I understood quantum interference as a heart of quantum computing, because it enables two possibilities to cancel each other. Quantum algorithms utilize this property to reduce the probability of returning wrong answers and thus give higher chance to returning correct answers compared to classical algorithms. And quantum interference is possible because wave function can have complex value, not just non-negative value like classical probability.

But according to many articles about quantum mechanics, wave function is not actual existence - and wave function itself does not have any physical meaning. Only its absolute value squared has a meaning of possibility to be in a specific quantum state. But still, quantum interference happen in real quantum world and quantum computers use this phenomenon to be effective.

How is this possible?

  • 2
    $\begingroup$ Without knowing what kind of articles you mean, I'd caution that "existence" and "physical meaning" are loaded terms that could be better approached by comparing $\psi$-ontic (the wavefunction "exists") vs. $\psi$-epistemic (the wavefunction characterizes our beliefs) models for quantum mechanics. However both kinds of models (usually!) need to be consistent with our observations, so both must allow for quantum interference. Relevant: physics.stackexchange.com/questions/290522/… $\endgroup$
    – forky40
    Feb 5, 2021 at 3:05
  • 1
    $\begingroup$ Whoever said that "only the absolute value squared of the wavefunction has a meaning" is simply wrong. The correct statement is that the global phase of the wavefunction has no physical meaning; however, the relative phases between different basis vectors are physically meaningful and can be (indirectly and statistically) measured. $\endgroup$
    – tparker
    Jan 2, 2022 at 22:00

2 Answers 2


The assertion that the wavefunction has no physical meaning might mean different things in different contexts.

Wavefunction is unobservable

In one interpretation the assertion simply means that the wavefunction is unobservable. This is a scientific fact that can be proven for example by showing that the ability to observe it would allow faster-than-light communication (e.g. using arguments similar to those that rule out single-shot fidelity measurements). The impossibility to observe the wavefunction together with its central role in quantum mechanics raises some tricky questions.

Wavefunction is similar to probability distribution

Quantum mechanics can be thought of as an extension of the theory of probability where wavefunctions correspond to probability distributions. This suggests that wavefunctions may be assigned similar ontological status to probability distributions. These may be interpreted as a description of an observer's imperfect knowledge of the state of a system or sometimes as an incomplete description of the state of a system. Therefore in another view the assertion about the wavefunction lacking physical meaning refers to the fact that the wavefunction may be interpreted in a similar way. In the hidden variables research program the respective types of models are sometimes called $\psi$-epistemic$^1$ and $\psi$-supplemented and are contrasted with $\psi$-complete models that interpret the wavefunction as a complete description of a physical state.

Wavefunction is a theoretical construct

Finally, someone making the assertion might be recognizing that physical theories are abstract systems of concepts and rules that aim to reproduce results of observations. Some of the concepts correspond to observations. Others, such as the wavefunction, make no direct contact with empirical reality but are included to make the theory work. In a sense, they represent intermediate steps in calculations. In this view, ascribing "actual existence" to them is moot. Consider for example two quantum engineers: Alice who is a "wavefunction skeptic" and Bob who is a "wavefunction believer". Faced with a task of predicting the output distribution of a quantum circuit they will both compute the wavefunction $|\psi\rangle$ as an intermediate step before applying the Born rule. They might tell themselves different stories about the quantity they computed. Alice might attach little meaning to $|\psi\rangle$, treating it as an intermediate step in her calculations. Bob on the other hand might regard $|\psi\rangle$ as something real even if unobservable. The key point is that regardless of the adopted narrative each computes the same outcome distribution.


Regardless of the ontological status we assign to the wavefunction, it is clear that to make the theory agree with observations the rules we use to compute it must account for the possibility of interference of different amplitudes. In other words, whatever the wavefunction is - something real, an incomplete picture of some other more fundamental reality or a reflection of our imperfect knowledge - it interferes.

$^1$ See PBR theorem and this paper for some issues with the epistemic interpretation of the wavefunction.

I don't think it is particularly meaningful to say that a wavefunction "does not have physical meaning". It absolutely does: it tells you the probabilities of observing different outcomes in different measurement scenarios.

Sure, it's not the (possibly complex) components themselves that are directly related to the outcome probabilities, but so what? Why is that a requirement for something to be physically meaningful? The components $c_j\in\mathbb C$ of a state $|\psi\rangle$ tell you the expectation value associated with an observable $A$ via $\sum_{ij}\bar c_i c_j A_{ij}$. That means that $(c_j)_j$ describe a very much physical reality.

Note that it's just straight-up false to say that only the squared moduli have physical meaning. Sure, $|c_j|^2$ give you the outcome probabilities in the computational basis, but the phase of $c_j$ is also required for a full description of the state, and gives you information about outcome probabilities in different measurement bases.

Now, this said, there might be "non-physical" aspects in a wavefunction, in the sense of parameters which do not directly translate into observable differences. For example, global phases don't matter in a wavefunction. But one can easily redefine the wavefunction to get rid of these redundancies anyway.


Your Answer

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

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