I have just started learning about Dirac Notation in Quantum mechanics and there have been some conflicting information about Dirac Notation.

Dirac Notation was introduced as a vector: $|a\rangle = (a_1\ a_2\ a_3)$ and $\langle a|b\rangle = (a_1\ a_2\ a_3)^T(b_1\ b_2\ b_3)$. Then later, the wave function in Dirac Notation is introduced, which have the equation: $\langle a|b\rangle =\int a^*b\ dx$. This is very confusing for me as how is the wave function a vector, shouldn't it be continuous from -infinity to infinity like in scattering states? How is it possible that an integral is equal to a vector?$(a_1\ a_2\ a_3)^T(b_1\ b_2\ b_3)=\int a^*b\ dx$? Please explain more about Dirac Notations as it seems that for the rest of the materials need to use Dirac Notations. Thank You.


Well, this question shows that linear algebra is just neccesary in physics.

There are so many issues in this question. First of all, it's true that $|a\rangle$ is a vector, but nobody said it was a three dimensional vector. It is a vector that can have any dimension, even infinite dimension (infinite rows), or even a continous of rows.

That's because a vector is just defined as "al element of a vector space", which is itself a "set of elements and two operations, an internal one (sum) and an external one (product of scalar times vector), such that the first one makes a group, and the second one verifies 4 axioms". That was a really short summary. Make yourself the research about "what a vector space is". It's in wikipedia for sure.

So, besides the usual arrows, there are many things that are vectors (or can be regarded as vectors). Some examples are

  • Matrices
  • The real numebrs themselves, or also complex numbers.
  • n-tuples of numbers ($\mathbb{R}^2,\ \mathbb{R}^3, ..., \mathbb{R}^n...)$
  • Linear functions

And let's higlight that one. Linear functions are also vectors. It can sound weird, but they are, because

  • There's a set of "linear functions", and two operations like above (sum of functions and product of number times function). And they satisfy all required properties. I'll sumarise them with formulas:

Let $f$, $g$, and $h$ be functions of the vector space $\mathcal{F}$, and $t,s$ scalars... then

  1. $\quad f+g\in \mathcal{F}$
  2. $\quad f+g=g+f$
  3. $\quad f+(g+h)=(f+g)+h$
  4. $\quad f+0=f$
  5. $\quad \exists (-f)\ \ \quad /\ \quad \ f+(-f)=0=(-f)+f$
  6. $\quad t\cdot(f+g)=tf+tg$
  7. $\quad (t+s)\cdot(f)=tf+sf$
  8. $\quad t\cdot(sf)=(t\cdot s)g$
  9. $\quad 1\cdot f=f$

Where 0 denotes the "void function".

So you can see that functions behave like arrow vectors. Linear functions can be seen as vectors. The fact that functions satisfy these prperties inmediately means that they satisfy all theorems of vectors.

What's more, we can define more operations. There exists a "scalar product" of two linear functions, and it is defined as

$$ \vec{f}\cdot \vec{g} = \int_{-\infty}^{+\infty}{f(x)\cdot g(x) dx} $$

It can be shown that this definition satisfies the three properties of a scalar product, so it is a valid scalar product for functions.

If functions are complex, then f must appear conjugated in the integrand.

Altough there are so many more issues concerning Hilbert space of kets (dirac notation), let's conclude here:

  • functinos behave like vectors. The scalar product of two functions is defined via the previous integral.

  • So vectors are not an integral. The scalar product of two vectors is a number, and so is a defintie integral.

Finally, what we do is "defining kets" so that their scalar product gives the same result as the scalar product of their associated wavefunctions.

  • $\begingroup$ Non-linear functions are also vectors. $\endgroup$
    – Dog_69
    Mar 22 '18 at 17:33
  • $\begingroup$ yes, but I didn't want to mess it up even more haha. I've actually said just "functions" many times. $\endgroup$
    – FGSUZ
    Mar 22 '18 at 17:34
  • $\begingroup$ @FGSIZ Right, OK. $\endgroup$
    – Dog_69
    Mar 22 '18 at 17:36

Dirac notation is often abused, leading to questions like yours. Let's be very specific:

A state vector $|a\rangle$ is an object in a Hilbert space $H$, which may have finite or infinite dimensions. We can generically find a set of basis vectors that span the Hilbert space (usually we pick an orthonormal set of basis vectors): let's denote the basis vectors $|e_1\rangle, |e_2\rangle, \ldots$ (up to the dimension of $H$). We can always expand our state vector $|a\rangle$ in the basis of $\{|e_i\rangle\}$: $$ |a\rangle = a_1 |e_1\rangle + a_2 |e_2\rangle + \ldots $$ We can equivalently use traditional vector notation to describe these states: expressed in the same basis, we can write: $$ |a\rangle \to \begin{pmatrix} a_1 \\ a_2 \\ \ldots \end{pmatrix} $$

The bra objects are the conjugate transpose of the original vectors (technically they live in a dual space to the main Hilbert space, but this is not important for this purpose). As you noted, the bras are represented as follows: $$ \langle a| \to \begin{pmatrix} a_1 \\ a_2 \\ \ldots \end{pmatrix}^\dagger = \begin{pmatrix} a_1^* & a_2^* & \ldots \end{pmatrix} $$ Again as you wrote, we can take the inner product as: $ \langle b | a \rangle = \begin{pmatrix} b_1^* & b_2^* & \ldots \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ \ldots \end{pmatrix} $

Now, how does the integral formulation arise for wavefunctions? The key is that we are now describing states in an infinite dimensional Hilbert space - the space of all spatial wavefunctions. A basis set that spans this Hilbert space are all the localized position vectors: $|x\rangle$, which is a state localized precisely at position $x$. We could expand our wave vector $|\psi\rangle$ in terms of the position basis states: $$ |\psi\rangle = \sum_x c(x) |x\rangle \to \int dx \ c(x) |x\rangle $$ where the integral is simply the limit of a sum over a continuous variable. The function $c(x)$ describes the amplitude of the vector $|\psi\rangle$ in each basis state $|x\rangle$. Note that this coefficient is obtained by taking the inner product with $\langle x|$: $$ \langle x | \psi \rangle = \int dx' \ c(x') \langle x | x' \rangle = \int dx' \ c(x') \delta(x-x') = c(x) $$ This is also often written as the 'wavefunction', $\langle x | \psi \rangle = c(x) \equiv \psi(x)$.

To compute the inner product of two wave vectors $|\psi\rangle$ and $|\phi\rangle$ in this infinite dimensional space, we can expand them each in the continuous basis of states $|x\rangle$: \begin{eqnarray} \langle \phi | \psi \rangle =& \bigg(\int dx \ \phi^*(x) \langle x| \bigg) \bigg(\int dx' \ \psi(x') |x'\rangle\bigg) \\ =& \int \int dx \ dx' \phi^*(x) \psi(x') \langle x | x' \rangle \\ =& \int \int dx \ dx' \phi^*(x) \psi(x') \delta(x-x') \\ =& \int dx \ \phi^*(x) \psi(x) \end{eqnarray}

This shows then how in an infinite dimensional space, the inner product of two vectors is expressed as an integral over the entire set of basis states.

  • 1
    $\begingroup$ I recommend annotating vector components with the basis to which they correspond. For example, $|a\rangle = a_1^e |e_1\rangle + a_2^e |e_2 \rangle + \cdots$. $\endgroup$
    – DanielSank
    Mar 22 '18 at 17:30

The inner product, $\langle a|b \rangle$ is not a vector, it is, $$ \langle a|b \rangle=(a_1 \; a_2 \; a_3)^T(b_1 \; b_2 \; b_3)=a_1b_1+a_2b_2+a_3b_3.$$ So your integral does not equal a vector.

(Let me know if I missinterpreted your question).


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