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Assuming superposition state

$$ \Psi = C_1 \psi_1 + C_2 \psi_2 $$ ,one can write the expectation value $\langle A \rangle$ of a physical magnitude A as follows

$$ \langle A\rangle = P_1 \langle A\rangle_1 + P_2 \langle A\rangle_2 + 2 \sqrt{P_1 P_2} A_{12} \cos(\phi_2 - \phi_1) $$

This last term can be interpreted as interference, and is an added term with respect to the classical probability formula:

$$ \langle A\rangle = P_1 \langle A\rangle_1 + P_2 \langle A\rangle_2 $$

I'm not really sure about how this added term can be interpreted. I'm well aware that the wave nature of systems at quantum level must be involved, but I don't get why.

My best guess comes from supposing plane wave behavior, that is,

$$ \psi_1 \propto \exp(i \phi_1) $$

$$ \psi_2 \propto \exp(i \phi_2) $$

So, if we calculate the expectation value in Dirac notation:

$$ \langle A \rangle = \langle \Psi |A|\Psi \rangle = ... = 2 C_1^* C_2 \langle \psi_1 |A|\psi_2 \rangle + C_1 C_1^* \langle \psi_1 |A|\psi_1 \rangle + C_2 C_2^* \langle \psi_2 |A|\psi_2 \rangle $$

These last two terms can easily be attributed to classical probability, whereas the first one, using our plane wave model, can be descomposed as

$$ C_1^* C_2 = |C_1| e^{-i \phi_1}|C_2| e^{i \phi_2} = e^{i(\phi_2 - \phi_1)} \sqrt{P_1 P_2} $$

And, as $\langle A \rangle \in \mathbb R$, one takes de real part through Euler's formula, so

$$ C_1^* C_2 \equiv \cos(\phi_2 - \phi_1) \sqrt{P_1 P_2} $$

So far so good, but, how can one take a classical limit to such interference, as to recover the classical formula? Cosine will vanish $\Longleftrightarrow \phi_1 - \phi_2 = \frac{(2n + 1) \pi}{2}$, but I don't really see a correlation between the difference of $\phi's$ being a multiple of $\frac{(2n + 1) \pi}{2}$ and a classical limit.

Note: $\psi_1$ and $\psi_2$ aren't necessarily eigenstates of operator A.

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    $\begingroup$ I'm not sure what sort of interpretation you're looking for. $\endgroup$
    – Prahar
    Mar 6 at 11:55

3 Answers 3

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During interference, the square amplitudes of quantum states don't respect the rules of probability. When information about an observable is copied out of a quantum system interference between different values of the observable is suppressed: this effect is called decoherence. As a result of decoherence the square amplitudes respect the rules of probability after interactions including measurements:

https://arxiv.org/abs/1911.06282

https://arxiv.org/abs/1111.2189

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Your $\langle A \rangle$ is the theoretical expectation value for a single prepared system. You can't measure that; you can only perform many trials and take the average. If you have enough fine control of the prepared systems that $\phi_2-\phi_1$ has a similar value in every trial, then the average will approach the quantum $\langle A \rangle$ with the interference term. If you don't, the effect of the interference term will average to zero over many trials and you'll see apparently classical behavior. This is the same reason that it's hard to see the classical wave nature of light in ordinary circumstances.

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Starting from two orthonormal vectors $\psi_{1,2}$ (with $\langle \psi_k |\psi_\ell \rangle =\delta_{k\ell}$) in some (complex) Hilbert space $\mathcal H$, their superposition $$\psi=C_1 \psi_1+C_2 \psi_2 \in \mathcal{H}, \quad C_{1,2} \in \mathbb{C}, \quad |C_1|^2+|C_2|^2=1 \tag{1} \label{1}$$ describes a pure state with the associated density operator $$\rho_\psi= \psi \psi^\dagger=|C_1|^2 \psi_1 \psi_1^\dagger+|C_1|^2\psi_2 \psi_2^\dagger+C_1C_2^\ast \psi_1 \psi_2^\dagger+C_2 C_1^\ast \psi_2 \psi_1^\dagger \tag{2} \label{2}$$ with the properties $\rho_\psi^\dagger=\rho_\psi \ge 0$ (positivity) , ${\rm Tr} \rho_\psi=1$ (normalization) and $\rho_\psi^2 =\rho_\psi$ (characterizing a pure state), where $(\alpha \beta^\dagger) \chi := \alpha \langle \beta |\chi\rangle$ for $\alpha, \beta, \chi \in \mathcal H$ (alternatively you could use the Dirac notation $\alpha \beta^\dagger \equiv |\alpha \rangle \langle \beta |$).

The expectation value of a self-adjoint linear operator $A=A^\dagger$ in this state is given by $$\begin{align} {\rm Tr}(\rho_\psi A)= \langle \psi |A \psi\rangle &= |C_1|^2 \underbrace{\langle \psi_1 |A\psi_1\rangle}_{A_{11}} +|C_2|^2 \underbrace{\langle \psi_2 |A \psi_2\rangle}_{A_{22}}\\[5pt] &+C_1 C_2^\ast \underbrace{\langle \psi_2|A\psi_1\rangle}_{A_{21}=A_{12}^\ast}+ C_2 C_1^\ast \underbrace{\langle \psi_1|A\psi_2\rangle}_{A_{12}}. \end{align} \tag{3} \label{3}$$ Using your favourite notation $P_{1,2}:= |C_{1,2}|^2$ with $C_{1,2}= \sqrt{P_{1,2}}\;e^{i \phi_{1,2}}$ and $A_{12}= |A_{12}|e^{i \delta}$, eq. \eqref{3} can be rewritten as $$ \langle \psi |A \psi \rangle = P_1 A_{11} +P_2 A_{22} + 2 \sqrt{P_1 P_2}\; |A_{12}| \cos(\phi_2-\phi_1 +\delta). \tag{4} \label{4}$$

On the other hand, the expectation value of $A$ in the mixed state described by the density operator $$\rho=P_1 \psi_1 \psi_1^\dagger +P_2 \psi_2 \psi_2^\dagger, \quad \rho^\dagger=\rho \ge 0, \; {\rm Tr} \rho=1, \; \rho^2 \ne \rho \tag{5}, \label{5}$$ is given by $${\rm Tr}( \rho A)= P_1 A_{11}+P_2 A_{22}, \tag{6} \label{6}$$ obviously without the interference term present in \eqref{4}.

Remarks:

  1. The formula given in your second equation does not hold generally, but only in the special case $A_{12}^\ast = A_{12}$. (See \eqref{4} for the general result.)

  2. Given the fact, that the pure state $\rho_\psi$ differs from the the mixed state $\rho$, it should not come as a surprise that they can be discriminated by measurents of observables and in particular by the associated expectation values.

  3. This has nothing to do with any "wave nature". (We did not talk about "waves" in this example.) The interference effect has its origin in the simple fact, that the superposition of two state vectors corresponds again to a possible state vector (if properly normalized).

  4. The complex phases $e^{i \phi_{1,2}}$ (completely unrelated to "plane waves") do not come from the original vectors $\psi_{1,2}$ but have their origin in the trivial mathematical fact that a complex number $C$ can be written in the form $C= |C| e^{i\phi}$.

  5. I have no idea what you mean by "classical limit". At best, this could be interpreted as the effect that a measurement destroys the coherent superposition in the pure state $\rho_\psi$ ending up with the mixed state $\rho$?

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