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Dirac$^1$, starting on pg. 18, introduces ‘Bra’ vectors. He starts with some existing ket vectors and introduces a new set of vectors, the bra vectors. He say’s of this new set of vectors, see pg. 19,

The new vectors are, of course, defined only to the extent that their scalar products with the original ket vectors are given numbers, but this is sufficient for one to be able to build up a mathematical theory about them.

He uses this point of view in presenting his general abstract mathematical theory of bras.

I wonder, if we have to adopt this standpoint, when we take a particular example of a complex vector space of kets to work with?

For example, if we want to work with the space $V$ of kets, where

\begin{equation*} V=\left\{ f|f:\mathbb{R}\to\mathbb{C},\,x \mapsto f(x), \int_{- \infty}^\infty f^\ast(x)~f(x)~dx~<\infty ~\right\} \end{equation*} and an arbitrary element from $V$, may be given in ket notation, as $|f\rangle$.

In this case, can we say that the bra that corresponds to the ket $|f\rangle$, i.e. $\langle f|$, is given by \begin{equation*} \langle f|=f^\ast \end{equation*} or do we merely know, that for any ket $|g\rangle=g$, that

\begin{equation*} \langle f|g\rangle=\int_{- \infty}^\infty f^\ast(x)~g(x)~dx \end{equation*}

My question is: In using Dirac’s approach to bra vectors, do we ever know what our bra vectors are?

Reference

1, P.A.M. Dirac, The Principles Of Quantum Mechanics 4th Ed., Clarendon Press, Oxford, 1958

user151522
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  • Yeah, the bra-ket notation confuses me too. But, to make it easier, consider a vector $v$ an element of a vector space $V$. You may write $v \in V$ or, I prefer arrows (and I especially get confused by bold-face) $\vec{v}$ to be clear. I see no difference between $\vec{v}$ and $| v >$. Now consider the notation $\overleftarrow{v}$ which I define to be the transpose conjugate of $\vec{v}$ (why you would want the transpose conjugate I don't know - but it's just a cool thing to have) So if $\vec{v}$ is the column vector of $1 + i, 2, 4 -i$, then $\overleftarrow{v}$ is the fancy notation – DWade64 Apr 17 '25 at 20:11
  • row vector $[1 - i, 2, 4 +i]$. I see no difference between $\overleftarrow{v}$ and $< v |$, but the answer given probably would disagree with me. So don't let $|f>$ confuse you from $\vec{f}$ and don't let $< f |$ confuse you from $\overleftarrow{f}$. I think the bra-ket notation is confusing; but maybe I'm just not good at it or don't understand – DWade64 Apr 17 '25 at 20:13
  • But I think your question partly comes down to (forgetting about bra's for a second) the notation $| ;; \rangle$ versus $\vec{;;}$. If they are the same (and I claim they are), then why even have such notation? Meaning, why would I write $v$ versus $\vec{v}$ (I prefer arrows! I enjoy arrows)? Because this is how we did math in other classes. When you do algebra, you write things such as $x^2 + 2x = 0$. Where are the arrows? When you do forces in physics, you write things such as $\vec{F} = m \vec{a}$. So you can't say $\vec{a} = a \hat{x}$, where $a$ is the magnitutde, and $\vec{a}$ – DWade64 Apr 17 '25 at 20:27
  • is the vector (magnitude and direction), you can't say $\vec{a} = a$. A vector must be equal to a vector. (now going to bras). You can't say $\overleftarrow{f}$ is equal something that doesn't have the vector symbol (and remember I wrote that $\overleftarrow{f} = \langle f|$, I'm using them interchangable - they are both vectors, the transpose conjugate of $\vec{f}$) So in your question, a vector (whatever direction the arrow is pointing) has to be equal to a vector at the very least – DWade64 Apr 17 '25 at 20:30
  • This is partly why the other answer has an issue with $\langle f| = f^*$ because you have a vector on the left, but not a vector on the right. Rather, $\langle f | = \langle f |$, or, $\langle f | := \text{transpose conjugate of} | f \rangle$. Or, if the $f$ and those symbols confuse you, $\overleftarrow{v} := \text{transpose conjugate of};; \vec{v}$. Notice, you have vectors on both sides – DWade64 Apr 17 '25 at 20:36
  • In short, I hope this site helps https://www.mathsisfun.com/physics/bra-ket-notation.html – DWade64 Apr 17 '25 at 20:37
  • (actually my 3rd and 4th comment are a little sloppy - let $|| \vec{a} ||$ mean the magnitude of the vector [which is always a positive number] - in one dimension, $\vec{a}$ and the variable $a$ basically encode the same information, because the sign of variable $a$ will give you the direction of the entire vector. They are only different, if you wanted to point out that $\hat{x}$ itself requires a definition which is contained in $\vec{a}$ but not in variable $a$ [some one has to show you what they meant, what direction they meant] - but other than that, $\vec{a}$ =/ $a$ but basically same) – DWade64 Jun 10 '25 at 09:59

2 Answers2

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Let $H$ be a Hilbert space. We can consider the vectors of $H$ as the ket vectors. So, for $v\in H$, we consider $v$ and $|v\rangle$ as the same thing.

The bra vectors are the functions of the form $$ \langle v|: H\to \mathbb{C}: x\mapsto \langle v,x\rangle. $$ These are clearly linear and bounded (by Cauchy-Schwarz).

Bounded linear functions that map $H\to \mathbb{C}$ are called bounded linear functionals. So all bra vectors are bounded linear functionals.

Conversely, the Riesz representation theorem (https://en.wikipedia.org/wiki/Riesz_representation_theorem) says that all bounded linear functionals are of this form. Thus the space of bra vectors is exactly the space of bounded linear functionals.

The reason for this notation, is that the function application works out nicely. Indeed applying the function $\langle v|$ to the vector $x$ gives $$\langle v|(x) = \langle v|\big(|x\rangle\big) = \langle v, x\rangle,$$ so writing $\langle v | x \rangle$ gives no ambiguity: it is both a function application and an inner product and in both cases it gives the same number.

So, in your example, the bra $\langle f|$ would not be equal to $f^*$, rather it is the function $$ \langle f|: V\to \mathbb{C}: g\mapsto \int_{-\infty}^{+\infty}f^*(x)g(x)\,\mathrm{d}x. $$

Note also that this is really the same as defining $\langle f|$ by the property that $$ \langle f| g \rangle = \int_{-\infty}^{+\infty}f^*(x)g(x)\,\mathrm{d}x $$ since functions are defined by defining how they act on their domain.

Mentastin
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  • Is a simpler explanation possible, within which we would have, $\langle f(x)|=f^\ast(x)$ ? – user151522 Apr 16 '25 at 18:15
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    Not really: $\langle f | f \rangle$ should not be the same as $f^*f$. The first is a number, the second a function! In the case of column vectors such an identification can be made (because matrix multiplication can give a number and column vectors have a notion of transpose, that is not obvious for functions), but this is in some sense accidental. It is better to consider $\langle f|$ as strictly a functional, in my opinion. – Mentastin Apr 16 '25 at 20:29
  • Setting aside questions of rigor for the moment, one can draw a connection with how physicists use the Dirac delta function to express this: writing the resolution of unity as $\mathbf{1}=\int_{-\infty}^\infty |x\rangle\langle x|,dx$, one obtains $$\langle f|g\rangle =\int_{-\infty}^\infty \langle f|x\rangle\langle x|g\rangle dx=\int_{-\infty}^\infty f^*(x)g(x) dx$$ So formally at least we get the same result. – Semiclassical Apr 17 '25 at 00:32
  • I would still not say $\langle f| = f^$. But it is true that the mapping $\langle f|\mapsto f^$ is a linear isomorphism: it is just the composition of $\langle f| \mapsto f$ (which is an antilinear bijection according to Riesz's theorem) and $f\mapsto f^*$ which is also an antilinear bijection. – Mentastin Apr 17 '25 at 12:04
  • To Mentastin, in one of your comments above you ignore that putting a bra and a ket next to each other means that you apply the rule for the scalar product. – user151522 Apr 17 '25 at 12:19
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    This is exactly why I say that it is not natural to say $\langle f| = f^*$! – Mentastin Apr 17 '25 at 12:22
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I assume two things about Dirac’s$^1$ algebraic scheme involving bras and kets.

1: it is a way of thinking and doing the algebra/analysis, that will reproduce the expressions one would obtain using the definitions of ‘Inner Products’ , and ‘Adjoint Operators’, one would find in, Lipschutz and Lipson$^2$, see pgs. 239, and 377, respectively.

2: has one or more advantages when used for quantum mechanical analysis.

Taking this point of view, we require that the inner product is reproduced somehow, by Dirac’s scalar product ( see reference 1, pg 21, for material on Dirac’s product ).

If we are working with the vector space $V$ of the question

\begin{equation*} V=\left\{ f|\,f:\mathbb{R} \to\mathbb{C},\, x \mapsto f(x), \int_{- \infty}^\infty f^\ast(x)~f(x)~dx~<\infty ~\right\} \end{equation*}

an inner product could be

\begin{equation*} \langle g,f\rangle=\int_{- \infty}^\infty g(x)~f^\ast(x)~dx \end{equation*} We would take a ket $|g\rangle$ to be $|g\rangle=g$, some function of ‘$x$’, and our Diracian scalar product could be taken to be \begin{equation*} \langle f|g\rangle=\int_{- \infty}^\infty f^\ast(x)~g(x)~dx \end{equation*} If we did this, then everytime we had an inner product, working with the “mathematical” notation, we could replace it by a scalar product as follows \begin{equation*} \langle g,f\rangle=\langle f|g\rangle \end{equation*}

We could take our bras, $\langle f|$, to be $\langle f|=f^\ast$.

So we can know what our bra vectors are.

Other Information

Related material, see at

Given a vector space of Dirac kets, could you give a way that the corresponding bras may be set up?

Please see some related material at

https://math.stackexchange.com/a/4412035/553318

I quote from that post, which I authored,

The linear functional $\phi$ referred to by Dirac$^1$, in the section '6. Bra and ket vectors' is not, the new vector, the dual vector, the bra, that Dirac is on about.

The following is in support of the above comment.

Dirac starts to explain his algebraic scheme on pg. 18, in the section entitled ‘6. Bra and ket vectors’

In the first three paragraphs he introduces the terms, ‘Dual Vectors’, ‘New Vectors’, ‘Bra Vectors’, and ‘Bras’, I take all of these terms to refer to the same set of vectors. Further, it seems that Dirac is saying, that the "$\langle B|$" part of the rule for ‘$\phi$’, where

\begin{equation*} \phi=\langle B|A \rangle \end{equation*}

is what is considered to be the bra, the complete $\phi$ is not the bra.

I think the bra’s of Dirac’s$^1$ book, are different to the “bra’s”, apparently favoured on ‘math.stackexchange’.

I guess there will be a vector space isomorphism, from the space of the

\begin{equation*} \langle v|: H\to \mathbb{C}: x\mapsto \langle v,x\rangle \end{equation*} of the answer at https://math.stackexchange.com/a/5056795/553318

, to some vector space, that I would consider to be the vector space of Diracian bras.

I think you can choose to work with either vector space of bras, just like you can choose to use the Dedekind cuts as the real numbers, rather than using the infinite decimals as the reals.

References.

1, P.A.M. Dirac, The Principles Of Quantum Mechanics 4th Ed., Clarendon Press, Oxford, (1958).

2, Seymour Lipschutz, PhD, Marc Lipson, PhD, SCHAUM’s outlines, Linear Algebra, Fourth Edition, McGraw Hill (2009).

user151522
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