Let a quantum register of size $n$ be in the following state, expressed in the computational basis for $n$-qubit states:

$$\left| \psi \right> = \sum_{i=0}^{2^n-1} \alpha_i \left| i \right>, \forall i \in \{0, 1, ..., 2^n-1\}, \alpha_i \in \mathbb{C}$$

Additionally, the normalization condition states that:

$$\sum_{i=0}^{2^n-1}|\alpha_i|^2 = 1$$

How many real-valued coefficients would it take to express these $2^n$ complex-valued coefficients (the probability amplitudes of the register's quantum state)?

The naïve answer would be two real coefficients per complex coefficient, one for the real part and one for the imaginary part, or one for the modulus and one for the phase, so in total $2^{n+1}$.

The correct answer, it seems, is however $2(2^n-1)$, as stated in the lecture notes I am following, on slide 5.

Obviously, the normalization condition can be used to reduce the required number of real coefficients, but I can only see it reducing the required number by $1$ (express a modulus using all of the other modulus values) instead of $2$.

I just can't find an explanation to the correct answer.

  • $\begingroup$ "The correct answer, it seems, is however 2(2^n−1). " --- Who says so? $\endgroup$ Jan 18, 2020 at 21:05
  • $\begingroup$ slideshare.net/DavidCian/week3-ap3421-2019part1 you can find here the lecture notes stating it $\endgroup$
    – David Cian
    Jan 18, 2020 at 22:20
  • 1
    $\begingroup$ Which slide? #5? What does it say there: "Global phase is not relevant." Thus, one parameter less. --- And please put this information (including which slide!) in your question! Comments are futile. $\endgroup$ Jan 18, 2020 at 23:30
  • $\begingroup$ I facepalmed, you're right, it was in front of me the entire time. I was actually wondering why they put that about the global phase there. $\endgroup$
    – David Cian
    Jan 19, 2020 at 1:57

2 Answers 2


You can also rephase. So that gets rid of 1 more real dimension. That is quotient by $\psi ~ e^{i \theta} \psi$. Caution: unlike normalization where you are taking a subset, instead you are identifying many points into one.


Let $| \psi \rangle = \sum_{i=0}^{2^{n}-1} \alpha_i | i \rangle$ There are identifications $| \psi \rangle \equiv z | \psi \rangle$. Rescaling $|z|$ means changing the normalization and rescaling by $\frac{z}{|z|}$ is the rephasing by $e^{i \theta}$ mentioned already. So suppose $\alpha_0 \neq 0$, then we can pick a represetative of the equivalence class of $|\psi \rangle$ by letting $z = \frac{1}{\alpha_0}$, then we get $| \psi \rangle \equiv | 0 \rangle + \sum_{i=1}^{2^n-1} \frac{\alpha_i}{\alpha_0} | i \rangle$. This is not normalized but it is okay because we are only talking about equivalence classes and you can find the normalized state in this equivalence class. There are no more free parameters to specify. There are $2^n-1$ complex numbers to specify by giving all the values of $\frac{\alpha_i}{\alpha_0}$. If $\alpha_0 = 0$, then repeat the same trick with $\alpha_1$ or $\alpha_2$ etc. The entire space is not a vector space so you can't just parameterize everything with one coordinate chart. But this shows that in each chart by given by saying $\alpha_0 \neq 0$ etc is made by specifying $2^n-1$ complex numbers. They all glue together into one manifold that has complex dimension $2^n-1$ or real dimension $2(2^n-1)$.

Take the construction section in Wikipedia: Complex Projective Space and substitute $n+1$ there is $2^n$ here. There it was states in a Hilbert space $\mathbb{C}^{n+1}$ but you want $\mathbb{C}^{2^n}$. Above I give you the coordinate chart $U_0$ as it is called there. The line below is probably what you are thinking of. First you normalized everything so you get some sphere by $\sum Re(\alpha_i)^2 + Im(\alpha_i)^2 = 1$ but then you still have to do a quotient by phase to get rid of one more real dimension.

  • $\begingroup$ What is the difference to normalization? One quotients a positive factor, the other a phase factor. $\endgroup$ Jan 18, 2020 at 19:59
  • $\begingroup$ Sorry, I'm still having some trouble understanding, would you mind providing a bit more detail? I'm confused as to what $\psi$ and $\theta$ stand for. $\endgroup$
    – David Cian
    Jan 18, 2020 at 20:49

The only thing that is physical about a quantum state $|\psi\rangle$ is its square overlaps with other states. In other words, given an arbitrary complete basis $\{|u_k\rangle\}_k$, what matter are the overlaps $p_k=|\langle u_k|\psi\rangle|^2$, and these are interpreted as outcome probabilities and must therefore be normalised as $\sum_k p_k=1$.

This tells you two things:

  1. The global phase of $|\psi\rangle$ doesn't matter: changing $|\psi\rangle\to e^{i\phi}|\psi\rangle$ doesn't affect the probabilities $p_k$. This means that one less degree of freedom is needed to characterise the state.
  2. The normalisation condition on the probabilities must always be satisfied, thus removing another degree of freedom.

If the state is $N$-dimensional, then it amounts to $2N$ numbers as a real vector. Removing the two degrees of freedom as discussed above thus tells you that the number of real degrees of freedom characterising $N$-dimensional qudits is $2(N-1)$.


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