Why are $|0 ⟩$ and $|1 ⟩$ perpendicular to each other?

Question

"The special states $|0 ⟩$ and $|1 ⟩$ are known as computational basis states, and form an orthonormal basis for this vector space."

Orthonormal means that both the qubits are perpendicular to each other. But if we look at a diagram of the Bloch-Sphere they are at $180^\circ$. I don't get how they are perpendicular to each other.

Answer 1

The Bloch sphere is not embedded in the state space. As it says in the Wikipedia link for the Bloch sphere:

The mapping from the unit 3-sphere in the two-dimensional state space $\mathbb C^{2}$ to the Bloch sphere is the Hopf fibration, with each ray of spinors mapping to one point on the Bloch sphere.

To put this in more mathematical/geometrical terms, the Bloch sphere is identified $\mathbb C P^1$, the 1-dimensional complex projective space, also known as the Riemann sphere in this abstract context.

Answer 2

This is because you are conflating geometrical space with the Hilbert space of states. They are different spaces and can have different dimensions.

In the Hilbert space of quantum states, states are orthogonal if they are states corresponding to distinct outcomes of some measurement. In the case of qubits, these states are: the state |0> and the state |1>. Depending on the particular physical system $\vert 0\rangle$ could be a spin state with spin down, and $|1\rangle$ could be a spin state with spin up, or $\vert 0\rangle$ could be the ground state of an atom and $\vert 1\rangle$ one if its excited state, etc. The outcomes here would be the outcomes of measuring the direction of spin, or the energy of the system, respectively. In particular, "spin up" and "spin down" appeal to the picture of spins being antiparallel in geometrical space, although the states are orthogonal in the 2d Hilbert space.

Two-level systems like the ones above can be visualized on the Bloch sphere by embedding the sphere in 3D, but they still “live” in a 2-dimensional complex Hilbert spaces. Thus, even if $\vert 0\rangle$ and $\vert 1\rangle$ happen to map to states at opposite end of the sphere in the 3D visualization of this sphere, they are still basis states for a 2d complex Hilbert space and so are orthogonal in this space.

It just so happens that it is possible to map the 2d complex Hilbert state of space to a sphere embedded in 3D, but the cost is that “visual” properties such as orthogonality are not carried over well.

Another example where visualization fails would be the Hilbert space of states for a simple harmonic oscillator. This space is infinite-dimensional but yet you can still visualize states by plotting a wave function in 1d. States like $\vert n\rangle$ and $\vert m\rangle$ are orthogonal if $n\ne m$, but there is no notion of geometrical orthogonality in 1d: all vectors must be either parallel or anti parallel one to the other.

Answer 3

The Bloch sphere is better suited to visualize binary quantum states that have a geometric interpretation in real 3d space, and not just in abstract Hilbert space. In particular, it is best suited to visualize spin states, where, for instance, a state with definite spin in $x$-direction is a superposition of states with definite spin in $z$-direction.

In general, for a qubit every possible state is of the form $a|0\rangle+b|1\rangle$, where $|a|^2+|b|^2=1$. But also, since the physical state does not depend on a global complex phase (any probabilities remain the same if we multiply the whole state vector with $e^{\mathrm i\alpha}, \alpha\in\mathbb R$), we can choose $a$ to be real. So the whole thing turns to $|a||0\rangle+e^{\mathrm i\varphi}|b||1\rangle$, where still $|a|^2+|b|^2=1$. But then we can express the coefficients as $|a|=\cos\beta$ and $|b|=\sin\beta$ for some angle $\beta$, and the arbitrary state vector is now $\cos(\beta)|0\rangle+e^{\mathrm i\varphi}\sin(\beta)|1\rangle$.

And it turns out that if we substitute $\beta=\frac\theta2$, and interpret $\varphi$ and $\theta$ as the angles in spherical coordinates, the state $\cos\frac\theta2|0\rangle+e^{\mathrm i\varphi}\sin\frac\theta2|1\rangle$ is exactly the one that has well defined spin in the direction specified by $\varphi,\theta$ on the Bloch sphere.

The main takeaway: The Bloch sphere visualizes in which direction the spin of a particle can point, not the direction in Hilbert space. In fact, states that are parallel on the Bloch sphere are orthogonal in Hilbert space and vice versa.