I have read Cohen's book, and various posts in this site; however, I'm still not convinced why we can't model a statistical mixture as a superpositions of pure states ?

For example, consider the Stern-Gerlach experiment; the book claims that when the atoms go outfrom the oven, we don't know their states; however, even when they pass from the apparatus, we still do now know the states of the particles coming from the $+z$ direction of the apparatus if we were to measure their $x$ component next. Of course, one can argue that, no we do know the state; it is $| +z \rangle$, but that state is composed of $|+x\rangle $ and $|-x\rangle$, so if were to measure their $x$ component, practically we just know that %50 percent of the atom will come out from the $+x$ direction.

Similarly, when we first pass the incoming beam coming out from the oven, we, statistically, knew that %50 percent of the beam was going to come out from $+z$ direction, so, in practice, I don't see any reason why can't we model the statistical mixture as a superposition of pure states.

For example, why don't we say that the state of the atoms in the beam coming out from the oven is $$\psi = 0.5 |+z \rangle + 0.5 |-z\rangle~?$$


The core difference between a mixture and a superposition is that, if we use orthogonal states, different parts of a mixture cannot interfere whereas different parts of a superposition will interfere.

In other word, if you define for instance $$ \vert \pm\rangle =\frac{1}{\sqrt{2}} \left(\vert\psi_1\rangle \pm \vert\psi_2\rangle\right) \tag{1} $$ with $\langle \psi_k\vert\psi_j\rangle =\delta_{kj}$ then an average value for a superposition $$ \langle +\vert \hat A\vert +\rangle = \frac{1}{2} \left( \langle \psi_1\vert \hat A\vert\psi_1\rangle +\langle \psi_1\vert \hat A\vert\psi_2\rangle +\langle \psi_2\vert \hat A\vert\psi_1\rangle +\langle \psi_2\vert \hat A\vert\psi_2\rangle \right) $$ and likewise for $\langle -\vert \hat A\vert -\rangle $ with cross terms occurring between different states $\vert \psi_1\rangle$ and $\vert \psi_2\rangle$ in the superposition, whereas the average value for a mixture $$ \hat \rho =\alpha \vert +\rangle \langle +\vert + (1-\alpha) \vert -\rangle \langle -\vert $$would be $$ \hbox{Tr}(\hat \rho\hat A) =\alpha \langle +\vert \hat A\vert +\rangle + (1-\alpha) \langle -\vert \hat A\vert -\rangle \tag{2} $$ with no cross-terms occurring between different states $\vert +\rangle$ and $\vert -\rangle$ of the mixture. Notice also that, in Eq.(2), the average value is the weight sum of $\langle \pm \vert \hat A\vert \pm \rangle$ with the coefficients $\alpha$ and $(1-\alpha)$, which are necessarily real, functioning as statistical weight and thus sum to 1.

On the other hand, there is no restriction on the possible coefficients of a linear combination of states beyond the normalization condition: these coefficients can be complex or real. The normalization condition involves a sum of their modulus squared and not a sum of the coefficients themselves.


Well, consider the probability distributions for all possible measurements. For the simplest case, suppose we did a full state tomography so we measured all the terms in the density matrix. Then if we have a mixture of $|+z\rangle$ and $|-z\rangle$ states we will measure only the symmetrical amplitudes, i.e. we will find that the state matrix is \begin{equation} \rho = \left(\begin{array}{ccc}0.5&0\\0&0.5\end{array}\right) \end{equation} whereas if state is really the pure state $\frac{1}{\sqrt{2}}|+z\rangle+\frac{1}{\sqrt{2}}|-z\rangle$ we would measure off-diagonal elements as well, resulting in the state matrix \begin{equation} \rho = \left(\begin{array}{ccc}0.5&0.5\\0.5&0.5\end{array}\right) \end{equation} This is a different state.

Another way to put it is that we will find different results for certain measurements in these different state matrices. For example, if you consider the first Pauli measurement operator $\sigma_x$, i.e. you would conduct a measurement along the x-axis instead of along the z-axis, then in the first (mixed) state you will find the average results is \begin{equation} \langle \sigma_x \rangle = Tr[\rho \sigma_x] = 0 \end{equation} (symmetrical results of +x and -x)

whereas in the second case (pure state) you would find the average result to be \begin{equation} \langle \sigma_x \rangle = Tr[\rho \sigma_x] = 1 \end{equation} (only +x results)

  • $\begingroup$ ok if the matrix representation of the density matrix becomes different, but so what ? experimentally, what does this tells us ? what does one predicts and what we observe ? what is the contradictory prediction of the pure state case ? $\endgroup$ – onurcanbkts Jan 31 '19 at 8:55
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    $\begingroup$ It predicts that if you measure in the x-direction you will always get +x in the pure state, whereas you will get 50% +x and 50% -x in the mixed state. $\endgroup$ – PhysicsTeacher Jan 31 '19 at 8:58

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