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In introductory quantum mechanics I have always heard the mantra

The phase of a wave function doesn't have physical meaning. So the states $| \psi \rangle$ and $\lambda|\psi \rangle$ with $|\lambda| = 1$ are physically equivalent and indiscernible.

In fact by this motivation it is said that the state space of a physical system shouldn't be a Hilbert space, but rather a projective Hilbert space, where vectors which only differ up to a multiplicative constant of magnitude 1 are identified.

But I also heard that one of the defining "feature" of quantum mechanics is the superposition principle: We can combine states $| \psi_1 \rangle, |\psi_2 \rangle$ to a new state $| \psi_1 \rangle + | \psi_2 \rangle$. This should for example explain the constructive / destructive interference we see in the double slit.

But if two states with the same phase are physically equivalent, so should the states $| \psi \rangle, -|\psi \rangle$. But their sum is zero. I have seen experiments which exploit this and measure the relative phase difference between two different states. But if relative phase difference is measurable, then surely the phase of a wave function does have physical meaning? This should mean that we can identify the phases of all states of a quantum system up to a $U(1)$ transformation by gauging some state to have phase $1$. Is this correct? How can this be solidified with the above mantra?

I have asked a second question here ("The superposition principle in quantum mechanics") regarding the superposition principle which is closely related to this question.

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    $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z May 18 at 5:53
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When people say that the phase doesn't matter, they mean the overall, "global" phase. In other words, the state $|0 \rangle$ is equivalent to $e^{i \theta} |0 \rangle$, the state $|1\rangle$ is equivalent to $e^{i \theta'} |1 \rangle$, and the state $|0\rangle + |1 \rangle$ is equivalent to $e^{i \theta''} (|0 \rangle + |1 \rangle)$.

Note that "equivalence" is not preserved under addition, since $e^{i \theta} |0 \rangle + e^{i \theta'} |1 \rangle$ is not equivalent to $|0 \rangle + |1 \rangle$, because there can be a relative phase $e^{i (\theta - \theta')}$. If we wanted to describe this very simple fact with unnecessarily big words, we could say something like "the complex projective Hilbert space of rays, the set of equivalence classes of nonzero vectors in the Hilbert space under multiplication by complex phase, cannot be endowed with the structure of a vector space".

Because the equivalence doesn't play nicely with addition, it's best to just ignore the global phase ambiguity whenever you're doing real calculations. Finally, when you're done with the entire calculation, and arrive at a state, you are free to multiply that final result by an overall phase.

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    $\begingroup$ So when we say "the phase doesn't matter" an analogous statement would be "mass doesn't matter" since we can always rescale our definition of mass by some constant. So the mass of an object isn't measurable, but its mass difference to another object is. The same way in QM: There is no notion of absolute phase since by a global $U(1)$-transformation we leave the physical content of the theory intact. $\endgroup$ – Jannik Pitt May 18 at 8:06
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    $\begingroup$ @JannikPitt Sure, but I would quibble a little bit with this. The mass of an object (in the units we choose) is indeed arbitrary, since the units can be rescaled. But this is a trivial kind of redundancy that applies to literally any quantity in physics that isn't dimensionless. I would say the invariance under global phase redefinitions is qualitatively different, but yes, it gives the same kind of result. $\endgroup$ – knzhou May 18 at 8:10
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    $\begingroup$ @JannikPitt I'm confused by your analogy. Why does the idea that you can express mass in different units mean that mass isn't measurable, but mass differences are? When you rescale your definition of mass by some constant, the mass differences also get rescaled by that same constant. I don't see how you can separate masses and mass differences in your analogy. $\endgroup$ – probably_someone May 18 at 12:15
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    $\begingroup$ @JannikPitt for completeness: we don’t measure masses: we measure ratios of masses, and these ratios do not depend on any rescaling. $\endgroup$ – ZeroTheHero May 18 at 13:33
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    $\begingroup$ @JannikPitt We don't measure phase ratios. We measure phase differences. Phase differences are what are important for the physics of interference, after all. $\endgroup$ – probably_someone May 19 at 22:56
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The global phase does not matter. In your example $\lambda(\vert\psi_1\rangle+\vert\psi_2\rangle)$ has the same physical contents as $\vert\psi_1\rangle+\vert\psi_2\rangle$ but this will be in general different from $\vert\psi_1\rangle-\vert\psi_2\rangle$ or more generally $\lambda’(\vert\psi_1\rangle+e^{i\varphi}\vert\psi_1\rangle).$

... and of course yes the relative phase can be measured, as indicated for instance in this answer and no doubt many others. In fact interferometry depends on such relative phases.

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While the other answers are correct, this is not a different answer but rather an illustration that the relative phase is indeed important in quantum mechanics. We know that bosons (particles with integer spin) have the following property: a rotation by $2\pi$ (around any fixed axis) leaves their states invariant, $R(2\pi)|{\rm boson}\rangle = |{\rm boson}\rangle$. This is obviously fine, since a rotation by $2\pi$ should be a symmetry operation. Fermions (particles with integer-and-a-half spin) have the property that a rotation by $2\pi$ changes their sign: $R(2\pi)|{\rm fermion}\rangle = -|{\rm fermion}\rangle$. This is also fine, since $-|{\rm fermion}\rangle$ belongs to the same ray as $|{\rm fermion}\rangle$ and hence describes the same state.

What, however, if we want to make a linear superposition of the form $|\Psi\rangle = \alpha|{\rm boson}\rangle+\beta|{\rm fermion}\rangle$, with $\alpha\neq\beta$? It is clearly seen that the operation of rotation by $2\pi$ on $|\Psi\rangle$ will not give a state proportional to $|\Psi\rangle$, and so is not a symmetry of that state. What went wrong?

The answer is that we simply should not be making such superposition. While it is well-defined mathematically, it is unphysical: it does not describe a state that can be physically prepared. Thus, we are forbidden from making a (physical) superposition of a boson and a fermion. This is an example of a powerful class of statements known as superselection rules.

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  • $\begingroup$ Very nice, something like this is exactly what I looked for. $\endgroup$ – Jannik Pitt May 18 at 12:36
  • $\begingroup$ +1: This is very interesting. How does this relate to the usual definition of superselection rules which goes something like "$\psi_1$ and $\psi_2$ are in different superselection sectors if $\langle \psi_1 | A | \psi_2\rangle=0$ for all observables $A$"? $\endgroup$ – Dvij D.C. May 18 at 14:51
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    $\begingroup$ -1: I don't think this is a correct explanation of superselection. Note that you could apply the same reasoning to almost any superposition: the momentum eigenstates $e^{ipx}$ and $e^{ikx}$ are "symmetric under translation" because translation just multiplies them by phases. Yet $e^{ipx} + e^{ikx}$ doesn't change by a phase. Does that mean that superpositions of different momenta are unphysical? $\endgroup$ – knzhou May 18 at 23:03
  • $\begingroup$ The point is that states that transform under different (i.e. non-equivalent) projective representations under a symmetry that is believed to be an exact symmetry belong to different superselection sectors. Bosons transform under ordinary single-valued representations. Fermions transform under double-valued spinorial representations. A linear combination "boson+fermion" does not have a well-defined transformation property under rotations. $\endgroup$ – printf May 19 at 3:31
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    $\begingroup$ You say the other answers are correct. All? It can change, so which? $\endgroup$ – Volker Siegel May 19 at 20:17
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But I also heard that one of the defining "feature" of quantum mechanics is the superposition principle: We can combine states |ψ1⟩,|ψ2⟩ to a new state |ψ1⟩+|ψ2⟩

...

But if two states with the same phase are physically equivalent, so should the states |ψ⟩,−|ψ⟩.

This seems quite confused to me. $|\psi_n\rangle$ isn't a state, it's a state vector. Grokking this difference is, I believe, crucial to untangling your question.

That is, it is the state vectors that are superposed, not the states (which don't form a vector space).

UPDATE: to address this comment (since comments are ethereal)

This should be a comment, nitpicking on terminology isn't an answer to the question. The physical "object" we aim model is a state, and we do that by assigning it a vector in some vector space. Then you could call this object a state vector, but calling it a state without differentiating between the object and the model works fine in most context (doing theoretical calculations, about which this is all about)

Weinberg is very careful in making the distinction between the state (ray) and the state-vectors in the ray when formulating Quantum Mechanics in section 2.1 of "The Quantum Theory of Fields". Here are some excerpts:

(i) Physical states are represented by rays in Hilbert space.

...

A ray is a set of normalized vectors (i.e., $(\Psi,\Psi)=1$) with $\Psi$ and $\Psi'$ belonging to the same ray if $\Psi'=\xi\Psi$, where $\xi$ is an arbitrary complex number with $|\xi|=1$.

...

(iii) If a system is in a state represented by a ray $\mathscr{R}$, and an experiment is done to test whether it is in any one of the different states represented by mutually orthogonal rays $\mathscr{R}_1,\,\mathscr{R}_2,\dots$, (for instance by measuring one or more observables) then the probability of finding it in the state represented by $\mathscr{R}_n$ is

$$P(\mathscr{R}\rightarrow\mathscr{R}_n)=|(\Psi,\Psi_n)|^2$$

where $\Psi$ and $\Psi_n$ are any vectors belonging to rays $\mathscr{R}$ and $\mathscr{R}_n$ respectively. (A pair of rays is said to be orthogonal if the state-vectors from the two rays have vanishing scalar products).

In your question, you seem (to me) to be mixing the concepts of state and state-vector together and the resulting confusion is, I think, at the root of your question.

As I read the section I quoted from your question above, you seem to be saying that since $|\psi\rangle$ and $-|\psi\rangle$ are physically equivalent states, it shouldn't be that their sum is zero (and then go on to conclude that phase should be physical).

But that doesn't follow if you carefully distinguish between the state (ray) and vectors. We form linear combinations of vectors, not states.

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  • $\begingroup$ This should be a comment, nitpicking on terminology isn't an answer to the question. The physical "object" we aim model is a state, and we do that by assigning it a vector in some vector space. Then you could call this object a state vector, but calling it a state without differentiating between the object and the model works fine in most context (doing theoretical calculations, about which this is all about). $\endgroup$ – Jannik Pitt May 18 at 7:58
  • $\begingroup$ @JannikPitt, I don't agree that it's nitpicking; your confusion seems to me to be conceptual in nature. That's all I have to say about that. $\endgroup$ – Alfred Centauri May 18 at 11:29
  • $\begingroup$ Then I don't know what you mean. How do you define the terms "state" and "state vector"? $\endgroup$ – Jannik Pitt May 18 at 12:34
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A fine example of physical behavior due to phase change is the Aharonov-Bohm Effect. A magnetic field that exerts no classical force on an electron nevertheless affects electron interference through the influence of the vector potential on the phase of the electron's wave function.

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Think of phase the same way you think of units. The universe doesn't care if we measure it in meters or furlongs, but that doesn't mean that we can mix the two in computations. The same is true here: instead of multiplying a real quantity by the meters-per-furlong conversion factor, we could multiply a complex quantity by a phase change, and just like in the real case, the physical meaning would still be the same. But we have to apply the same phase change to other quantities if we want them to be compatible and not mix "complex units".

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  • $\begingroup$ the problem with this argument as stated is this doesn't single out the global phase but could also be applied to the relative phase. $\endgroup$ – ZeroTheHero May 18 at 18:52
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The global phase does not matter because of charge-current conservation. According to the Noether theorem, charge-current conservation is related to invariance under a global phase change. As long as the problem at hand obeys charge-current conservation only phase differences have physical consequence.

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As the OP points out, many people say

the phase of a wave function doesn't have physical meaning

and there are also some who say

the state space of a physical system shouldn't be a Hilbert space, but rather a projective Hilbert space

But an alternative view that the OP might find less confusing is

The state of a physical system is a vector in a Hilbert space, but is a global U(1) symmetry that prevents us from observing the overall phase factor of the state of the universe. More generally, an observer confined to an isolated system cannot observe the overall phase factor for the system.

One might object that this statement fails to convey some of the information that the previous statement is trying to convey: namely, the fact that any observable properties of a subsystem don't depend on the subsystem's phase. However, this fact just becomes a "theorem" as opposed to a "postulate". It follows from the fact that the phase factor of $|\psi\rangle$ cancels out when we evaluate $\langle \psi | A | \psi \rangle$ for any Hermitian operator $A$. Whenever the phase of $|\psi\rangle$ can be observed, it is only through evaluating something like $\langle \psi' | A | \psi \rangle$ where $|\psi'\rangle$ is some other prepared state that $\psi$ can interact with; this is not an observable of $\psi$ alone.


Another person who answered this question mentioned Weinberg's The Quantum Theory of Fields. Because Weinberg does discuss projective space and rays for a particular reason, I want to address this in case the reader is not convinced that we can do away with the concept.

In Chapter 2 of the text, Weinberg discusses (among other things) the Poincaré transformation and mentions projective representations of the Lorentz group. It's thanks to the existence of these projective representations that we have half-integer spin. But one can use the theory of projective representations without using the projective space formalism as the ontology of quantum mechanics.

There are specific reasons why I think that we shouldn't say states are rays (elements of projective space). I think it's an attempt to make the unobservable phase factor disappear entirely from the ontology (sort of like how you often must do calculations on vectors using their components, but the vector itself has a coordinate-independent existence). However, one can do this only for an isolated system. As soon as we start examining subsystems that can interact with each other, we see that the state of the combined system contains more information than the states of the subsystems, since there are also $N-1$ phase factors if there are $N$ subsystems. I believe that because this violates some people's intuition about what the word "state" is supposed to mean, it leads to confusion.

None of what I'm saying alters the substance of Weinberg's text, though (far be it from me to challenge the actual physics). In particular, the reasoning still stands that each field should correspond to a projective representation of $SO^+(3, 1)$, which may fail to be an ordinary representation (in which case the spin is half-integer). Suppose we have $\Lambda_1, \Lambda_2 \in SO^+(3, 1)$ (the proper orthochronous Lorentz group). We can still assign to each Lorentz transformation an operator that acts on physical states, and we must have, like (2.2.10) in Weinberg,

\begin{equation} U(\Lambda_2) U(\Lambda_1) |\psi\rangle = e^{i\phi(\Lambda_2, \Lambda_1)} U(\Lambda_2 \Lambda_1) |\psi\rangle \end{equation}

Weinberg's interpretation of this relation is that performing the Lorentz transformation $\Lambda_1$ then $\Lambda_2$ must yield the same state as performing the Lorentz transformation $\Lambda_2 \Lambda_1$, but since a state is a ray, that means there may be a phase factor when comparing the vectors.

The alternative interpretation I suggest using, which doesn't involve rays, is to say: doing $\Lambda_1$ then $\Lambda_2$ doesn't always yield the same state as doing $\Lambda_2 \Lambda_1$. However, the two states should yield the same values for all observables, therefore they can only differ by a phase factor.

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  • $\begingroup$ Speaking of a global $\mathrm{U}(1)$ symmetry in this case is not accurate. Phase ambiguity is not a symmetry, but rather a redundancy in the description of the system. In a system with a global symmetry configurations related by a symmetry transformation are physically distinct. There exist physical observables which are not invariant under the symmetry. Measurement of these observables allows to distinguish states related by a symmetry transformation. If a group of transformations describes redundancies of the description, no such observables exist. This is the case of quantum mechanics. $\endgroup$ – Blazej May 19 at 9:02
  • $\begingroup$ @Blazej I don't really get what yo're trying to say. Multiplying every state vector by some phase does leave the physics invariant in the same way as "rotating" (in fancy words acting on them by a $SO(3)$-action) leaves the physics invariant. You just view the system from a different perspective. $\endgroup$ – Jannik Pitt May 19 at 10:30
  • $\begingroup$ No, this is not the same. If you multiply a wavefunction by a phase, the new wavefunction represents exactly the same physical situation. On the other hand in the case of rotational symmetry of physical space it is not true that rotated system is the same physical system as the unrotated one, even though it happens to be true that looked upon in a different reference frame the new situation looks the same as the original situation viewed in the original reference frame. $\endgroup$ – Blazej May 19 at 14:12
  • $\begingroup$ @Blazej If the whole universe hypothetically had a different rotational orientation, we wouldn't be able to tell, since we're inside the universe. In that sense the analogy with the quantum mechanical phase holds. On the other hand, when a subsystem of the universe is either rotated or altered in phase, we can tell from outside the system. $\endgroup$ – Brian Bi May 19 at 14:30
  • $\begingroup$ If $\psi$ is a normalized vector representating some quantum state, $\psi' = R \psi$ where $R$ is the unitary operator representing some rotation in physical space and $\psi'' = e^{i \alpha} \psi$, then there exists an observable represented by an operator $\mathcal O$ such that the expectation values $(\psi, \mathcal O \psi)$ and $(\psi', \mathcal O \psi')$ are different. There is not such operator for $\psi$ and $\psi''$. $\endgroup$ – Blazej May 19 at 16:24

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