$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$ In Quantum Computation and Quantum Information the authors make the following statement "[G]iven any basis states $\ket{a}$ and $\ket{b}$ for a qubit, it is possible to express an arbitrary state as a linear combination $\alpha\ket{a} + \beta\ket{b}$ of those states." They go on to say that if the states are orthonormal, then one can perform a measurement with respect to the $\ket{a}\!, \ket{b}$ basis.

My question is does this imply that any two distinct pure states $\ket{a}$ and $\ket{b}$ form a basis for the Bloch sphere? (I apologize if this might make more sense in reference to a Hilbert space or some other vector space but I haven't gotten to that point in the book yet and have no experience with it otherwise).

It appears that any two distinct pure states are linearly independent, so I would suspect yes, but I do not believe I know enough to prove it.

For example, I attempted to try to transform the state defined by $\ket{\psi} = \frac{\sqrt{2 + \sqrt{2}}}{2}\ket{0} + \frac{\sqrt{2 - \sqrt{2}}}{2}\ket{1}$ into the basis $\{\ket{0}\!, \ \ket{+}\}$ where $\ket{+} = \frac{\ket{0} + \ket{1}}{\sqrt{2}}$. I obtained the following: $$ \begin{align*} \ket{\psi} &= \alpha\ket{0} + \beta\ket{1} = \alpha\ket{0} + \beta(\sqrt{2}\ket{+}-\ket{0}) = (\alpha - \beta)\ket{0} + \sqrt{2}\beta\ket{+}\\ &= \frac{ \sqrt{2+\sqrt{2}} - i\sqrt{2-\sqrt{2}} }{2}\ket{0} + \frac{i\sqrt{2}\sqrt{2 - \sqrt{2}}}{2} \ket{+} \end{align*} $$

This basis is neither orthogonal nor normal, so we would not be able to make a measurement with respect to it, however I believe that this is still a valid way to represent a state.

As the requirement that the basis be orthonormal in order to perform a measurement with respect to it implies that such bases exist, I would be interested in an example of orthogonal but not normal basis. I believe that taking two orthogonal bases (antipodal with respect to the bloch sphere), then scaling them, would be sufficient to realize this type of example.

I would also be interested in an example of a normal but not orthogonal basis, however no simple examples come to mind (it may be that there are no simple examples).


3 Answers 3


Your statements about bases are correct. For a qubit, any two distinct states can be used as a basis such that linear combinations of them can describe any state that you want. (If you want to describe something larger, you need more states.)

First, a comment about the term "distinct". Are states $$ \frac{|0\rangle+|1\rangle}{\sqrt{2}},\qquad{\text{and}}\qquad \frac{1+i}{2}|0\rangle+\frac{1+i}{2}|1\rangle $$ distinct? Clearly, they are not equal (which is the way I would usually take the meaning). However, they should not be considered distinct in the way that you need to take the meaning here. This is because the two states are the same up to an irrelevant global phase.

With that taken care of, let me expand on the proper explanation (which is basically just a generalisation of the argument you gave): Let's say I have a state of a qubit $|\psi\rangle$. I would like to express it in terms of two basis states $|\phi_0\rangle$ and $|\phi_1\rangle$ which are not orthogonal. To help, I can define a state $|\phi^\perp\rangle$ which is orthogonal to $|\phi_0\rangle$. This means that I can write $$ |\psi\rangle=\alpha|\phi_0\rangle+\beta|\phi^\perp\rangle $$ and also $$ |\phi_1\rangle=\gamma|\phi_0\rangle+\delta|\phi^\perp\rangle, $$ where $\delta\neq 0$. So we can trivially substitute $$ |\psi\rangle=\alpha|\phi_0\rangle+\frac{\beta}{\delta}\left(|\phi_1\rangle-\gamma|\phi_0\rangle\right), $$ and hence we can always express $|\psi\rangle$ in terms of $|\phi_0\rangle$ and $|\phi_1\rangle$, as desired, and this is a perfectly valid way to decompse the state. Your case of $|0\rangle$ and $|+\rangle$ is the perfect example of this.

You start, then, to talk about measurement in your question. Measurement is an entirely separate topic. Yes, writing a state with regards to a particular decomposition can be helpful when calculating measurement outcomes, but it's by no means a necessary procedure. (I'd argue that most of the time, the reason why we use an orthonormal basis is that it makes all calculations, such as inner products, easier, not specifically those associated with measurement.) Equally, the states for that don't need to be orthogonal. The orthogonality is a feature of projective measurements but if you were using POVMs, for example, the most natural basis could well be non-orthogonal.

An example with an orthogonal but not normal basis is trivial - whatever normalisation factor you put on the basis states you compensate for with the coefficient. However, in quantum, you should (essentially) always be using normalised states.

  • $\begingroup$ Thanks for your answer! Am I correct in understanding that you are saying that two distinct states form a basis for a qubit only if we define distinct as not being equal up to the global phase? If so, this would seem to imply that not all two non-equal points on the Bloch sphere define a basis for a qubit. This is slightly counter intuitive when compared to the real case, but I imagine this is due to the fact that $\alpha$ and $\beta$ are complex, and multiplying by a complex number can change direction in more ways than multiplying by a real number. $\endgroup$ Commented Apr 24, 2020 at 1:15
  • $\begingroup$ I'm also not certain I got it correct when I said that $\left|0\right>\!,\ \left|+\right>$ are not normal. They are both normal as I understand it, however in using them to represent a state that resulting state may not be normal with respect to the basis. Is that due to the fact that they are not orthogonal? $\endgroup$ Commented Apr 24, 2020 at 1:16
  • $\begingroup$ $|0\rangle$ and $|+\rangle$ are normal but not orthogonal. If you use a not-orthogonal basis, then the amplitudes do not have to have a sum-mod-square equal to 1. $\endgroup$
    – DaftWullie
    Commented Apr 24, 2020 at 7:45

Generally set of $n$ linearly indepedent vectors forms basis of a space with dimension $n$. However, in QC the basis has to be orthonormal to be able to distinguish basis states in a measurement and to have fulfilled condition $\sum_{i=1}^n|x_i|^2=1$, where $x_i$ is a vector member.


I will start by saying that if you have a vector space $\mathcal{V}$ with dim $V = N$, then this space is spanned by a basis comprised of $N$ linearly independent vectors. And any vector in this space can be described as a linear combination of these basis elements. So yes, I am inclined to say that $|a\rangle, |b\rangle$, standing for $|0\rangle, |1\rangle$ can form the basis for the 2-level system.

I am a bit confused on the statement regarding mapping $ \mathcal{B} =\{ |0\rangle , |1\rangle\}$ into $ \mathcal{B'} =\{ |+\rangle , |- \rangle \}$. My understanding is that in order to map your original basis into this new basis you need $H$ gate applied to your qubit.

Provided: \begin{eqnarray} H|0\rangle &=& |+\rangle \\ H|1\rangle &=& |-\rangle \\ H \Big ( \alpha|0\rangle + \beta |1\rangle \Big ) &=& \alpha H|0\rangle + \beta H|1\rangle\\ H |\psi\rangle &=& \alpha |+\rangle + \beta |-\rangle \end{eqnarray}

Given that $ |+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, |-\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}$ you can see that given that the initial basis was indeed orthonormal, so is this new basis.

Which is to say, I am not sure you can just change old basis for the new basis without applying the rotation which effectively does that change.

But of course, I can be wrong.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.