# If global phase doesn't matter why is relative phase important?

As I understand the state of any $1/2$ spin particle can be expressed as:

$$\chi = \dbinom{\cos(\beta/2)e^{-i \alpha/2}}{\sin(\beta/2)e^{i \alpha/2}} \, .$$

Why is it stated that "a phase common to both the upper and the lower components is devoid of physical significance"? Why is the different phases of each term important then?

• In the square of the absolute value of the wavefunction, a global phase just gives you 1. But a relative phase matters, because the different states can interfere. – NickD Apr 28 '17 at 16:44
• @Nick Is the relative phase a measure of this interference? Is this what Guillermo describes as coherence in his answer below? – Alex Apr 28 '17 at 17:17
• Yes - see the answer by ZeroTheHero: the $cos(\alpha)$ term is the interference term. I don't understand why Guillermo says that $\alpha=0$ implies decoherence. We are talking about pure states here - at least, I thought so: $\alpha$ is not some randomly changing phase: it is fixed. – NickD Apr 28 '17 at 18:37
• possible duplicate of physics.stackexchange.com/questions/177588/… – Ofek Gillon Apr 28 '17 at 18:55

Nothing like a simple example to illuminate the situation. Take $$\psi(x)=\sqrt{2} e^{i \phi } \left(\cos \left(\frac{\beta }{2}\right) \sin (\pi x)+e^{i \alpha } \sin \left(\frac{\beta }{2}\right) \sin (2 \pi x)\right)$$ as a linear combination of two eigenfunctions for an infinite well of width $L=1$. Then $\psi(x)\psi^*(x)$ will obviously not depend on the overall phase $\phi$ but it will depend on $\alpha$ and $\beta$. Taking $\beta=\pi/2$ for simplicity we get for instance, the average values \begin{align} \psi^*(x)\psi(x)&=\sin ^2(\pi x) (4 \cos (\alpha ) \cos (\pi x)+2 \cos (2 \pi x)+3)\, ,\\ \langle x\rangle &= \int_0^1 dx x \psi^*(x)\psi(x) =\frac{1}{2}-\frac{16\cos\alpha}{9\pi^2}\, ,\\ \langle x^2\rangle &= \int_0^1 dx x^2 \psi^*(x)\psi(x)= -\frac{16 \cos (\alpha )}{9 \pi ^2}-\frac{5}{16 \pi ^2}+\frac{1}{3} \end{align} showing - very inelegantly - that the overall phase $\phi$ is irrelevant but the relative phase does play an important role.

Even for $\alpha=0$, $\langle x \rangle$ does display some interference between the two components of $\psi(x)$.

• Yeah I see thanks. How would you define coherence of say spin states as given in the Bloch sphere notation $\chi$ below? Also, am I correct in stating that $| \psi \rangle$ is in a state of superposition regardless of whether $\alpha = 0$ as I commented below? – Alex Apr 28 '17 at 18:18
• "Coherence" is a word I only use in the context of mixed states, when refering to the off-diagonal elements of the density matrix. Possibly others have a different use but the components in a superposition will always interfere if the state is pure, i.e. the system is defined by a wave function $\psi(x)$ or a ket $\vert \psi\rangle$. – ZeroTheHero Apr 28 '17 at 18:27
• Okay but give me that one please. I happen to be studying that section now. – Alex Apr 28 '17 at 18:30
• Not sure I understand what "that one" refers to... v.g. the eigenstates of $\sigma_y$ are $[\vert + \rangle \pm i \vert -\rangle]/\sqrt{2}$; there is a phase there but I don't see how these are more "coherent" than any other eigenstates... I usually use "coherence" to refer to off-diagonal elements of the density matrix, but your $\chi$ is a pure state so there is no coherence here. – ZeroTheHero Apr 28 '17 at 18:30
• I though you said you have a definition of coherence with regard to density matrices. Oh you mean that coherence is the off-diagonal elements of the density matrix. Could I ask one thing you might be a able to assist with regarding density matrices or should I post a new question. It's quite short so I don't really want to post a new question. What do you think? – Alex Apr 28 '17 at 18:34

Remember that for a 1/2 spin particle the only states we have are spin up $|\uparrow >$ and spin down $|\downarrow >$, or a general linear combination of both: $$\chi = \cos{(\beta/2)} e^{-i\alpha/2} \binom{1}{0}+\sin{(\beta/2)} e^{i\alpha/2} \binom{0}{1} = c_1 |\uparrow > + c_2 |\downarrow >$$ Through the angles $\beta$ and $\alpha$ we express exactly the value of the coefficients $c_1$ and $c_2$, i.e., we specify the linear combination. It's more clear if we imagine the state $\chi$ located on the surface of a unit radius sphere (the so called Bloch sphere). The polar angle $\beta$ and the azimutal angle $\alpha$ specify the location on the sphere. For example, $\alpha=\beta=0$ (the north pole of the sphere) would be state $\chi =|\uparrow >$. In the same way, the south pole would represent $|\downarrow >$.

The angle $\beta$ gives information about the $\textbf{population}$ of the quantum state: when we have many spin 1/2 particles, it will tell us in which proportion the particles are on state up and which proportion are in state down. For a single particle, it will just tell if it's more likely that this particle is in state up or state down. The phaseangle $\alpha$ gives information about the $\textbf{coherence}$ of the quantum state: it's only relevant when we have a superposition of both states: it will tell how strong is the superposition. If this angle is zero, there's no superposition at all (decoherence, state can only be in state up or down).

Therefore they're relevant angles. But when measuring, a common phase factor will not affect at all, because the probabilities remain the same (that's what they mean with physical significance). A new state $\chi'$: $$\chi'= e^{i\alpha/2} \chi = \cos{(\beta/2)} \binom{1}{0}+\sin{(\beta/2)} e^{i\alpha} \binom{0}{1}$$

$$|<\chi' | \chi'>|^2 = \cos^2{(\beta/2)} + \sin^2{(\beta/2)} = 1 = |<\chi | \chi>|^2$$
(since $|\uparrow >$ and $|\downarrow >$ form an orthonormal basis), and will lead to the same probabilities as $\chi$. For example, the probability of measuring state $|\uparrow >$ in state $\chi$ is the same as measuring in state $\chi'$: $$|<\uparrow| \chi'>|^2 = \cos^2{(\beta/2)} = |<\uparrow | \chi>|^2$$
• Thanks for your answer. I thought the fact that $\chi$ is a linear combination of two states implies that $\chi$ is in a superposition of states. Now you are saying that id $\alpha = 0$ implies $| \psi \rangle = \cos(\frac{\beta}{2})| \uparrow \rangle + \sin(\frac{\beta}{2})| \downarrow \rangle$ is not a superposition of states? Could you elaborate on this please? – Alex Apr 28 '17 at 16:31
• Why do you think that $\alpha=0$ implies decoherence? – NickD Apr 28 '17 at 18:38
In the notation you have provided the set of angles $\alpha$ and $\beta$ describe the orientation of the Bloch vector on the Bloch sphere; $$|\chi\rangle\;=\; \begin{pmatrix} e^{-i\tfrac{\alpha}{2}}\cos\left(\tfrac{\beta}{2} \right) \\ e^{i\tfrac{\alpha}{2}}\sin\left(\tfrac{\beta}{2} \right) \end{pmatrix}$$ However the general description of the qubit includes the global phase. It would be written as, $$|\chi\rangle\;=\;e^{i\frac{\gamma}{2}} \begin{pmatrix} e^{-i\tfrac{\alpha}{2}}\cos\left(\tfrac{\beta}{2} \right) \\ e^{i\tfrac{\alpha}{2}}\sin\left(\tfrac{\beta}{2} \right) \end{pmatrix}$$ While the $\alpha$ and $\beta$ angles describe the orientation of the Bloch vector on the 2-sphere, the third angle $\gamma$ is important as it describes the position of the qubit "globally" on the 3-sphere. For example, if we consider some closed loop on the 2-sphere, and calculate that the global phase of one orbit is $\gamma=2\pi$, then the total value of the global phase coefficient is; $$e^{i\tfrac{2\pi}{2}}=-1$$ The negative coefficient tells us that we have only traveled half of the total path on the 3-sphere. While it appears we have reached our starting point after one orbit, in actuality we require a second orbit of the path on the Bloch sphere to return to the initial point. This is the nature of the spin-1/2 particles, as 2 orbits in 3-dimensions corresponds to a single orbit in 4-dimensions.
In the 3-dimensional picture the global phase is a "non-observable" (so-called). This is due to the fact that it is a hidden variable in the 3-dimensional frame. This well known from the outer product, $$|\chi\rangle\langle\chi|\;=\;\frac{1}{2}\bigg(\sigma_1+\sin(\beta)\cos(\alpha)\sigma_x+\sin(\beta)\sin(\alpha)\sigma_y+\cos(\beta)\sigma_z\bigg)$$ The quaternion viewpoint of quantum mechanics purports that the global phase is a natural hidden variable. This was recently shown to be the case in a J. Phys. A article (2015) which you can find on the arXiv at: https://arxiv.org/abs/1411.4999