Suppose an ensemble of a spin $\frac{1}{2}$ particle in the $S_z$ basis is described by a diagonal density operator $$\rho=w_{11}|+\rangle\langle+|+w_{22}|-\rangle\langle-|$$ where $w_{ii}$ are real and non-negative. In this ensemble, measurements of $S_z$ will give the value $S_z=+\frac{1}{2}$ with a probability $w_{11}$ and the value $S_z=+\frac{1}{2}$ with a probability $w_{22}$, and $w_{22}=1-w_{11}$.

What is the interpretation of the coefficients $w_{ij}$ of a nondiagonal density matrix of the form $$\rho=w_{11}|+\rangle\langle+|+w_{22}|-\rangle\langle-|+w_{12}|+\rangle\langle-|+w^*_{12}|-\rangle\langle+|$$ where $w_{ij}$ for $i\neq j$ can in general be complex?

  • $\begingroup$ It is always possible to find a basis in which the density matrix is diagonalized. So, being diagonal or non-diagonal doesn't really convey any useful interpretation. However, the eigenvalues do tell us about a state being pure or mixed (or pure or entangled). $\endgroup$ – abhijit975 Oct 8 '19 at 16:20
  • $\begingroup$ @abhijit975 Through the eigenvalues or through the condition $\rho^2=\rho$ tell us whether we deal with a pure state? And please also explain what will I get if I measure $S_z$ in the first ensemble and in the second ensemble. $\endgroup$ – mithusengupta123 Oct 8 '19 at 16:25
  • $\begingroup$ Page 4&5 of this pdf may be helpful. wwwphy.princeton.edu/~verlinde/PHY305/density2.pdf $\endgroup$ – abhijit975 Oct 8 '19 at 16:31
  • $\begingroup$ To find what does a measurement of $S_z$ gives, find the trace of the matrix $\rho S_z$. This is the expectation value of the measurement of $S_z$ $\endgroup$ – abhijit975 Oct 8 '19 at 16:36
  • 2
    $\begingroup$ @annav the diagonalization has nothing to do with classical or quantum mechanics per se. It is a theorem from linear algebra that any Hermitian matrix is diagonalizable. But what really matters is the measurement. Is the basis of the density matrix the same as the basis of the operator you are measuring? The nondiagonal terms are only important when you are taking the apparatus and the spin as a composite system. This article is insightful for this discussion: vvkuz.ru/books/zurek.pdf $\endgroup$ – abhijit975 Oct 8 '19 at 18:05

The answer is implicit in the general principles, so I'll review the general principles first.

The observable for any component of the particle's spin can be written $\hat u\cdot\vec S$ where $\hat u$ is a unit vector and $\vec S=(S_x,S_y,S_z)$. For any unit vector $\hat u$, this observable has eigenvalues $\pm 1/2$, so we can always write this observable as $$ \hat u\cdot\vec S = \frac{1}{2}P + \left(-\frac{1}{2}\right)(1-P) \tag{1} $$ where $P$ is the projection operator onto this observable's spin $+1/2$ subspace and $1-P$ is the projection operator onto its spin $-1/2$ subspace. When the observable $\hat u\cdot\vec S$ is measured, the probabilities of the possible outcomes are \begin{align} \text{probability}\Big(\hat u\cdot\vec S \to +1/2\Big) &= \text{trace}(P\rho ) \\ \text{probability}\Big(\hat u\cdot\vec S \to -1/2\Big) &= \text{trace}\big((1-P)\rho \big), \tag{2} \end{align} where the left-hand sides use a notation that is hopefully clear. Equations (1)-(2) are basis-independent.

The OP uses a basis that diagonalizes $S_z$. Equations (2) imply that if we measure $S_z$, then the probabilities of the outcomes $+1/2$ and $-1/2$ are $w_{11}$ and $w_{22}$, respectively. The same equations imply that if we measure $S_x$ or $S_y$, then the probabilities of both outcomes depend on all of the components of $\rho$, including the complex off-diagonal component $w_{12}$.

What is the interpretation of the coefficients $w_{ij}$ of a nondiagonal density matrix of the form [shown in the OP] where $w_{ij}$ for $i\neq j$ can in general be complex?

One way to answer this is that the probabilities (2) generally depend on $w_{12}$. The only exception is when $\hat u=(1,0,0)$ so that $\hat u\cdot\vec S=S_z$.

If a more specific "physical" interpretation of $w_{12}$ by itself is really desired, here's one example of such an interpretation: \begin{align} 2w_{12} &= \text{probability}\Big(S_x \to +1/2\Big) \\ &- \text{probability}\Big(S_x \to -1/2\Big) \\ &+ i\, \text{probability}\Big(S_y \to +1/2\Big) \\ &- i\, \text{probability}\Big(S_y \to -1/2\Big). \tag{3} \end{align} To check this, use the fact that (in one convention) the matrices $$ \frac{1}{2} \left(\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}\right) \hskip2cm \frac{1}{2} \left(\begin{matrix} 1 & i \\ -i & 1 \end{matrix}\right) \tag{4} $$ project onto the $+1/2$ eigenspaces of $S_x$ and $S_y$, respectively, together with $$ \rho = \left(\begin{matrix} w_{11} & w_{12} \\ w^*_{12} & w_{22} \end{matrix}\right). \tag{5} $$ The interpretation (3) looks unnatural because attempting to interpret specific components in a specific basis is technically always unnatural. The important message is that any valid physical interpretation must follow from equation (2). The physical content of those equations is basis-independent, just like the physical content of general relativity is coordinate-independent.

A comment below the question asked about a relationship between eigenvalues and purity, so I'll address that, too. The state is pure if and only if $\rho^2=\rho$, which is equivalent to the condition that $\rho$ have one eigenvalue equal to $1$ and another equal to $0$ (because the trace of $\rho$ must be $1$). The eigenvalues of $\rho$ depend on all of its components, not just on the diagonal ones, so the purity/impurity of the state also depends on all of its components.

| cite | improve this answer | |

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.