# Can Objects in A Product State Jump to a Mixed State without an Intermediate State of Entanglement?

I read that irreversibly going from a pure state to a mixed state requires a two step Von Neumann measurement process: 1) entanglement, via a unitary operator 2) wave function "collapse" (or "observation", or some other interpretation of what to call step 2 that makes the result irreversible). Also, generally, the range of states of the object being measured (the range with non-trivial probability) is reduced as a result of the measurement. For instance, a wave function that is a Gaussian might have a smaller standard deviation after step 2. However, below is an example where step 1 seems to be skipped. In this example, the quantum system undergoes an non-unitary transformation (non-quantum transformation), which is not a measurement as described above, but still moves the composite system, and its subsystems, into a mixed state. Also, the range of non-trivial states is not reduced. Therefore, it seems that going from a pure to a mixed state need not always be the result of a Von Neumann measurement process.

Consider the double slit experiment, where we shoot a single electron, from $x = 0$, toward a screen, that is in the $yz$ plane at $x_1 > 0$, and the screen contains two appropriately placed slits. The detector is at $x_2 > x_1$. At each slit, is a device that sets the spin of the electron along the $\epsilon$ axis. One device sets it to $|\uparrow\rangle$ and the other sets it to $|\downarrow\rangle$. So, this is a "which way" kind of set up.

Assume within $[0,x_1]$:

• electron position $|\Psi\rangle = |\Psi_1\rangle$, a plane wave
• electron spin $|\epsilon\rangle =(2^{-1/2}|\uparrow\rangle + 2^{-1/2}|\downarrow\rangle)$
• product state $|\Psi\rangle \otimes |\epsilon\rangle = |\Psi_1\rangle \otimes (2^{-1/2}|\uparrow\rangle + 2^{-1/2}|\downarrow\rangle)$

Within $[x_1, x_2]$:

electron is irreversibly in the mixed state of

• $\hat{\rho} = (2^{-1}(|\Psi_2\rangle \langle\Psi_2|)\otimes (|\uparrow\rangle\langle \uparrow|) + 2^{-1}(|\Psi_3\rangle \langle\Psi_3|)\otimes (|\downarrow\rangle\langle \downarrow|)$)

where $\langle x|\Psi_2\rangle$ is a radial wave function, for electron position, with origin at one slit, and $\langle x|\Psi_3\rangle$ is a radial wave function, for electron position, with origin at the other slit.

We also can trace out the spin to arrive at the reduced density matrix for electron position:

• $\hat{\rho_{position}} = \langle\uparrow|\hat{\rho}|\uparrow\rangle +\langle\downarrow|\hat{\rho}|\downarrow\rangle = 2^{-1}|\Psi_2\rangle\langle\Psi_2| + 2^{-1}|\Psi_3\rangle\langle\Psi_3|$

where electron position wave functions $\langle x|\Psi_2\rangle$ and $\langle x|\Psi_3\rangle$ do not add, which would produce interference patterns at the detector, because the $2^{-1}$ factors are classical probabilities. That is, the two position state vectors are in an either/or situation, not a superposition of both existing at the same time.

Comparing the product state in the $[0,x_1]$ interval to $\hat{\rho}$ in the $[x_1,x_2]$ interval, it seems that the electron went from a pure state to a mixed state without the intermediate step of entanglement, via a unitary operator, of the states of $|\Psi\rangle$ (its position state vector) with the states of $|\epsilon\rangle$ (its spin state vector). (One cannot correlate the $|\Psi_2\rangle$ term of a state of $|\Psi\rangle$ with the $|\uparrow\rangle$ state of $|\epsilon\rangle$ and the $|\Psi_3\rangle$ term of the same state of $|\Psi\rangle$ with the $|\downarrow\rangle$ state of $|\epsilon\rangle$, via a unitary operator). That is, the interaction of the electron with the spin devices was not a quantum interaction via a unitary operator. It also wasn't a traditional measurement that would narrow the the ranges of states of $|\Psi_2\rangle$ and $|\Psi_3\rangle$, where their probabilities are non-trivial, compared to what they would be if we did not insert the spin setting devices. So, perhaps, any kind of action on a quantum system, that is not a unitary transformation of the system, forces the system out of being a quantum system, whether or not it is a traditional Von Neumann (two step) measurement?