I Inner products and wavefunctions

[dyn]

Hi
As far as i am aware taking the inner product of a bra and a ket results in a (complex) number or scalar. So why does < x | ψ > give ψ ( x ) , the wavefunction in position space ? Surely < x | ψ > should give a number not a function ?
Thanks

 

[PeterDonis]

The notation $\braket{x | \psi} = \psi(x)$ is actually a slight abuse of notation. What it says is: the function $\psi(x)$, for a given ket $\ket{\psi}$, is that function that maps each position $x$ to the number $\braket{x | \psi}$, where $\bra{x}$ is the bra corresponding to the eigenstate of position in which the position is $x$.

 

[dyn]

Thanks for your reply. Is there a mistake in the Latex as it does not seem to be displaying properly ?

 

[PeterDonis]

Is there a mistake in the Latex as it does not seem to be displaying properly ?

Do you mean in my post? It displays fine for me. You might have to reload the page, log out/back in, restart your browser, clear your cookies for physicsforums.com, or some combination of those.

 

[dyn]

Yes , it displays fine now. Thank you. So is it correct to say that an inner product of a bra and a ket always gives a number not a function ?

 

[PeterDonis]

is it correct to say that an inner product of a bra and a ket always gives a number not a function ?

If you're talking about a single bra and a single ket, yes. The notation $\braket{x | \psi} = \psi(x)$ is actually using $\bra{x}$ to refer to an infinite collection of bras (one for each position $x$), not just one.

 

[dyn]

Thank you. I'm trying to picture what the kets | ψ > and | x > look like. I know they are infinite-dimensional vectors. Is the ket | ψ > an infinite column vector consisting of the value of ψ( x ) at each of the infinite values of x ? Is the general ket | x > an infinite column vector consisting of the number 1 at each of the infinite values of x . ie an infinite collection of basis vectors ? When calculating < x | ψ > the bra is an infinite collection of zero's with the exception of a 1 at the exact position of the value of x in ψ ( x ) . have i got that right ?
Thanks

 

[PeterDonis]

have i got that right ?

All of these things are highly heuristic (i.e., be careful how you use them) ways of describing kets and bras. Or more precisely, describing a (vaguely defined) representation of kets and bras.

Kets (and bras) are in general best left unvisualized, so to speak: when trying to prove anything about them in general, you should only use their known general mathematical properties, which will be valid regardless of any choice of representation. The only time you should choose a representation is when you have to--i.e., when there is no other way to do some particular calculation you want to do.

For the specific kets and bras you are talking about, i.e., those in the Hilbert space of a single particle (and its dual), there actually is no well-defined "infinite vector" representation. The only really well-defined representation you will see in most textbooks is the wave function representation, where $\psi(x)$ is a function (and operators on the Hilbert space are represented as operators on functions, mostly but not always differential operators). The technical issue with this representation is that, strictly speaking, the Hilbert space only consists of square integrable functions, and the eigenfunctions of the position and momentum operators, namely the "delta function" $\delta(x)$ and the plane wave $e^{i k x}$, which are often used as two possible choices of basis for the Hilbert space, are not square integrable, so they aren't actually in the Hilbert space. There are ways of dealing with this, which is why it is only a technical issue and does not invalidate the wave function representation.

Many other Hilbert spaces, such as those for spin, are finite dimensional, so there is a valid vector representation (with operators being matrices), which is easier to use.

 

[PeroK]

Hi
As far as i am aware taking the inner product of a bra and a ket results in a (complex) number or scalar. So why does < x | ψ > give ψ ( x ) , the wavefunction in position space ? Surely < x | ψ > should give a number not a function ?
Thanks

Technically, $\psi(x)$ is a number. $\psi$ is the function and $x$ is the variable.

 

[vanhees71]

This confusion arises, because physicists are sloppy in never making a difference between a function $\psi$ and the value $\psi(x)$ evaluated at the argument $x$. So they say "$\psi(x)$ is the wave function". What they mean is that $\psi:\mathbb{R} \rightarrow \mathbb{C}$ is a function $x \mapsto \psi(x)$, i.e., it maps the real position coordinate to a complex number.