# Is quantum interference only observed for observables that have a continuous spectrum?

The probability distribution for an observable with a discrete spectrum is defined by an inner product operation (and implemented as a dot product) on its wave function. $$\langle \psi|\psi\rangle$$ If $\psi$ is expressed as a superposition of eigenvectors $\left\{|\psi_1\rangle,|\psi_2\rangle \right \}\$ $$|\psi\rangle=\alpha|\psi_1\rangle + \beta|\psi_2\rangle$$ then the corresponding probability distribution is given by $$\langle\psi|\psi\rangle=\alpha^2|\psi_1|^2+\beta^2|\psi_2|^2$$ With the restriction that $\alpha^2+\beta^2 = 1$. The probability distribution for an observable with a continuous spectrum is also formed by an inner product but in this case the inner product is defined as an integral of the product of the wave functions $$\int_{-\infty}^{\infty}\psi^*(x)\psi(x)dx$$ If $\psi(x)$is a superposition of two states, such as $$\psi(x) = \alpha\psi_1(x)+\beta\psi_2(x)$$then $$\psi^*(x)\psi(x)=\alpha^2|\psi_1(x)|^2 + \beta^2|\psi_2(x)|^2 + 2\alpha\beta\psi_1(x)\psi_2(x)$$ I should probably integrate this $$\int_{-\infty}^{\infty}\alpha^2|\psi_1(x)|^2 + \beta^2|\psi_2(x)|^2 + \int_{-\infty}^{\infty}2\alpha\beta\psi_1(x)\psi_2(x)$$There is a interference term (the second integral) present for the continuous spectrum case that is not present for the discrete spectrum case. Is this true? Is quantum interference only observed for observables that have a continuous spectrum? Does the interference term integrate to zero if $\psi_1$ and $\psi_2$ are orthogonal? And if so, what does that mean with respect to interference?

NOTE: Based on the comments I can rephrase my question to the following. Is the following equation true? $$\langle\psi|\psi\rangle = |\psi|^2$$ I think this where I have a misunderstanding. I suspect it is not true but I do not immediately see why.

• If $\psi(x)$is a superposition of two states, such as $\psi(x) = \alpha\psi_1(x)+\beta\psi_2(x)$ then $\psi^*(x)\psi(x)=\alpha^2|\psi_1(x)|^2 + \beta^2|\psi_2(x)|^2 + 2\alpha\beta\psi_1(x)\psi)2(x)$ the interference term is always there irrespective of the spectrum is discrete or continuous. – hyportnex Apr 16 '17 at 15:51
• Are you sure about that? For a discrete spectrum we do not multiply the two wave functions, we take the dot product, which is different. – hooch Apr 16 '17 at 16:05
• Yes, I am sure besides the typo in the equation (irrelevant for this question) I just copied form yours, it should really read $\psi^*(x)\psi(x)=|\alpha|^2||\psi_1(x)|^2 + |\beta^2||\psi_2(x)|^2 + 2\Re{(\alpha^* \beta \psi_1(x)^*\psi_2(x))}$ – hyportnex Apr 16 '17 at 16:10
• Well, if you are sure I would like you to demonstrate it, because I do not think you are correct. A dot product cannot have the interference terms. – hooch Apr 16 '17 at 16:12
• $\psi^*(x)\psi(x)$ is not a dot product, it needs to be integrated to be one, and the spectrum has nothing to do with it. The continuity of the spectrum has to do with the normalizability of the wave function; interference terms will appear in the linear sum whether you can normalize it or not. – hyportnex Apr 16 '17 at 16:15

There is no problem with interference here, there is a problem that you are misinterpreting the symbols you are using:

1. There is no difference between the discrete and the continuous case, there is always a "dot product", more formally a Hermitian inner product, on our Hilbert space. That in the case of finite-dimensional Hilbert spaces $\mathbb{C}^n$ it looks like the ordinary dot product of vectors and in the case of the infinite-dimensional space $L^2(\mathbb{R}^n)$ of wavefunctions it looks like an integral is immaterial, it is an inner product in all cases. The inner product between two wavefunctions $\psi(x),\phi(x)$ is $\langle \phi\vert\psi\rangle = \int \phi^\ast(x)\psi(x)\mathrm{d}^n x$. In particular this means that if your two wavefunctions $\psi_1,\psi_2$ are orthogonal eigenfunctions of something, then the "interference" term vanishes in the continuous just as in the discrete case.

2. Given any quantum state $\lvert \psi\rangle$, $\langle \psi\vert\psi\rangle$ is not a "probability distribution". It is a physically meaningless normalization factor that is usually fixed to be 1. So is $\int \psi^\ast (x) \psi(x)$, when the integral is over all of space.

3. The object $\rho(x) = \psi^\ast(x)\psi(x) = \lvert \psi(x)\rvert^2$ is not the analogue of $\langle \psi \vert \psi\rangle$, but the analogue of $\langle \psi\vert x\rangle\langle x \vert \psi\rangle$, where $\lvert x\rangle$ is an eigenstate of the position operator (whose existence is rigorously a delicate question, but which we should just assume for now). By the Born rule, this is a probability (density) to detect the state $\lvert \psi\rangle$ as having position value $x$. So integrating the density $\rho(x)$ over a certain interval $[a,b]$ is also (sloppily) $$\int_a^b \rho(x)\mathrm{d}x = \langle \psi \vert \left(\int_a^b \lvert x\rangle\langle x\rvert\right)\vert \psi\rangle,$$ and yields the probability to detect the object in the region $[a,b]$. Note that for $a,b\to -\infty,\infty$, this becomes just $\langle \psi\vert\psi\rangle$ and so is just 1 - as the integral of a probability density over all possible events should be.

• So you are saying that the interference terms drop out for the continuous case just as it is missing in the discrete case. So how do we get a probability distribution that contains the interference terms? – hooch Apr 17 '17 at 13:43
• Is this equality incorrect? If so, ,what am I missing...$$\langle\psi|\psi\rangle=|\psi|^2$$ – hooch Apr 17 '17 at 13:45
• @hooch As I wrote, you have $\langle \psi \vert \psi\rangle = \int_{\mathbb{R}^n} \lvert \psi(x)\vert^2$ and you get interference terms by, well, actually computing a probability distribution instead of the normalization factor. Alas, "interference term" is a bit of a vague notion, so I'm not sure what exactly you're looking for. – ACuriousMind Apr 17 '17 at 13:48
• I think the state vector $|\psi\rangle$ represents a probability amplitude distribution in some n-dimensional space of n possible outcomes. In this n-dimensional space how do I represent the probability distribution for the n-possible outcomes? Is it also a n-dimensional vector? – hooch Apr 17 '17 at 13:58
• Adding to my previous comment...Assuming the probability distribution is also a vector, is it a vector in which the components are the squares of the components of the probability amplitude vector? – hooch Apr 17 '17 at 14:08

Regarding your rephrased question, suppose you have some observable $\hat A$ with a discrete spectrum $a_1,a_2,a_3,\ldots$, with corresponding eigenstates $\{|n⟩:n=1,2,3,\ldots\}$ forming an orthonormal basis. Then any state $|\psi⟩$ can be represented in the form $$|\psi⟩ = \sum_n c_n |n⟩, \quad\text{where }c_n = ⟨n|\psi⟩.$$ Here the $c_n = ⟨n|\psi⟩$ are the probability amplitudes for each of the measurements, and they are analogous to the wavefunction $\psi(x)=⟨x|\psi⟩$ (i.e. the probability density amplitude) of the continuous case.

The analogue of the probability density $|\psi(x)|^2=⟨\psi|x⟩⟨x|\psi⟩$ of the continuous case in discrete QM is then simply $$|c_n|^2 = ⟨\psi|n⟩⟨n|\psi⟩,$$ and it is known as the occupation probability of the state $|\psi⟩$ for that eigenstate.

If you have a superposition state, $$|\psi⟩=|\psi_1⟩+|\psi_2⟩, \text{ where }|\psi_1⟩ = \sum_n c_n^{(1)} |n⟩ \text{ and } |\psi_2⟩ = \sum_n c_n^{(2)} |n⟩,$$ so that $$|\psi⟩ = \sum_n c_n |n⟩ = \sum_n \left(c_n^{(1)} + c_n^{(1)} \right) |n⟩$$ and therefore $c_n |n⟩ = c_n^{(1)} + c_n^{(1)}$, then the corresponding measurement probability of outcome $a_n$ is given by $$|c_n|^2 = \left|c_n^{(1)}+c_n^{(2)}\right|^2 = \left|c_n^{(1)}\right|^2+\left|c_n^{(2)}\right|^2+ 2\operatorname{Re}\left({c_n^{(1)}}^*c_n^{(2)}\right),$$ where the last term represents interference, exactly as in the continuous case.

• Yes, that is what I was looking for in the discrete case. Now back to my question regarding interference. When $|\psi\rangle$ is considered to be in a superposition of its eigenstates, how do we get interference terms in the probability distribution? – hooch Apr 17 '17 at 15:27
• @hooch For future reference, this kind of switching back and forth between different questions is strongly discouraged. – Emilio Pisanty Apr 17 '17 at 15:35
• What is your reasoning for your final equation? – hooch Apr 17 '17 at 15:41
• @Emilo don't worry about my switching back and forth, you and your friends will take away all my points so I cant post another question, just like you did before. – hooch Apr 17 '17 at 15:44
• @hooch See edit. I won't do any more back and forth on this answer, though. – Emilio Pisanty Apr 17 '17 at 15:45