1
$\begingroup$

In quantum mechanics, one often encounters complex interference patterns.

Here, I am curious about geometric interference patterns. For instance, let me define two wave-functions:

$$ \psi_1 = \exp (A_1+\mathbf{X}_1+B_1i)\\ \psi_2 = \exp (A_2+\mathbf{X}_2+B_2i) $$

where $A_1,A_2,B_1,B_2 \in \mathbb{R}$, and where $\mathbf{X}_1=x_1\sigma_x+y_1\sigma_y,\mathbf{X}_2=x_2\sigma_x+y_2\sigma_y$ are vectors ($\sigma_x,\sigma_y$ are the Pauli matrices).

If I define a probability rule as $P=\det \psi$, then I can create an interference pattern as follows:

$$ \det ( \psi_1+\psi_2 )= \underbrace{\det \psi_1+\det\psi_2}_{\text{sum of probabilities}}+ \underbrace{\det \psi_1\operatorname{Tr}\psi_1^{-1}\psi_2}_{\text{interference pattern}} $$

I note that if $\mathbf{X}_1\to 0,\mathbf{X}_2\to0$ then the probability rule reduces to the Born rule.

My question is what am I describing? Is it the geometric interference of planar waves with a complex phase, or is it something else? This is a generalization of the Born rule to include geometric terms.

$\endgroup$

1 Answer 1

0
$\begingroup$

This question was closed on Physics StackExchange but as I had aready written an answer I will post it here.

Interpretation

I am assuming that your notation for $\left\{\psi_j\right\}$ translates to the more conventional notation:

$$\psi_j=e^{A_j+iB_j}e^{\mathbf X_j}$$

i.e. we are not adding scalars and matrices but instead multiplying the matrix $e^{\mathbf X_j}$ by the scalar $e^{A_j+iB_j}$.

Please do let me know if this interpretation is correct? Dispite this the below proof holds only on the assumption that $\left\{\boldsymbol\psi_j\right\}$ are matrices.

Hilbert Spaces

Hilbert spaces are vector spaces with an inner product. In quantum mechanics, a pure state is represented by a vector $|\psi\rangle$ in the Hilbert space of the system. And the probability of observing the system in a state $|\phi\rangle$ if the system is in a state $|\psi\rangle$ is given by the square magnitude of the inner product of $|\psi\rangle$ and $|\phi\rangle$:

$$P\equiv\left|\langle\phi|\psi\rangle\right|^2\tag{1}$$

Representation

The most common representation of the state vectors used is that of column vectors for finite-dimensional Hilbert spaces, or functions for infinite-dimensional Hilbert spaces. However, we can represent the vectors as matrices, one uses case for $2\times2$ matrices is when we have a pair of two-state systems (for more details on use cases please see the comments on this question I asked). A valid inner product using a matrix representation is:

$$\langle\mathbf A|\mathbf B\rangle=\operatorname{Tr}\left(\mathbf A^\dagger\mathbf B\right)$$

Probability Rule in Question

Consider the probability rule $P=\operatorname{det}\boldsymbol \psi$. Now let us work backwards and see what the inner product in this representation would need to be to give this probability rule. Comparing this to equation (1) we note that this probability rule only has a single argument and so is not the general probability rule for this representation but is a probability rule giving the probability of the system being measured in a some fixed state $\boldsymbol\phi$. Next we need to take the square root and add some general phase to get the inner product: $\langle\phi|\psi\rangle=e^{i\theta}\sqrt{\operatorname{det}\boldsymbol\psi}$ where $\theta$ is real and can depend on both $\boldsymbol\phi$ and $\boldsymbol\psi$.

Now an inner product must satisfy three conditions. The condition we will consider is the inner product must be linear in $\boldsymbol\psi$. Thus, let $\boldsymbol\psi=a_1\boldsymbol\eta_1+a_2\boldsymbol\eta_2$ for any complex coefficients $a_1$ and $a_2$. The only way for $\operatorname{det}\boldsymbol\psi$ to be linear in $2\times2$ matrices is for $\left\{\boldsymbol\eta_j\right\}$ to be traceless as shown in this post. But even if all the matrices in the Hilbert space were traceless the square root of a scalar is not linear and for this reason $\langle\phi|\psi\rangle=e^{i\theta}\sqrt{\operatorname{det}\boldsymbol\psi}$ is not a valid inner product and so $P=\operatorname{det}\boldsymbol \psi$ is not a valid probability rule, at least within the formulation of quantum mechanics.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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