# Confusion about position state kets

• I

## Summary:

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

## Main Question or Discussion Point

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!

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hilbert2
Gold Member
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.

• etotheipi and 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
Gold Member
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
Gold Member
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.

• etotheipi
PeroK
Homework Helper
Gold Member
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##.

• etotheipi
hilbert2
Gold Member
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
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