing corrections to the spectrum of Hydrogen in an applied electric field), a key step is identifying a complete set of observables that commute with the perturbation Hamiltonian that can be used to find a "good basis" of states in order to solve the perturbation equations.
• If you study relativistic quantum field theory, you will eventually find that the Schrodinger equation is actually not used at all in calculations because it becomes too unwieldy, and other techniques are needed. The same thing is true for some areas of condensed matter physics.

There are also many formal advantages.

• The Heisenberg uncertainty principle can be derived in a few lines from the commutator of the position and momentum operators in the operator language. The derivation can be generalized to give an uncertainty relation between any two Hermitian operators that don't commute.
• Symmetries in quantum mechanics can be understood by using representation theory of symmetry groups in the operator language. This lets us classify particles depending on how they transform under symmetries (this kind of logic led to the discovery of quarks).
• In relativistic quantum field theory, the symmetries of special relativity (the Poincaire group) are used to define what is meant by a particle. You can also derive that particles can be classified by their mass, spin, and internal quantum numbers like charge.
• When you move to multi-particle systems, having a good understanding of the operator representation is essential. The Schrodinger equation gives many people an incorrect idea that the wavefunction is defined over "space", while in the operator picture it is clearer that the state is a function of the position (and other degrees of freedom) of every particle. This leads to an incredible amount of avoidable confusion.
  • Properties of identical particles can also be derived from exchange operators.
  • It's also important to understand that the Hilbert spaces that bosons and fermions "live in" are different from the Hilbert space of non-identical quantum particles because the states have to be totally symmetric or anti-symmetric.
• Density matrices, which are easiest to define in the operator language, let you deal with situations where there is both "quantum" and "classical" uncertainty in the state of the system, letting you treat quantum systems at finite temperature.
• In bra-ket notation, one can prove the equivalence between the Schrodinger picture, the Heisenberg picture, intermediate (or "interaction") pictures, and the path integral formalism. These are all different but equivalent formulations of quantum mechanics that are each useful in different circumstances. Being able to pass from one to the other is an essential skill as you become more advanced, and would be very difficult without the operator and Hilbert space language.

This is not a complete list at all. But just to say that there are many ways in which the operator language is used in quantum mechanics, and solving the Schrodinger equation is not sufficient for a deep understanding.

Answer 3

I think that the issue is that sometimes in Physics people are much more concerned with the results of calculations instead of the nature of the underlying mathematical structures. Just to give a quick example of that, in General Relativity we are often interested in studying the motion of a particle in some spacetime. One then often denotes the coordinates of the particle's worldline as $x^\mu(\tau)$ and studies the geodesic equation $$\ddot{x}^\mu+\Gamma^\mu_{\nu\sigma}\left(x\left(\tau\right)\right)\dot{x}^\nu \dot{x}^\sigma=0.\tag{1}$$

Well, $x^\mu(\tau)$ are then just four functions of a single variable, while $\Gamma^\mu_{\nu\sigma}(x)$ is just a collection of functions of the four variables $x^\mu$.

As far as the calculation is concerned it boils down to (1), but behind the scenes $x^\mu(\tau)$ is really the coordinate representative of a curve in a smooth manifold with semi-Riemannian metric, while $\Gamma^\mu_{\nu\sigma}$ are the coordinate representatives of an object called the Levi-Civita connection on said manifold.

The same can be said of Quantum Mechanics. One may just look to the Schrodinger equation $$-\dfrac{\hbar^2}{2m}\nabla^2\Psi+V\Psi=i\hbar \dfrac{\partial \Psi}{\partial t}\tag{2}$$

and see there a second order partial differential equation for a complex-valued function $\Psi(t,x)$. This would be the quantum mechanical, non-relativistic analogue of (1).

Again, as far as the calculation is concerned it boils down to (2), but now behind the scenes $\Psi(t,x)$ is actually a curve in a Hilbert space $t\mapsto \Psi(t,\cdot)$ whose elements are complex valued, square integrable functions. One may invoke one even more abstract point of view in which we have a basis $|x\rangle$ and we view the quantities $\Psi(t,x)=\langle x|\Psi(t)\rangle$ as the coordinates of the vector $|\Psi(t)\rangle$ in that basis.

This kind of thing is everywhere: we have one abstract mathematical structure underlying all calculations in Physics. The reason why this abstract mathematical structure is important, instead of just the calculation, is that it organizes things in a much more logical way. This allows for a better comprehension, for the removal of ambiguities, and sometimes it gives insight on simplifying calculations.

Since you are studying Quantum Mechanics you'll soon find this out: when you study the harmonic oscillator, going through the components calculation (2) will lead you to solve Hermite's equation. But harnessing the abstract Hilbert space viewpoint will lead you to annihilation and creation operators and a relatively simpler (and IMO more elegant) method of obtaining your solutions.

The same will happen with orbital angular momentum where (2) will lead you to study the Laplacian eigenvalue problem on the sphere (and eventually Legendre's associated equation) whereas the abstract Hilbert space point of view will lead you to study the algebra of rotations which ultimately gives you an alternative construction of the spherical harmonics.

Answer 4

Since you wanted an example of where the abstract formulation of state vectors can be used here is sort of a 'derivation' (read as motivation) of the momentum operator in position space. All of the following is in 1-D

Suppose $\left|\Psi(t)\right>$ is the state vector of the system and $\left|x\right>$ is a position eigenstate (strictly speaking these are not in the Hilbert space). Since $\left|x\right>$ forms the basis for our space we can write the state as a linear combination of the $\left|x\right>$ as follows $$\left|\Psi(t)\right>=\int \mathrm{d}x \left|x\right> \left<x| \Psi(t)\right>$$ We can now borrow from classical mechanics and use the fact that momentum is a generator of translations. Thus we can write the infinitesimal translation operator $T(dx)$ as $$T(dx)=1-i\frac{\hat p dx}{\hbar}$$ The $-i$ appears since $\hat p$ is assumed to be hermitian (the $\hbar $ is for matching units) and that

a) $T(dx)$ be unitary
b) $T(-dx)=T^{-1}(dx)$ and that
c) $T(dx)T(dx')=T(dx+dx')+O(dx^2)$

We can apply $T(\Delta x)$ to our state above and we get $$\begin{aligned}(1-i\frac{\hat p \Delta x}{\hbar})\left|\Psi(t)\right>&=\int\mathrm{d}x'T(\Delta x)\left|x'\right> \left<x'| \Psi(t)\right> \\&=\int\mathrm{d}x'\left|x'+\Delta x\right> \left<x'| \Psi(t)\right> \\&=\int\mathrm{d}x'\left|x'\right>\left<x'-\Delta x| \Psi(t)\right> \\&=\int\mathrm{d}x'\left|x'\right>(\left<x'| \Psi(t)\right>-\Delta x\frac{\partial}{\partial x'} \left<x'| \Psi(t)\right>) \end{aligned}$$ The second line uses the definition of $T(dx)$ on a position eigenket, the third line does a substitution of variables $x'+\Delta x \rightarrow x'$ and the last line performs a taylor expansion. Expanding both sides and cancelling $\left|\Psi(t)\right>$ we get $$\hat p \left|\Psi(t)\right>=\int\mathrm{d}x'\left|x'\right>(-i\hbar \frac{\partial}{\partial x'} \left<x'| \Psi(t)\right>)$$

Multiplying on the left by $\left<x\right|$ and using the fact that $\left<x|x'\right>=\delta(x-x')$ we get $$\left<x\right|\hat p \left|\Psi(t)\right>=-i\hbar \frac{\partial}{\partial x} \left<x| \Psi(t)\right>$$

Since $\left<x| \Psi(t)\right>$ is the wavefunction the momentum operator in position space is $$\hat p_{position\ space}=-i\hbar \frac{\partial}{\partial x}$$

Answer 5

I see two important questions being asked here:

Question 1: What is the point of linear algebra in QM when all the problems I solve are typically working with wavefunctions and done use any linear algebra?

Question 2: Conceptually, what is the meaning of "Hilbert spaces"? Why are "Hilbert spaces" mentioned so much like their is something special about them, when the definition is almost never actually mentioned or used in any real problem solving I've encountered.

The first question is what is answered by @user7896, @andrew, @jMurray, and @gold. I think these answers are fine, but have another suggestion as well.

I recommend you look at spin-1/2 systems. This is a commonly used system which can be in a discrete superposition of only 2 discrete possibilities (spin up or spin down). The quantum state of this system can be represented by a simple 2x1 matrix (an array for the probability amplitudes associated by each state). This is the simplest quantum state to work with and is often referred to as a qbit. Often just linear algebra is enough to tell you what will occur with this state. If for-instance you want to evolve it in time, you can multiply it by $e^{i\hat{H} t}$ where $\hat{H}$ is a 2-by-2 matrix representing the Hamiltonian of the system. I find the linear algebra method is much more useful for understanding how probability amplitudes interfere. Now any problem with a discrete number of states can just work with linear algebra, and you can see how the probability amplitudes add and subtract very clearly.

Additionally, all continuous systems (such as a particle in a box) can actually be thought as a what occurs when you take a vector and give it an uncountably infinite number of possibilities (with probability amplitudes associated with each possibility). So imagine a particle in a box as being in N discrete bins, and then take the number of bins to infinity. You can represent it with linear algebra for the discrete part, and the full continuous representation is just when you take it to infinitely many bins.

Now only one person (@J. Murray) so far has tried to answer the second question:

Question 2: Conceptually, what is the meaning of "Hilbert spaces"? Why are "Hilbert spaces" mentioned so much like their is something special about them, when the definition is almost never actually mentioned or used in any real problem solving I've encountered.

Technically on wikipedia it is written that:

Hilbert spaces (named for David Hilbert) allow generalizing the methods of linear algebra and calculus from the two-dimensional and three dimensional Euclidean spaces to spaces that may have an infinite dimension. A Hilbert space is a vector space equipped with an inner product operation, which allows defining a distance function and perpendicularity (known as orthogonality in this context). Furthermore, Hilbert spaces are complete for this distance, which means that there are enough limits in the space to allow the techniques of calculus to be used.

Which is similar to what I was saying in the first part of the answer. (Take the bins to infinity and rework the algebraic inner product in that limit).

But there's one conceptual component which I think is missing here. Colliqually we often use the term "Hilbert space" to refer to the "possibility space" that our probability amplitudes live in. This "possibility space" for simple problems is usually very obvious for simple problems. So for a qbit system (take a spin1/2 particle for example), our probability amplitudes can only exist in 2 possible states (so we might have a state $\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle$ for example, which could be written as $\left[\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\right]$ as a vector.).

But for a 2qbit system the amount of possibilities are 4. And a 3-qubit system the amount of possibilities are 8 (every possible configuration of possible ways the 3 particles could be up or down, like how many configurations there are for flipping 3 coins). Each possible outcome can be assigned its own independent probability amplitude.

For example, the Hilbert space for the 3-qubit system is:

$ \begin{align} |0\rangle_1|0\rangle_2|0\rangle_3\\ |0\rangle_1|0\rangle_2|1\rangle_3\\ |0\rangle_1|1\rangle_2|0\rangle_3\\ |0\rangle_1|1\rangle_2|1\rangle_3\\ |1\rangle_1|0\rangle_2|0\rangle_3\\ |1\rangle_1|0\rangle_2|1\rangle_3\\ |1\rangle_1|1\rangle_2|0\rangle_3\\ |1\rangle_1|1\rangle_2|1\rangle_3 \\ \end{align} $

The Hilbert space for a 3-qubit system is the set of all possible configurations each of these states can be in, and we assign probability amplitudes to these combinatoric configurations. So for example a 3-qubit system could be in a superposition state $|0\rangle_1|0\rangle_2|0\rangle_3- |0\rangle_1|1\rangle_2|0\rangle_3$. This fact that probability amplitudes are assigned to possible outcomes is, in my opinion, the essence of QM.

This means that our "Hilbert space" (which we really mean is the possibility space that all of our configurations exist in that can be assigned probability amplitudes) gets much, much larger.

It is this colloquial use of the term "Hilbert space" in which we are referring to the exponentially-increasing possibility space that I think is very rarely explained and is often the source of conceptual confusion.

Answer 6

Several great answers so far, but it seems everyone is only hinting at a basic key point that I feel should be made explicit. The question I have in mind is: what is the reason why Hilbert space $L^2$ is the appropriate space for wavefunctions, instead of any other space? And it's really simple.

A quantum state is expressed by a complex-valued wavefunction $\psi$. If we admit that its square magnitude, $$ |\psi|^2 $$ can always be interpreted as a probability, then it follows that this probability distribution is normalizable, $$ \int_{-\infty} ^\infty |\psi(q)|^2\, dq =1. $$ There are plenty of functions that cannot meet this criteria, e.g. $\sin$ and $\cos$, unless we integrate only across finite limits. We should only admit solutions which could actually define probability distributions. For a wavefunction to be considered physical then, it must not go to infinity when its square norm is integrated. In other words, the wavefunction must be in the space of bounded square-integrable functions $L^2$.

The principle of superposition requires that this space be a linear vector space. (Or alternatively, the space of wavefunctions satisfies the criteria to be considered a vector space.)

The inner product has obvious utility, thus we can say the space is equipped with an inner product. In fact, it is the inner product which allows us to define things like orthogonality, and to determine whether a family of functions is a total set.

For notions of continuity and infinitesimal changes (necessary from a physical point of view) the space must be complete, i.e. every sequence which is Cauchy actually converges (loosely, every sequence of functions that looks like it converges, actually does; compare the rationals to the reals).

The study of complete inner product spaces is named for David Hilbert. Detailed study provides us with the same insights that real and complex analysis provide to elementary calculus. In the same way that we can perform calculations without explicitly mentioning Hilbert space, we can do calculus without referring to the completeness of the reals. But any mathematician will tell you, the only reason we can do calculus at all is because the field of real numbers is complete. This distills the unique features of real numbers down to the most basic distinguishing element, and provides the necessary framework for various interesting and useful theorems and counterexamples which are can used to "stress test" the theory and navigate abstract and difficult problems.

In my opinion, Hilbert spaces being foundational to QM is like complete fields being foundational to classical mechanics - it is, but I'm not sure a student needs to hear that, unless you're going to follow it up with more discussion about the role and utility of formalism.

Answer 7

Your confusion stems from the difference in how math is taught to physicists and mathematicians.

For example take an expression like $$\int_{-\infty}^\infty\mathrm d x\, |f(x)|^2.$$ A physicist would immediately start to try compute this using the tools that he knows but a mathematician would first be asking questions. This is because this integral is actually a limit: $$\lim_{R\rightarrow\infty}\int_{-R}^R\mathrm d x\, |f(x)|^2$$ So the mathematician might ask does this limit even exist? If it doesn't you might get nonsense answers which change depending on how you take the limit. There might still be other valid questions you could ask but I'm a physicist so I don't know them.

One of the results that led to Hilbert spaces is the fact if $f,g$ are both square integrable i.e. $$\int_{-\infty}^\infty\mathrm d x\, |f(x)|^2\text{ is finite}$$ then we can define an inner product given by $$\langle f|g\rangle=\int_{-\infty}^\infty\mathrm dx f^*(x)g(x).$$ This inner product then has many nice properties which make it behave like the regular dot product and this allows us to use the power of linear algebra. See also https://en.wikipedia.org/wiki/Hilbert_space#History.

So doing quantum mechanics without knowing about Hilbert spaces is like driving a car without knowing how the engine works. You'll be fine most of the time but you should appreciate all the hard work that went into making and maintaining your car. I passed my QM class with only a vague understanding of Hilbert spaces. In fact one of the first wavefunctions you learn about is $\exp(ipx/\hbar)$ which is an eigenfunction of the momentum operator and also of the free particle Hamiltonian. This wavefunction isn't even square integrable and so it isn't a proper solution of the free particle (a superposition in the form of an integral is though). This is often swept under the rug and for a good reason. You want to learn about QM and learning the entire theory behind Hilbert spaces would be lengthy and/or confusing.