First course in QM, so understand that I know next to nothing. I have been reading and watching videos from a number of sources but I am confused a bit in certain aspects of the Dirac notation due to various sources giving me seemingly contradictory information.

For example, I saw the polarization state as $$ \left|\theta\right> = \cos\theta\left|x\right> + \sin\theta\left|y\right> $$ and the spin state as $$ \left|s\right> = \Psi_u \left|u\right> + \Psi_d \left|d\right> . $$

To me it implies that some observable's state is written as a linear combination of basis kets, where basis ket has a 1 in a designated, unique place and zeros everywhere else. And the inner product of some state with the other should give the probability amplitude of the given state collapsing (?) into another: for example, $ \Psi_u = \left<u\middle|s\right> $ gives the probability amplitude of the system being detected as spin up.

But this seems to get confusing if we extend the same thing to things like position, momentum, or energy. For a particle in a one-dimensional space, shouldn't the overall state be $$ \left|x\right> = \sum_i \Psi(x_i) \left|x_i\right>, $$ where $\left|x_i\right>$ is a Dirac delta function i.e. a position eigenfunction, and $\Psi(x_i)$ is the probability amplitude at that particular position eigenvalue $x_i$?

Why do so many texts write this summation as $\left|\Psi\right>$? It's confusing: $\Psi$ is for wavefunction. Why don't we represent a state of an observable by writing its name in the ket? What is $\psi$ supposed to mean here? I thought $\Psi$ was the probability amplitude.

Confusion about Dirac notation. When represent basis kets of position state by dirac delta function, when by matrices? These are supposed to be eigen-function right?

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    – rob
    Commented Sep 17, 2016 at 15:45

2 Answers 2


The equation you're trying to write goes like this: $$ \left|\psi\right> = \sum \left|\psi\right>\left|x\right>\left<x\right| = \sum \left<x\middle|\psi\right>\left|x\right> = \sum \psi(x) \left|x\right> $$

Here $\left|x\right>$ are the position base elements, $\left|\psi\right>$ describes the state of the system, and I have used that $1=\sum \left|x\middle>\middle<x\right|$. As you can see, I can write the state in the position base, via what's called the wave function $\psi(x)$. As you know $|\psi(x)|^2$ gives the probabilty of finding the system at a given x.

You could do the same thing in the momentum base: $$ \left|\psi\right> = \sum \left<p\middle|\psi\right> \left|p\right> = \sum \psi(p) \left|p\right> $$

Neither $\left|x\right>$ or $\left|p\right>$ are Dirac delta functions by themselves. The delta functions appear when you project onto the same base, for example:

$$ \left<x\middle|x'\right> = \delta (x-x') $$


$$ \left<p\middle|p'\right> = \delta (p-p') $$

But you could also project a vector from one base onto the other, getting the famous plane waves, which you can use to change bases:

$$ \left<x\middle|p\right> = (1/\sqrt{2\pi \hbar}) e^{ipx/\hbar} $$

Maybe that helps a bit. I recommend you read the first chapters of any standard college textbook on Quantum Mechanics (Sakurai, for example).

  • $\begingroup$ but i am wondering why are you representing these position states in as a ket labelled psi ? i mean its telling us about the x position of the system so why not call it x ? psi shoudl be the probability amplitude. $\endgroup$ Commented Sep 17, 2016 at 14:41
  • $\begingroup$ and that summation you wrote is essentially trying to make the point that we add all the position states (given by ket x) with each one having a coefficient psi(x) implying the probability amplitude of that position state ? and arent those position states the position eigen functions ? $\endgroup$ Commented Sep 17, 2016 at 14:50
  • $\begingroup$ and if those position eigen functions are not dirac delta funtions in that summation, how else should they be represented ? i have smth like colomn vectors with 1 in a unique place and zeros everywhere else. And really, why cant they be dirac delta functions ? $\endgroup$ Commented Sep 17, 2016 at 14:57
  • $\begingroup$ that was pretty helpful of you.Just clear a few more of these queries i posted in the comments. $\endgroup$ Commented Sep 17, 2016 at 15:30

Quantum Mechanics uses the abstract concept of Linear Vector Spaces. If you look at any elementary Linear Algebra textbook, you will see that a vector can be any mathematical object which satisfies a particular set of conditions. This includes the set of column matrices, as well as many sets of functions.

Now the linear vector space which consists of all state vectors in QM is known as a Hilbert Space. As with any other linear vector space, there are various sets of basis vectors that can span it (the 3D real space has cartesian, spherical, etc. coordinates). In the case of a Hilbert space, it can be spanned in terms of column matrices and dirac delta functions (and more!). So your question regarding the representation of the position kets, in fact is answered as follows: Use whatever is convenient! In some cases, using the delta function leads to easier calculations and in others a different basis set leads to easier computation.

So in short, the state vectors written in terms of the delta function and column matrices are merely representations of an abstraction which is a vector in a vector space.

  • $\begingroup$ hmm...some people seem to imply that the position kets by themsleves are not dirac delta functions.Can you shed anyy light ? $\endgroup$ Commented Sep 17, 2016 at 15:32
  • $\begingroup$ In QM, all the information about your system is in the wavefunction, denoted by $\mid \psi \rangle$. In order to find the position of the particle, you must act on this wavefunction with the position operator. This will give you an eigenvalue equation, in which the eigenvalue is the position (which you can observe) and the eigenket (or eigenstate) is one of the operator's eigenkets corresponding to the particular eigenvalue you obtain. Now in position space, the set of eigenkets of the position operator are dirac delta functions. $\endgroup$ Commented Sep 18, 2016 at 1:18

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