# Separating Math From Physics In Quantum Mechanics [closed]

So, I'm going through my first quantum mechanics class in university and I feel like there is not a strong distinction made between a purely mathematical system and purely physical system, which are not exactly the same.

To me, for all the classes I've took so far, it seems like we model physical phenomenon by creating a "mathematical world/geometry" that fits with our understanding of the universe. But in a way this mathematical world and this physical world are separate and, quite often, in science, we end up deciding that a certain mathematical world does not actually describe well our own, so we find another one.

In the past, I've taken a calculus class and then took a physics class to realize that you could model the real-world with the math I've learned. But, in QM, we're not talking about the mathematical tools and models, we are using, as a subset of a larger set of possible models (I'm not trying to reinvent a new QM theory, just to understand how it fits in math). Say, in classical mechanics, we model the trajectory of a free falling object with a parabola, but this parabola is not only physical: it's a mathematical object before anything else. But now we might add drag and then use a different mathematical object that accounts for this factor.

We've been using Hilbert Spaces for the entire semester, and, as a purely mathematical object, I have no idea what it represents. I can only tell you that it contains my quantum states.

What I'm trying to understand: How does quantum mechanics fit in a non-physical, purely mathematical creation? So, here are a couple of satellite questions that I've been asking myself

• Would you need an $$\hbar$$ everywhere in this imaginary world (and actually I have no idea where these come from in our mathematical discussions of QM, apart from simply saying: "well we got that experimentally, so let's just shove them in there somewhere")?

• Also, how does the uncertainty principle fit within calculus? If I have a function for the position of an object, by taking its derivative, I can find both its position and momentum with absolute certainty. This makes sense in math, but apparently, it does not in physics, because we take the fourier transform to find the momentum. What is the difference between these two models.

I apologize if my train of thought is a bit confused, I have difficulty making sense of all of it. Perhaps, time is the only remedy.

• I sounds like the questions you're struggling with really come down to: how do we relate the abstract mathematical objects we are manipulating in quantum mechanics back to the things we are used to working with. This is always a problem for students first learning QM, and unfortunately the most common book, Griffiths, explicitly avoids discussing this. In fact, the only book I know off the top of my head which works hard to motivate/derive the structure of QM is by Ballentine (also undergraduate level). I recommend looking there. – Richard Myers Nov 22 '20 at 0:58
• Asking “how does quantum mechanics fit in a non-physical, purely mathematical creation? seems the opposite of “separating math and physics”. – G. Smith Nov 22 '20 at 1:02
• Would you need an $\hbar$ everywhere in this imaginary world? That is a vague and meaningless question with no sensible answer. – G. Smith Nov 22 '20 at 1:04
• we are using, as a subset of a larger set of possible models We are using what? You left out the object of the verb “to use”. – G. Smith Nov 22 '20 at 1:08
• Can you elaborate on your current knowledge of variational calculus and linear analysis? Do you know what a vector space is? Are you comfortable with Fourier transforms? Are you aware of the Fourier transform as a 'change of basis'-- do you know what that means? etc. – Myridium Nov 22 '20 at 4:49

This answer will hopefully complement what you are already learning in university. It won't be a replacement however. I will show you how the Heisenberg uncertainty principle between position and momentum emerges as a statement of Fourier transforms. It is nothing more or less than that.

Personally I feel that this topic is so difficult to understand because it requires a heavy heaping of Linear Analysis knowledge and also of variatonal calculus in order to appreciate how the development of Quantum Theory came about. Physics degrees should be longer in order to develop this knowledge properly, but they aren't, and students are forced to 'learn' something they are not yet equipped to understand.

I would advise you not to worry too much about not understanding QM at this stage. I know that is difficult for people who feel a need to understand, I feel that too. You will be able to come back to this subject to truly appreciate it if you continue your studies. And the more knowledge you have about surrounding topics (Linear Algebra, variational mechanics, qualitative description of Quantum Phenomena) the more easily you will be able to understand the mathematics. It is a tough subject because the historical context is also of great importance. QM is basically an extension of classical mechanics, where you assume particles are waves instead of points. The maths was brute-forced via the quantum postulates in order to create something that matches classical behaviour in the limit. I still don't understand it myself. It's amazing how much theory emerged from the Ultraviolet Catastrophy and the quantisation of atomic electron energies. I think that's where it all came from.

## State vector space

Historically, I guess it was noticed that particles behave like waves-- e.g. an electron through a single slit interferes with itself, in exactly the same way as an electromagnetic wave would. At some point it was figured out that you could represent the particle state as a complex wave $$\psi(\mathbf x)$$. A complex wave has both a phase and a magnitude as nicely separated properties. (phase = $$\arg(\psi(\mathbf x))$$ and magnitude = $$|\psi(\mathbf x)|$$) This makes it an appealing model for a wavelike particle.

We need to make this new model correspond with classical physics in the limit. A classical particle would be something which is completely localised in $$\mathbf x$$. In other words, a Dirac-delta function $$\psi(\mathbf x) = \delta(\mathbf x - \mathbf x')$$ for a classical particle located at $$\mathbf x'$$.

Next it is important to understand that the space of possible complex 'waves' is a vector space. More precisely, the space of square-integrable functions from $$\mathbb R$$ to $$\mathbb C$$ we'll call $$L^2[\mathbb R, \mathbb C]$$. This is the space of possible states of a particle we could hope to represent in one spacial dimension. We would also have to modulo out the varying total integral-- we only want states that integrate to $$1$$. For simplicity we will ignore this though. Any function in this $$L^2$$-space (corresponding to a particle space) can be represented as an infinite sum of Dirac-delta functions: $$\psi(x) = \int \psi(x) \delta(x - x') \; \mathrm dx'$$ You may say 'well duh' but this is important. We can express any wavefunction $$\psi(x)$$ as a linear combination of these basis states $$\mathbf e_{x'} = \delta(x - x')$$. The $$L^2$$-space is a vector space. It is also a Hilbert space which means it is additionally a complete space (bit of a technicality, not really important here) and also **is equipped with an inner product $$\langle \cdot , \cdot \rangle$$. It is this inner product which allows you to compute how you decompose some arbitrary state into a sum of vectors in an orthonormal basis $$\{\mathbf e_i\}_i$$.

## Braket notation

The particle state described by $$\psi(x)$$ is a vector in described $$L^2$$-space. When we write it as a function of $$x$$ we are really just choosing to express it in a particular basis-- the basis of definite-position states (a.k.a. position eigenvector states). In order to be clear about referring to the vector itself, not its components in any given basis, we rewrite the state as $$| \psi \rangle$$. The inner product allows us to express it in any basis we choose like so: $$| \psi \rangle = \int_i \langle \mathbf e_i , | \psi \rangle \rangle \mathbf e_i \; \mathrm d i$$ where we have expressed $$| \psi \rangle$$ as a sum of scalar multiples ($$\langle \mathbf e_i , | \psi \rangle \rangle$$) of the basis functions ($$\mathbf e_i$$). We denote the definite-position basis function, the Dirac-delta, as: $$| \delta(x - x') \rangle \equiv | x' \rangle.$$ Once again, this is an instance of re-expressing a particular state in a coordinate-free way. We are not choosing to express it as a function of $$x$$ because that selects a special basis. As a notational aside: now that $$\mathbf e_x = | x \rangle$$ we write the inner products like: $$\langle | x \rangle , | \psi \rangle \rangle = \langle x | \psi \rangle.$$ This is called the Bra-ket notation. It is purely a notational convenience. It is easier to manipulate beautiful equations, I think we will all agree. For a vector in this Hilbert space, we can find what it is 'as a function of $$x$$', by computing its decomposition in the basis of definite-position states i.e. Dirac-deltas. Precisely speaking, this equation holds: $$\psi(x) = \langle x | \psi \rangle = \langle \psi | x \rangle^*.$$ In fact this is the proper way to think about what a wavefunction of $$x$$, i.e. $$\psi(x)$$ actually means in Quantum Mechanics. The state vector $$| \psi \rangle$$ is rewritten: $$| \psi \rangle = \left(\int_{x'} \underbrace{| x' \rangle}_{\text{Position-basis vector}} \overbrace{\langle x' | \psi \rangle}^{\text{Complex scalar multiple}} \; \mathrm d x'\right).$$

## Heisenberg Uncertainty Principle as property of Fourier Transforms

We can re-express $$| \psi \rangle$$ as a sum of definite-momentum states. I won't get into the why, but such a state is given as a function of $$x$$ by: $$\langle x | p \rangle \equiv e^{- \frac i \hbar p x}.$$ Also known as a De Broglie wave. Given $$\psi(x)$$, how do we compute its expression as a sum of definite-momentum-states $$\psi_p(p)$$? You know from Cartesian vector spaces that if you can express two vectors in the same orthonormal basis, then you can simply multiply their components together (and sum) to get the inner product. So it is here, except we do need to take the complex conjugate of the first thing in the inner product. $$\psi_p(p) = \langle p | \psi \rangle = \int p(x')^* \psi(x') \; \mathrm d x' = \int \underbrace{\langle p | x' \rangle \langle x' | \psi \rangle \; \mathrm d x'}_{\text{x' may be replaced with any orthonormal basis}}.$$ Expressed in the $$x$$ basis it becomes a Fourier transform: $$\psi_p(p) = \int p(x')^* \psi(x') \; \mathrm d x' = \int e^{- \frac i \hbar p x'} \psi(x') \; \mathrm d x'.$$ You may then refer to theorems/intuition on the Fourier transform to understand why there is mutually exclusive narrowness in expressing $$| \psi \rangle$$ in the $$x$$ basis or the $$p$$ basis.

I must add a disclaimer to this section that the following advice is not from a vantage point of understanding quantum mechanics. I am still learning it myself. These are just my present thoughts on how one should think about the development of quantum mechanics and where to look to develop a rigorous understanding. I welcome suggestions for improvement of this section.

The 'first quantisation' of Quantum Mechanics was developed to reproduce classical behaviour. It is found that if you express the expectation value of any operator $$\hat p$$ as $$\langle \hat p \rangle = \langle \psi | \hat p | \psi \rangle$$ then you can recover the same equations of motion of those expectation values if you make those operators have commutators that look like the Poisson brackets of the corresponding classical operators.

For example, in the Hamiltonian formulation of classical mechanics, we have the Poisson bracket: $$\left\lbrace x, p \right\rbrace = 1 .$$ It turns out that if you choose your quantum operators $$\hat x$$ and $$\hat p$$ to have a commutation relation that is equal in value, multiplied by $$i$$ and some positive real factor, $$[\hat x, \hat p] = \hat x \hat p - \hat p \hat x = i \hbar ,$$ then the equations of motion of the expectation values $$\langle \hat p \rangle , \qquad \langle \hat x \rangle$$ matches the equations of motion of the classical $$x$$ and $$p$$. You may think of this as the the 'center' of a wavepacket following the same dynamics as a classical particle would. Indeed if you look at quantum mechanics at a large enough scale then the particles look pointlike anyway, and so we do need the equations of motion to match in this way.

My (limited) understanding is that the operators of quantum mechanics, e.g. $$\hat x$$ and $$\hat p$$ among the others, were deliberately chosen so that they would satisfy these commutation relations, and in turn produce classical behaviour in the limit.

In order to fully appreciate this, I think it is necessary to consider quantum mechanics as it was developed from classical analytical mechanics (e.g. Lagrangian, Hamiltonian formulations). For a text on classical mechanics, I can recommend The Variational Principles of Mechanics, Cornelius Lanczos which I am currently reading. There is a also a popular book I haven't read, Classical Mechanics, Herbert Goldstein. I think the material in these books is prerequisite to understanding the formulation of quantum mechanics, which was, to my understanding, a continuation of classical analytical mechanics.

After one is comfortable with classical mechanics, I think the next step would be to read an introductory book in quantum mechanics that pays respect to the historical development. I have found that introductory quantum mechanics texts seem to take it for granted that the reader will readily accept the transition from classical variables to quantum operators (the 'first quantisation'). Perhaps they consider the topic too large to cover. However, with the appropriate background in the Hamiltonian formalism, it should be less of a logical leap to accept that the quantum postulates are designed to produce a wave-behaviour that emulates classical physics in the limit.

• ...not sure it’s entirely right to think of HUP as related to Fourier transform. It is closely related to that for $\hat x$ and $\hat p$ but not so well related for other observables such as angular momenta (which are not conjugate). – ZeroTheHero Nov 22 '20 at 7:38
• @ZeroTheHero I'm not familiar with angular momentum uncertainty but it will be the same principle. A dirac-delta in one observable basis will be spread out in another observable basis, creating an uncertainty between them. In the position/momentum case it happens to be the Fourier transform between the two bases. – Myridium Nov 22 '20 at 12:21
• yes and no. The bases are not related by fourier transform and the product of uncertainty is state-dependent so the HUP is certainly not like conjugate variables. in particular the rhs can be 0. – ZeroTheHero Nov 22 '20 at 13:14
• @ZeroTheHero the product of position-momentum uncertainty is also state-dependent. – Myridium Nov 23 '20 at 2:47
• Of course not since the commutator is the unit: the rhs of $\Delta x\Delta p\ge \hbar/2$ is state-independent, whereas for angular momenta $\Delta L_x\Delta L_y\ge \frac{1}{2}\vert \langle L_z\rangle\vert$ explicitly depends on the average value of $L_z$ in the state used to evaluate the product of uncertainties. In particular it’s perfectly possible to have a state for which $\langle L_z\rangle=0$ but the product of variances is non-zero. – ZeroTheHero Nov 23 '20 at 2:55

It's too much to repeat here, but I suggest you find a copy of my late friend Vic Stenger's book, The Comprehensible Universe, and look at look at "Mathematical Supplement D, Gauge Invariances" pp 229 - 241. Vic gives a good exposition of why quantum mechanics is a natural consequence of assuming a kind of invariance of viewpoint. It's not a rigorous proof, but it's intuitive and it can be made rigorous. The rest of the book is pretty good to.

There is next to $$0$$ physics in quantum mechanics: when you walk into a lab, you don’t walk into a Hilbert space or find a Hermitian operator on the table.

Quantum mechanics is a set of mathematical rules to predict outcomes of experiments, period. The dimension of the Hilbert space depends on the number of mutually exclusive outcomes (aka eigenvalues) of operator, not on any physical property of this operator: it’s perfectly legitimate to have a Hilbert space of dimension $$5$$ if you a considering a quantum system with angular momentum $$L=2\hbar$$, even if angular momentum as we think of it is actually a vector in 3D. If the system has angular momentum $$3\hbar$$, the Hilbert space has dimension $$7$$, so the dimension of the Hilbert space has nothing to do with the operator but everything to do with outcomes of measuring the associated observable; the dimension of the Hilbert space reflects in part the preparation procedure of your system.

The rules are not unique: there are different formulations to reproduce the experimental data, some of which are summarized (with some overlaps and not all completely independent) in

Styer DF, Balkin MS, Becker KM, Burns MR, Dudley CE, Forth ST, Gaumer JS, Kramer MA, Oertel DC, Park LH, Rinkoski MT. Nine formulations of quantum mechanics. American Journal of Physics. 2002 Mar;70(3):288-97,

and we’re still struggling to understand what to make of the rules or what they mean: the problem of the interpretation of rules of quantum mechanics is not solved. All these include the uncertainty relations in some form as it appears to be a fundamental attribute necessary of any quantum theory.

Thus, it’s not clear at all what the Hilbert space represents in the formulation that you refer to, beyond stating that we have rules to manipulate objects in this abstract space and extract meaningful quantities that we can verify against experiments. It has the advantage of being accessible via linear algebra, and the rules that we have are known to work.

Maybe the phase space approach (it is closer to classical mechanics) is better for some problems: the dimension of the associated phase space is independent of the dimension of the Hilbert space and of the number of outcomes of measuring observables, but it is plagued with technical difficulties in all but the simplest cases (a practical expression for the $$\star$$-product for anything but HW algebra comes to mind). Moreover this formulation is not so useful to find eigenvalues, or maybe this quantity does not even make sense in this formulation.

At least we’ve made progress. The current formulations are clearer than the old Born-Sommerfeld-like approaches.

• There is next to 0 physics in classical mechanics: when you walk into a lab, you don’t walk into a configuration space or find a force vector on the table. – alexarvanitakis Nov 22 '20 at 5:06
• ... classical mechanics is not even accessible via linear algebra! – alexarvanitakis Nov 22 '20 at 5:16
• @alexarvanitakis ? the position of an object is pretty classical, and force I use to open the door is pretty classical too. – ZeroTheHero Nov 22 '20 at 6:10
• I mean: a mass on a spring does require coordinates and the pushing on the spring is something that’s pretty classical. – ZeroTheHero Nov 22 '20 at 6:22
• Quantum mechanics is a set of mathematical rules to predict outcomes of experiments, period. - it's strange how QM has this way of making physicists turncoat, and decide the universe is how they imagine it, rather than what its mathematical properties say about it. – Myridium Nov 22 '20 at 6:29