In the two-state formalism of Yakir Aharonov, the weak expectation value of an operator $A$ is $\frac{\langle \chi | A | \psi \rangle}{\langle \chi | \psi \rangle}$. This can have bizarre properties. If $A$ is Hermitian, the weak expectation value can be complex. If $A$ is a bounded operator with the absolute value of its eigenvalues all bounded by $\lambda$, the weak expectation can exceed $\lambda$.

If $A = \sum_i \lambda_i P_i$ where $\{P_i\}_i$ is a complete orthonormal set of projectors, the strong expectation value is $$\frac{\sum_j \lambda_j |\langle \chi |P_j | \psi \rangle|^2}{\sum_i | \langle \chi | P_i | \psi \rangle |^2}$$ which is also confusing as the act of measurement affects what is being postselected for.

More specifically, $$|\langle \chi |\psi\rangle|^2 = \sum_{i,j} \langle \psi | P_i| \chi \rangle \langle \chi | P_j | \psi \rangle \neq \sum_i |\langle \chi |P_i|\psi \rangle |^2$$ in general.

  • $\begingroup$ What happens when $\chi$ and $\psi$ are orthogonal? $\endgroup$
    – Ron Maimon
    Commented Aug 2, 2012 at 3:33
  • $\begingroup$ @RonMaimon, it seems that you can obtain arbitrarily large weak value, but in the limit of weak coupling, for which the formalizm is only valid, the probability of success in the postselecting measurement $P\sim\left|\left<\chi|\psi\right>\right|^2$ will vanish. $\endgroup$
    – straups
    Commented Aug 2, 2012 at 7:53

3 Answers 3


Let's first discuss what one usually means by "weak measurement". In a standard von Neuman measurement scheme the measurement apparatus (so called "pointer") is also treated quantum mechanically and described by the state $\left|\varphi\right>$. The pointer is coupled to the measured system such that the interaction hamiltonian is $H=gAp\delta(t-t_0)$, where $A$ is the observable to be measured, and $p$ the pointer momentum. After the interaction the pointer position becomes correlated with eigenvalues of the observable $A$: $$\left|\psi\right>\left|\varphi\right>\rightarrow e^{-igAp}\left|\psi\right>\left|\varphi\right>=\sum a_i\left|\psi_i\right>\left|\varphi(x-ga_i)\right>,$$ so by measuring the pointer position we can infer information about $a_i$. The measurement is "strong" if $\left<\varphi(x-ga_i)|\varphi(x-ga_k)\right>\sim\delta_{ik}$, i.e. the different pointer states have negligible overlap. This corresponds to a standard projective measurement (a nice description is given here).

Measurement is said to be weak in the opposite limit, when the coupling is weak enough for the pointer states to have large overlap. If after the interaction we postselect the system in the state $\left|\chi\right>$, the pointer state is: $$\left|\varphi_\chi\right>=\left<\chi\right|e^{-igAp}\left|\psi\right>\left|\varphi\right>\approx\left<\chi|\psi\right>e^{-igA_wp}\left|\varphi\right>,$$ where $A_w$ is the weak value.

The real part of $A_w$ corresponds to translation of the pointer coordinate as in the strong measurement: $$\left<x\right>=\left<x\right>_0+g\mathrm{Re}(A_w),$$ while the imaginary part corresponds to the change of pointer momentum: $$\left<p\right>=\left<p\right>_0+2g\mathrm{Im}(A_w)\mathrm{Var}_p,$$ where $\mathrm{Var}_p=\left<\varphi|p^2|\varphi\right>-\left<\varphi|p|\varphi\right>^2$ is the initial variance of pointer momentum. Proof in the most general case may be found in this paper by Jozsa.

The key moment is postselection of the final state of the system: weak values increase as the postselected state becomes nearly orthogonal to the initial state of the system. This can be considered as a kind of amplification of small pointer dispalcements due to the weak interaction on the expense of discarding almost all outcomes in the postselection.


The requirement for the validity of weak measurements is $$ga_i |\langle \chi | P_i | \psi \rangle | \ll |\langle \chi | \psi \rangle|$$ and $$ga_i \ll 1$$ where g is the weak coupling strength and $a_i$ is the ith eigenvalue. So, the weak expectation can only be as large as $1/g$ at most.


The problem with complex expectation values is easily fixed. If A is Hermitian, replace $\langle A \rangle$ with $\Re \left\{\frac{\langle \chi | A | \psi \rangle}{\langle \chi |\psi \rangle}\right\}$. If not Hermitian, $\frac{1}{2}\left[ \frac{\langle \psi | A | \chi \rangle}{\langle \psi |\chi \rangle} + \frac{\langle \chi | A | \psi \rangle}{\langle \chi |\psi \rangle} \right]$ will do.

As for expectation values exceeding the highest eigenvalue, well it's possible for $\langle \chi |P_i| \psi\rangle$ and $\langle \chi | P_j| \psi \rangle$ to have opposite signs, for $i\neq j$, while simultaneously, $\lambda_i$ and $\lambda_j$ also have opposite signs. This is a problem inherent with extended probabilities.


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