I Confusion about position state kets

[Oliver321]

TL;DR Summary

Why is it that |-x> ≠ -|x>, even though kets are vectors and thus obey linearity?

I am a bit confused about how kets in dirac notation are working.
I read on wikipedia, that kets are linear, so |a*Φ>=a*|Φ>.
Also I read (https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_04.pdf) that this is not true for the position state ket ( or non denumerable bases in general): |-x> ≠ -|x>

My question is now: Why is this the case? If I am understanding it right, the position state ket is representing the position of a particle in that way, that a particle with ket |2> is at the point 2 on the x axis. Or expressed otherwise: |2> =(0,0,...,1,0,0,...), where the second representation is a vector in an infinite space where every component (or base vector) is representing a real number (in this example the number 2). So it would be valid to say
|-2> =(0,0,...,-1,0,0,...) = -(0,0,...,1,0,0,...)= -|2>

Probabliy I have a wrong understanding of what kets really are. I would appreciate every help!

 

[hilbert2]

If you refer as $\left|\right.\psi\rangle$ to the ket vector related to a wave function $\psi$, then you can write $c\left|\right.\psi\rangle = \left|\right.c\psi\rangle$. But the $x$ in $\left|\right.x\rangle$ does not refer to a wave function, it's a label that indexes the different position eigenstates. The $\left|\right.x\rangle$ and $\left|\right.-x\rangle$ are actually orthogonal vectors if $x\neq 0$, so it's impossible to write one of them as a multiple of another.

 

[Oliver321]

Thank you very much! I think I understand it. So the x is a Index and not the position itself. I could also use an index n if it’s continuous?

 

[anuttarasammyak]

Let us see orthogonality and normalization
<\alpha|\beta>=\delta(\alpha-\beta).

<\alpha|-\alpha>=\delta(2\alpha)=0\ for\ \alpha \neq 0
Orthogonal.
<\alpha|(-|\alpha>)=- <\alpha|\alpha>=-\delta(0)
Same but different sign.

 

[hilbert2]

Thank you very much! I think I understand it. So the x is a Index and not the position itself. I could also use an index n if it’s continuous?

It is the position eigenvalue, but you can't absorb a constant multiplier into it, like you can when labeling the states with the corresponding wave functions.

 

[PeroK]

Thank you very much! I think I understand it. So the x is a Index and not the position itself. I could also use an index n if it’s continuous?

Yes. One probem with the Dirac notation in this case is that it is so slick that symbols can be overloaded. What is $|x \rangle$ really? It's the eigenstate of the position operator $\hat x$ with eigenvalue $x$. So, let's write it as something that ends the confusion: $|\alpha(x) \rangle$. Where:
$$\hat x |\alpha(x) \rangle = x |\alpha(x) \rangle$$
And, similarly $|-x \rangle$ is the eigenstate of $\hat x$ with eigenvalue $-x$:
$$\hat x |\alpha(-x) \rangle = -x |\alpha(-x) \rangle$$
This should help you see that we cannot have $|\alpha(-x) \rangle = - |\alpha(x) \rangle$. For one thing, $- |\alpha(x) \rangle$ is just the same state as $|\alpha(x) \rangle$, with a phase factor of $-1$. I.e. that is also an eigentstate of $\hat x$ with eigenvalue $x$.

 

[hilbert2]

And if you could write $\displaystyle i\left|\right.x\rangle = \left|\right.ix\rangle$, then you'd have produced a state with an imaginary position eigenvalue, which is not possible.

 

[vanhees71]

I think it's a drawback of this "extreme form" of the Dirac bra-ket formalism, often leading to confusion. It's common to write $|\vec{x} \rangle$ as the generalized position-operator eigenvector of eigenvalue $\vec{x}$ (more precisely: the common eigenvalues of the three commuting self-adjoint operators representing the Cartesian components of the position vector).

This can be easily avoided by writing something like $|u_{\vec{x}} \rangle$ for it. Then it's clear that $|u_{(-\vec{x})} \rangle$ is the eigenvector for the position eigenvalues $-\vec{x}$ rather than the vector $-|u_{\vec{x}} \rangle$.