While using dirac bra-ket notation for quantum mechanics, what's the difference if the minus sign is inside or outside the ket ? I know that $\left| -x\right\rangle$ and $- \left|x \right\rangle$ doesn't mean the same thing but I don't really understand the difference between those 2 kets.

  • 3
    $\begingroup$ $+\left| -x\right\rangle$ is the state of a particle at position -x, with phase +1. $-\left| +x\right\rangle$ is the state of a particle at position +x, with phase -1. The overall phase doesn't matter, but keeps track of whether two states will interfere constructively or destructively. (you simply add coefficients of states, so opposite sign phases cancel out) $\endgroup$
    – adipy
    Nov 7, 2014 at 13:34
  • $\begingroup$ If you like this question, you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Nov 7, 2014 at 13:42
  • $\begingroup$ The thing within the bracket |thing> can be anything. It's just a label. You can put a picture in there if you want. $\endgroup$
    – Nick
    Nov 7, 2014 at 20:42
  • $\begingroup$ @adipy that should be an answer $\endgroup$
    – David Z
    Nov 8, 2014 at 3:09

4 Answers 4


In a nutshell:

plot of wavefunctions

With a couple caveats: what is plotted here is really the wavefunction $-\langle x\lvert 2\rangle$ or $\langle x\lvert -2\rangle$ (so actually this graph is highly misleading in one way), and the spikes should be thought of as delta functions. But this is the basic idea.


$\def\ket#1{\left|#1\right>}$Each of those kets represents the state of a system. In quantum mechanics the constant multiplying a wavefunction is not physically significant (multiplying a wavefunction by a number does not change the physical meaning of the wavefunction). So if $\ket{x}$ represents a wavefunction completely localized to the point $x$, then by what I just said $-\ket{x}$ also represents a wavefunction completely localized to the point $x$, its just that the wave function has been multiplied by $-1$.

However $\ket{-x}$ represents a wavefunction completely localized at $-x$ so it really is a physically different state from $\ket{x}$, because the wavefunction is concentrated at $-x$ instead of $x$.


When dealing with the bra-ket notation, we must be clear that what we put inside the bras or kets is just a label and not something we make math with, although good labels makes visualization easier. When you write $|x\rangle$ you are considering the state of the particle in the $x$ position. (There is a problem here that this is not an acceptable state, since it's not an element of the Hilbert space $\mathcal{H}$, but we can ignore this subtlety for the explanation working with the rigged Hilbert Space $\Phi^*$, but I'll ommit this assumption). What comes next is the fact that in Quantum Mechanics, states are represented by equivalence classes of vectors of the Hilbert space and not just vectors. Two vectors $|\psi\rangle$ and $|\phi\rangle$ are equivalent $(|\psi\rangle\sim|\phi\rangle)$, i.e., they represent the same state iff $$|\psi\rangle=\alpha|\phi\rangle, \qquad\alpha\in\Bbb{C},$$ that is, if they differ only by a scalar multiple. That being said, we can see that although $|x\rangle$ and $-|x\rangle$ are different vectors in the space, they represent the same state in Quantum Mechanics, since in this case we have $\alpha=-1$. On the other hand, the vector $|-x\rangle$ represents another state, that is a particle localized in the $-x$ position, so $$|x\rangle\nsim|-x\rangle$$ because they do not differ only by a scalar multiple.


this is just a change of basis. Its like a finding a new set of base vectors after rotating a coordinate system in 3 space. For your example you have infinite base vectors not three. For the negative of in front of the ket its just like multiplying a base vector by some constant. -1 in this case. it would make more sense to write -1 times the ket then writing a |-1 x i>. Don't forget that |x> is a set of base vectors.


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