This question is about a probable confusion of definitions that I may have somewhere. Also my math knowledge is not too big so I'll try not to get too abstract.

Let's say I have a vector space made of objects called Kets (${|\psi\rangle}$), the ones used in Quantum Mechanics. They have an inner product ${\langle\phi|\psi\rangle}$, and they have continuous (uncountable) dimension. Take an Orthonormal Basis of the space, for example, the eigen-kets of the position operator, ${|x_j\rangle}$, where ${x_j}$ sweeps all the real numbers (as they are all the possible positions).

-Orthonormal means (I think) that the kets satisfy:

${\langle x_a|x_b\rangle=0}$ if a ${\neq}$ b and ${\langle x_a|x_b\rangle=1}$ if a = b.

-Basis means that every ket ${|\psi\rangle}$ can be written as a linear combination of the basis kets. I'm going to take a shot and say that this can be expressed as

$${|\psi\rangle=\int_{-\infty}^\infty dx_i.C(x_i)|x_i\rangle}$$

However, if I turn that into the following expression

$${\langle x_a|\psi\rangle=\int_{-\infty}^\infty dx_i.C(x_i) \langle x_a|x_i\rangle}$$

and then use the Orthonormality condition, I am left with something akin to

$${\langle x_a|\psi\rangle=\int_{x_a}^{x_a} dx_i.C(x_i).1 }$$

which I am pretty sure yields ${\langle x_a|\psi\rangle\neq C(x_a)}$ -in fact it probably yields ${\langle x_a|\psi\rangle=0}$ since the area under a single point of finite height is zero-. This seems totally wrong for the orthonormal basis of a vector space, which makes me think that I there is something wrong somwhere.

-A possible solution to this would be to redefine the concept of orthonormality for such an uncountable vector basis, so that Orthonormal means:

${\langle x_a|x_b\rangle=0}$ if a ${\neq}$ b and ${\langle x_a|x_b\rangle=\infty}$ if a = b, where said ${\infty}$ makes sense inside an integral and yields 1 when integrated. This would indeed then return ${\langle x_a|\psi\rangle= C(x_a)}$.

However, redifining orthonormality also seems kind of iffy. Plus it would imply that ${\langle x_a|x_a\rangle=\infty}$ which is not very encouraging.

So could you tell me how do I make these statements work?


Bonus: A third option could be to say that the expression is not an integral, but a summation of the form

$${|\psi\rangle=\sum_{x_j=-\infty}^\infty C(x_j)|x_j\rangle}$$

where ${x_j}$ sweeps all the real numbers.

Yet I'm sure that such an expression is not correct either.

  • $\begingroup$ Pretty sure all Hilbert Spaces in QM are separable. So they always have a countable basis. $\endgroup$
    – user73352
    Nov 27, 2016 at 7:20

1 Answer 1


As you correctly concluded, there is something wrong with your original orthonormality condiction and indeed it has to do with the fact that you are using a discrete basis to span a continuous space.

So if you have a continuous infinite dimensional Hilbert space, then your basis would also need to be continuous and infinite. (Actually, one can sometimes use a discrete infinite basis, but let's ignore this caveat for the moment.) In that case the orthonormality condition would be given by $$ \langle x_a| x_b \rangle = \delta(x_a-x_b) , $$ where $\delta(x_a-x_b)$ is a Dirac delta function.

When you now want to expand an arbitrary ket, $$ |\psi\rangle = \int |x\rangle C(x)\ dx , $$ you can apply the inner product and get $$ \langle x_a|\psi\rangle = \int \langle x_a|x\rangle C(x)\ dx = \int \delta(x_a-x) C(x)\ dx = C(x_a) , $$ which follows form the properties of the Dirac delta function.

In the case where you have a discrete basis you can replace the Dirac delta with a Kronecker delta function, which is basically what you had originally. Then the expansion would be a summation as you have shown at the end.


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