I'm having trouble accepting, intuitively, that $-\rvert1\rangle \otimes \rvert1\rangle = -\rvert11\rangle = \rvert1\rangle \otimes -\rvert1\rangle$.

It's my understanding that $ -\rvert1\rangle$ is just $\rvert1\rangle $ but time or space delayed in phase, where what "phase" means here depends on the hardware realization.

But $-\rvert1\rangle \otimes \rvert1\rangle = -\rvert11\rangle $ suggests that, if I take 2 independent systems, with the first completely out-of-phase relative to the second because (by, e.g., a Z gate), and I look at these two systems together, the composite system is completely out of phase with $\rvert11\rangle $ Somehow, I delayed one qubit, looked at it together with others, and suddenly all N of them are delayed.

All the more, if I have $\rvert111\rangle$, phase delay the first qubit with a Z gate, I get $-\rvert1\rangle \otimes \rvert11\rangle$, but now I can reassign this phase delay to any qubit and say I have $\rvert11\rangle \otimes -\rvert1\rangle$ . I.e., it's like saying qubits A-B are in phase together and collectively out of phase with qubit C, but that's the same as qubits B-C are in-phase and collectively out of phase with qubit A. This is a transitivity-violation contradiction of what we expect from equality!

My picture of phase is electrons precessing, or some other periodic process where physical objects corresponding to qubits are time delayed with definite time to each other. In this picture, phase can't be reassigned as will in the seeming contradiction I mentioned above. Is this picture incorrect?

  • $\begingroup$ "...but time or space delayed in phase"...that's probably where your misconception arises from. You're thinking of qubits in terms of plane wave propagation, which is misleading. How would you interpret the phase factor of $-1$ if your qubits are only (say) electrons? $\endgroup$ May 30, 2019 at 13:56

2 Answers 2


First of all note that, strictly speaking, quantum states are defined up to multiplication by complex scalars (that is, they are elements of the associated projective space $\mathbb{CP}^n$). This means that it is not true that "$-\lvert1\rangle$ is just $\lvert1\rangle$ but time or space delayed in time". The vectors $-\lvert1\rangle$ and $\lvert1\rangle$ are simply two different ways to write the same exact physical state.

Regardless, that is not the correct way to describe the sort of physical scenario you seem to be having in mind. If a particle is freely propagating in space, its state will change, and you can describe this change as an accumulating phase factor. Note that this is a phase factor between the different spatial modes occupied by the particle. In other words, you have a state with two possible degrees of freedom (say, position one and position two), and you accumulate a phase factor between them.

However, if you consider two particles, the situation is a bit different. If particle $1$ is in position/state $a$ and particle $2$ is in position/state $b$, you can write your state as $\lvert 1a\rangle\otimes\lvert2b\rangle$. If particle one goes through a delay line, or other kind of change of its internal state that results in a phase accumulation, then what you would get is something like $$\lvert1a\rangle\otimes\lvert 2b\rangle\to\lvert1a'\rangle\otimes\lvert 2b\rangle,\tag1$$ where $\langle a\rvert a'\rangle=e^{i\phi}$. In other words, the accumulated phase factor is in the internal state of the first particle, not in the overall state. This phase factor will be detectable if you then make the particles interfere, for example by joining the two positions $\lvert1\rangle$ and $\lvert2\rangle$ with something such as a beamsplitter$^{(1)}$.

Another way to understand the question might be: why is the following true?

enter image description here

Again, if you interpret $Z\lvert1\rangle=-\lvert1\rangle$ as "$Z$ evolving the state $\lvert1\rangle$ into the state $-\lvert1\rangle$" then this identity would not make much sense. Instead, what is going on is that $\lvert1\rangle$ (as well as $\lvert0\rangle$) is a state that is not affected by $Z$. The phase factor brought by $Z$ only has physical meaning when it acts on a different state $\alpha\lvert0\rangle+\beta\lvert1\rangle$, which is not stable under the action of $Z$.

${}^{(1)}$ In this kind of situation, you would most probably also have to take into account the type of particles you are dealing with. If these were photons (or more generally bosons), then the correct state would be the symmetrised version of Eq. (1). Equivalently, you could use second quantisation formalism to describe the state and its evolution, which would read as $$a_{1a}^\dagger a_{2b}^\dagger\lvert\mathrm{vac}\rangle \to a_{1a'}^\dagger a_{2b}^\dagger\lvert\mathrm{vac}\rangle.$$ Interfering the spatial modes $1$ and $2$ on a beamsplitter would correspond to writing \begin{align} a_{1\alpha}^\dagger&\to \frac{1}{\sqrt2}(a_{1\alpha}^\dagger+a_{2\alpha}^\dagger), \\ a_{2\alpha}^\dagger&\to \frac{1}{\sqrt2}(a_{1\alpha}^\dagger-a_{2\alpha}^\dagger). \end{align}


Maybe this is a suboptimal answer, which doesn't fully answer your question, but maybe it helps you understant why the global phase doesn't matter from a computational point of view (actually, it does matter from a physical point of view). In quantum computing, we manipulate states in the Hilbert space, represented by the Bloch sphere. And this sphere is not sensitive to global phases, i.e. phases multiplying by whole state. This is surely better explained here.

As for the visualization of that global phase introduced, picture the case of one qubit in the block sphere, in the |1> state. A Z gate introduces a rotation over the Z axis, but in this case, the state is already fully projected over the Z axis, so no effect is noticeable (it would be like the state vector rotating over itself, which doesn't change the position in the Bloch sphere).

Something similar would happen for a 2 qubit system, except it is harder to visualize because it is high-dimensional.

As referred in the link I shared, this doesn't mean that the global phase doesn't have physical meaning. It does. But that physical meaning isn't captured by the Hilbert space formalism ($C^2$ for a single qubit), on which quantum computation relies. To detect a global phase you would have to look at the $C^3$ formalism, I think.

Disclaimer: I work in quantum computing, and my advanced quantum physics knowledge is not as good as before...

  • 2
    $\begingroup$ What would you say the significance of global phase is, physically? $\endgroup$ May 30, 2019 at 11:27

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