Why do physicists use the Bra and Ket notation when mathematicians tend not to?

Question

Please forgive me that this is a lay question. I realise it will not add any profound physics to stack exchange, and I ask only out of layman's curiosity.

I saw a question on google from a physicist asking (on Quora I think) why mathematicians do not use <bra| and |ket> notation? I thought an equally interesting question would be, why do physicists like to use the <bra| and |ket> notation? This isn't a criticism of the physics notation, I am sure both notations are good for the purpose they are being used for: mathematicians want to prove theorems, physicists want to perform computations. I am just genuinely curious what bras and kets provide. I thought an example might help me identify the reason why bras and kets are preferred. Below I am trying to present what are the mathematician's and physicist's way of expressing the same problem. At the end I try to ask what the bra and ket vectors are delivering (which the mathematicians view might be missing).

My simple example case is of the infinite potential well "particle in a box" in 1D. The Schrödinger equation inside the box $0<x<L$ where the potential V is zero is given by

$$-\frac{\hbar^{2}}{2m}\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}}\Psi=E\Psi$$

Since the potential is infinite we say that the wave function $\Psi$ is zero at boundaries of the box: $x=0$ and $x=L$

The solutions are then

$$\Psi=A\sin\left(k_{n}\frac{x}{L}\right)$$

with

$$k_{n}=n\pi$$

A mathematician would say that the $\Psi_{n}$ are the eigenvectors which belong to the $L^{2}$ of Hilbert periodic complex functions. They would say that there is an inner product between two vectors $\Psi_n$ and $\Psi_m$, which they write as$(\Psi_n,\Psi_m) $, and which they define as:

$$(\Psi_n,\Psi_m) = \int_{0}^{L} \Psi_{n}^{*}\Psi_{m}dx$$

A physicist will say (I think), that the $\Psi_{n}$ are not eigenvectors at all, but are wavefunctions. I think physicists will say that the eigenvectors are the quantities which they call $|\Psi_n\rangle$. I think physicists will say that the $|\Psi_n \rangle $ belong to the Hilbert space and that the $\Psi_{n}$ do not. Have I got that right?

I think a physicist will say that they take an "inner product" between a $\langle \Psi_n|$ and a $|\Psi_n\rangle $ as follows:

$$\langle \Psi_n,\Psi_n\rangle $$

I think they say that

$$\langle \Psi_n,\Psi_n\rangle = \int_{0}^{L} \Psi_{n}^{*}\Psi_{m}dx$$

Have I got that right?

My main question is, what does the $|\Psi\rangle$ give physicists that is missing in the mathematicians notation? Something that might help me understand would be to see what is the the ket notation is adding here would be to see what is the concrete expression for $|\Psi\rangle$ in the example given. What I mean is, is there a concrete expression for $|\Psi\rangle$ of the form below? Or is it $|\Psi\rangle$ just an abstract thing that is never assigned a concrete form?

$|\Psi\rangle = $ what concrete equation?

Answer 1

I've always thought it was partly the NIH (not invented here) syndrome --- but this is perhaps unfair. Mathematicians were very chary of $\delta(x)$ when Dirac invented it, and for good reason. Dirac's $\delta(x)$ makes many calculations easier, but it sweeps much under the rug and takes some effort to make rigorous., The same is true for $|\psi\rangle$ versus $\psi(x)\in L^2[{\mathbb R}]$. Dirac notation cleverly conceals all issues of convergence, domains of definition, and hermitian versus self adjoint and so on, so that one can see the general structure of a calculation. The very things that Dirac notation conceals are often the central interest of a mathematician.

Answer 2

$| \Psi \rangle$ really is just the vector corresponding to the wavefunction $\Psi$, there really is nothing more to it than that. Think of the $| \cdot \rangle$ as just a bit of notational dressing, indicating it is a state vector. The thing you put inside the $| \cdot \rangle$ is just a label. Here, that label is the wavefunction $\Psi$, as the function is itself the vector. It is exactly akin to how one might write a vector as $v_i$. Here, the subscript $i$ is just a label saying which vector you're talking about.

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I think physicists like to do this because it gives you a natural way to discuss the states $| \Psi \rangle$ and dual states $\langle \Psi |$ on the same footing. It certainly makes sense to me. For instance, say you have the Hamiltonian operator $H$, which has a set of eigenstates $| i \rangle$ with eigenvalue $E_i$. $$ H | i\rangle = E_i | i\rangle $$ One is able to write this operator easily as $$ H = \sum_i E_i | i \rangle \langle i | $$ assuming we are using an orthonormal basis $$ \langle i| j \rangle = \delta_{ij}. $$ such expressions often allow for the quick derivation of many identities. In the mathematician notation you use, there is no easy way to write $H$ in this way.

Completeness relations are also really easy to express with the bra-ket notation, for the same reason For instance, $$ 1 = \int dy |y \rangle \langle y | $$ is an expression which gets used a lot.

Ultimately, there is no reason to privilege states over dual states, and the notation reflects this fact.

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Notice that the states $| y \rangle$ correspond to the states with the wave function $\delta(y - x)$. So $$ \langle y | \Psi \rangle = \int dx \delta(y - x) \Psi(x) = \Psi(y). $$ Notice that, if we insert the identity, $$ | \Psi \rangle = 1 | \Psi \rangle = \int dy | y \rangle \langle y | \Psi \rangle = \int dy \Psi(y) | y \rangle $$

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Edit: As pointed out in mike stone's insightful answer, I think the difference in usage between mathematicians and physicists can be accounted for as follows.

Physicists are happy to make an isomophism between the states and dual states (as I myself did in the above answer). However, in infinite dimensions, that is often not the case. For instance, $\delta(y- x)$ is not really a function, but the map $|\Psi\rangle\mapsto \Psi(y)$ is certainly a bonafide linear map from $L^2 \mapsto \mathbb{C}$. A physicist would denote this map as $\langle y|$, and also argue that $| y \rangle$ should itself be thought of as a legitimate state, one with a wave function of $\delta(y-x)$. Now, obviously, $\delta(y-x)$ is not actually a function, but morally speaking it is. So physicists are happy to make this identification, switching back and forth between $| y \rangle $ and $\langle y|$, even though mathematicians might frown at this.

Answer 3

I would like to play Devil's advocate here (Devil being "the mathematicians"). Dirac's bra-ket notation is only useful when considering a Hermitian operator sandwiched between two state vectors --even ignoring matters of domain, topological properties, etc. It just so happens that that's the case most of the time in quantum mechanics.

Here are some examples of very simple situations for which Dirac's notation only stands in the way:

1. Fundamental definitions. E.g., what the adjoint of an operator $Q$ is: $$ \left(\psi_{1},Q\psi_{2}\right)=\left(Q^{\dagger}\psi_{1},\psi_{2}\right) $$
2. Operators acting on both variables (time inversion in quantum mechanics being an important case) $$ \left\langle T\psi_{1},T\psi_{2}\right\rangle =\left\langle \psi_{1},\psi_{2}\right\rangle ^{*} $$
3. Products of operators (even if they're all Hermitian) that do not commute: $$ \left(\psi_{1},ABCD\cdots\psi_{2}\right)=\left(\cdots DCBA\psi_{1},\psi_{2}\right) $$ Take a look at Weinberg's book on QFT, that deals extensively with symmetries and properties of states. Very rigorous book. Weinberg shies away from Dirac's notation very often, siding much closer to mathematicians in that respect. And he has very good reasons for it. Try to express any of the former examples with Dirac's notation. It's just not possible.

It's only because most of the time in QM we consider observables (Hermitians) $Q$ that, $$ \left(\psi_{1},Q\psi_{2}\right)=\left(Q\psi_{1},\psi_{2}\right) $$ and so we can write without any ambiguity, $$ \left\langle \psi_{1}\left|Q\right|\psi_{2}\right\rangle $$ There are other very important points related to the OP's question that have been more than satisfactorily addressed, but I thought this was a very important matter related to the question that somehow was missing.

Answer 4

One additional answer to add onto everyone else's insightful answers: a great reason for physicists to use bra-ket notation is because the notation is completely agnostic about the Hilbert space. To a mathematician, the states $ | \uparrow >$ for a spin-half system, $| n \rangle$ for a simple harmonic oscillator, $| k \rangle$ for a free particle momentum eigenstate, and the ground state $| 0 \rangle$ of a quantum field theory are all extremely different beasts, living in vastly different Hilbert spaces (which might not even be so well-defined, or at least require significant scaffolding to make into an appropriate Hilbert space). But to a physicist, they're all just "states". We know from context which sort of states we're dealing with, but we aren't too picky about exactly defining which Hilbert space we're working with, because it almost never matters for practical computations. As soon as you see the dressing $| \cdot \rangle$ you know you're working with a unit vector in the relevant Hilbert space, whatever it is. On the other hand, if you write a vector in your Hilbert space as $\psi$, or $v$, it usually needs to be accompanied by "$\in \mathcal{H}$" to specify what sort of object $\psi$ or $v$ is, which naturally prompts the reader to follow up by asking what $\mathcal{H}$ is.

Answer 5

To answer this succinctly, it’s because physicists are doing something conceptually cleaner, at the cost of notational precision and rigor.

On pp. 2-3 of Folland's "Quantum Field Theory: A Tourist's Guide for Mathematicians", he begins (remembering here that a Hilbert space, albeit rather abstract, is a place where you can have things like a sequence of vectors which are mutually orthogonal, and there's a topology so you can take limits of sequences, etc. Typically they consists of functions on some space and are fairly familiar.)

Now, some more subtle matters. In mathematicians' dialect, $\langle u|v\rangle$ is the inner product of two vectors $u$ and $v$ in the Hilbert space $\mathcal{H}$. Moreover, if $u \in \mathcal{H}$, the map $\phi_{u}(v)=\langle u | v \rangle$ is a bounded linear functional on $\mathcal{H}$; the correspondence $u \leftrightarrow \phi_u$ gives a conjugate-linear identification of $\mathcal{H}$ with its dual $\mathcal{H}'$, which mathematicians generally take for granted without employing any special notation for it. Physicists, on the other hand, distinguish between elements of $\mathcal{H}$ and elements of $\mathcal{H}'$, which they respectively call ket vectors and bra vectors and denote by symbols of the form and $|u \rangle$ and $\langle u|$. If $|u \rangle$ is a ket vector (what mathematicians might simply call $u$), the corresponding bra vector (linear functional) $\langle u |$ on the ket vector (element of $\mathcal{H}$) $|u \rangle$ is the inner product (i.e., "bracket" or "bra-ket") $\langle u | v \rangle$.

Be forgiving, symbols and abstract concepts are often all mathematicians have (there isn’t necessarily a concrete real world object corresponding to, say, a scheme), so one must use them very carefully.

The bra is a linear functional which acts on a vector to give a number. The bra vectors live in a dual space which is typically linearly and topologically isomorphic to the space they’re acting on. Physicists are happy to use any convenient label, often skipping subscripts so that $|a_n \rangle$ becomes $|n \rangle$.

So, rather than writing $\phi_n(|m\rangle)$ to denote something like a projection operator acting on a state to give an observable/measurable number, physicists are more comfortable saying that when you smoosh two states corresponding to, say, energy levels $n$ and $m$ together you get a number, which is neat.

It also appears to be handy when you want to fluidly build and manipulate practical operators (linear functionals) that move states (vectors) around a Hilbert space, possibly annihilating them by sending them to 0 (not be confused with the vacuum, $|0 \rangle$).

Answer 6

The Bra-Ket notation is really a physicist notational trick, down to Dirac (?), otherwise I don't think there is any real difference between the subject areas.

A Hilbert space is just a vector space with inner product; in QM those vectors tend to be functions or spin vectors. I.e. $|\Psi\rangle$ can be any relevant vector and depends on the problem. E.g. $|\Psi\rangle = A\exp(k.x - w.t)$ or $|\Psi\rangle = \big(\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\big)$

Note that Mathematicians often use <.|.> for inner products as well as (.,.).

I was told 30 years ago by my physics lecturer that the bra-ket notation came about by pulling apart the inner product into its constituent parts, so the inner product $\langle a|b\rangle$ of the vectors $|a\rangle$ and $|b\rangle$ looks like the application of the dual of $\langle a|$ to the vector $|b\rangle$. (Mathematicians love that kind of stuff). Also, using the bra-ket notation, the tensor $|a\rangle\langle b|$ looks like $a \otimes b$.

What does Bra-Ket notation give? The notation <$\Psi$|Q|$\Phi$> is very evocative of the physical process

$$Initialstate = \Phi \rightarrow process \rightarrow \Psi = finalstate.$$

Bra-Ket is also widely used and as good as anything. That's probably all you can ask for a notation.

Answer 7

Culturally, I have the impression that notation is a lot more "loaded" in physics: for example, there are many letters with specific, hard-coded, unchangeable meanings. This happens less in mathematics. Using bra/ket notation, it's immediately clear what kind of object you're dealing with (a state, or a functional on states, or an expectation value), and no-one has to spend time defining things and writing sentences like "Consider the state $v_{\Psi}$ corresponding to the position-space wavefunction $\Psi$" or "let $(,)$ be (some specific) inner product on the Hilbert space of states".