Mistake in using dirac notation when applying $X$ gate to vector

Question

The X gate is given by $big(begin{smallmatrix}
  0 & 1 
  1 & 0
end{smallmatrix}big)$ in the computational basis. In the Hadamard basis, the gate is $X_H = big(begin{smallmatrix}
  1 & 0
  0 & -1
end{smallmatrix}big) = |+ rangle langle +|  - |-rangle langle-|$. When I apply the gate to the Hadamard basis vectors, the vectors should flip, and they do when I use matrix notation but not when I'm using dirac notation. I know I'm making a mistake somewhere.
$X_H |+rangle = (|+ rangle langle +|  - |-rangle langle-|)|+rangle = |+ rangle langle +|+rangle  - |-rangle langle-|+rangle = |+rangle(1) - |-rangle(0) = |+rangle$ and
$X_H |-rangle = (|+ rangle langle +|  - |-rangle langle-|)|-rangle = |+ rangle langle +|-rangle  - |-rangle langle-|-rangle = |+rangle (0) -|-rangle(1) = -|-rangle$
Meanwhile, in matrix notation,
$X_H|+rangle = big(begin{smallmatrix}
  1 & 0
  0 & -1
end{smallmatrix}big) frac{1}{sqrt{2}}big( begin{smallmatrix}
1 
1
end{smallmatrix}big) = frac{1}{sqrt{2}}big( begin{smallmatrix}
1 
-1
end{smallmatrix}big) = |-rangle
$
$X_H|-rangle = big(begin{smallmatrix}
  1 & 0
  0 & -1
end{smallmatrix}big) frac{1}{sqrt{2}}big( begin{smallmatrix}
1 
-1
end{smallmatrix}big) = frac{1}{sqrt{2}}big( begin{smallmatrix}
1 
1
end{smallmatrix}big) = |+rangle
$

JSdJ · Accepted Answer

The basis states should not flip, as these two basis states are the eigenstates of the $X$ gate. The $X$ gate flips the computational basis states, the $Z = begin{bmatrix} 1 & 0  0 & -1 end{bmatrix}$ gate flips the Hadamard basis states.
Expressing everything in the computational basis
In the computational basis, we have $X = begin{bmatrix}0 & 1  1 & 0end{bmatrix}$. Thus the states (expressed in the computational basis) $|+rangle = begin{bmatrix}1  1end{bmatrix}$ and $|-rangle = begin{bmatrix}1  -1end{bmatrix}$ are the $+1$ and $-1$ eigenstates respectively, as you can readily check.
Expressing everything in the Hadamard basis
If you express everything in the Hadamard basis, the $X$ gate becomes $X_{H} = begin{bmatrix}1 & 0  0 & -1end{bmatrix}$.
But, the $|+rangle$ and $|-rangle$ states should now also be expressed in this basis. That is, $|+rangle_{H} = begin{bmatrix}1  0end{bmatrix}$ and $|-rangle_{H} = begin{bmatrix}0  1end{bmatrix}$. It's now obvious that these states are indeed the $+1$ and $-1$ eigenstates of $X$, expressed in whatever basis.
To summarize your notation
So you dirac notation is correct, and in your matrix notation you expressed the $X$ operator in the Hadamard basis, but the states in the computational basis.
But wait, then what are these states if not the $|+rangle$ and $|-rangle$ states?
So what are the states $begin{bmatrix}1  1end{bmatrix}_{H}$ and $begin{bmatrix}1  -1end{bmatrix}_{H}$, i.e. these states in the Hadamard basis? As you showed with your matrix notation, they are those states that are flipped under operation of the $X$ gate - they are the computational basis states/eigenstates of the $Z$ operator!
Of course, you can write this out mathematically as well:
$$
begin{bmatrix}1  1end{bmatrix}_{H} = begin{bmatrix}1  0end{bmatrix}_{H} + begin{bmatrix}0  1end{bmatrix}_{H} = begin{bmatrix}1  1end{bmatrix} + begin{bmatrix}1  -1end{bmatrix} = begin{bmatrix}1  0end{bmatrix}
$$
As you can see, I've been very sloppy with the normalization factor - the above equation is a factor of $2$ off.

Mistake in using dirac notation when applying $X$ gate to vector

One Answer

Expressing everything in the computational basis

Expressing everything in the Hadamard basis

To summarize your notation

But wait, then what are these states if not the $|+rangle$ and $|-rangle$ states?

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