Is there a work-energy theorem in quantum mechanics? The work-energy theorem is an important result of classical mechanics stated as follows:
$$ \int \vec F \cdot d\vec s = \Delta E $$
However, this theorem depends on $d \vec s$ which requires the existence of a definite trajectory of any material body. 
When we move to quantum mechanics, the energy is still well-defined as eigenvalues of the Hamiltonian operator. But quantum trajectories simply do not exist. So is it still possible to define work? Is there an analogous work-energy theorem? 
 A: This might be derived with the momentum operator $\hat p~=~-i\hbar\nabla$. Consider the force operator $\hat F~=~\partial\hat p/\partial t$. The expectation of any operator $\langle\psi|{\cal O}|\psi\rangle$ under time derivative is
$$
\frac{\partial}{\partial t}\langle\psi|\hat p|\psi\rangle~=~\langle\frac{\partial}{\partial t}\psi|\hat p|\psi\rangle~+~\langle\psi|\frac{\partial}{\partial t}\hat p|\psi\rangle~+~\langle\psi|\hat p|\frac{\partial}{\partial t}\psi\rangle
$$
which by using the Schrodinger equation $i\hbar\frac{\partial}{\partial t}|\psi\rangle~=~H|\psi\rangle$ gives
$$
\frac{\partial}{\partial t}\langle\psi|\hat p|\psi\rangle~=~\frac{i}{\hbar}\langle\psi|[H,~\hat p]|\psi\rangle~+~\langle\psi|\frac{\partial}{\partial t}\hat p|\psi\rangle.
$$
Now consider the derivative
$$
\delta\vec r\cdot\frac{\partial}{\partial t}\langle\psi|\hat p|\psi\rangle~=~\frac{i}{\hbar}\langle\psi|[H\delta\vec r,~\hat p]|\psi\rangle~+~\langle\psi|\delta\vec r\cdot\hat F|\psi\rangle.
$$
$$
\delta\vec r\cdot\frac{\partial}{\partial t}\langle\psi|\hat p|\psi\rangle~=~\frac{i}{\hbar}\langle\psi|H[\delta\vec r,~\hat p]|\psi\rangle~+~\frac{i}{\hbar}\langle\psi|\delta\vec r[H,~\hat p]|\psi\rangle~+~\langle\psi|\delta\vec r\cdot\hat F|\psi\rangle.
$$
The first commutator is $H[\delta\vec r,~\hat p]~=~i\hbar\delta H$, and $[H,~\hat p]~=~\hat p$. This then becomes 
$$
\delta\vec r\cdot\frac{\partial}{\partial t}\langle\psi|\hat p|\psi\rangle~=~-\langle\psi|\delta H|\psi\rangle~+~\frac{i}{\hbar}\langle\psi|\delta\vec r\cdot \hat p|\psi\rangle~+~\langle\psi|\delta\vec r\cdot\hat F|\psi\rangle,
$$
where we recognize a part of the answer here. The middle term moved to the left then produces
$$
\delta\vec r\cdot\frac{d}{dt}\langle\psi|\hat p|\psi\rangle~=~-\langle\psi|\delta H|\psi\rangle~+~\langle\psi|\delta\vec r\cdot\hat F|\psi\rangle.
$$
This is an Ehrenfest form of the work energy theorem. The momentum operator in raising and lowering operator $p~\sim~a~-~a^\dagger$ is off diagonal and the expectation $\langle\psi|\hat p|\psi\rangle~=~Tr~\hat p~=~0$.
A: There is no analogous statement in quantum mechanics because the l.h.s. does not make sense. "Force" should be the (expectation value of the) operator $F := \frac{\mathrm{d}}{\mathrm{d}t}p$, but this doesn't depends on any "position" against which we could integrate. Ehrenfest's theorem tells us that
$$ \langle F\rangle_\psi = -\left\langle\frac{\partial V}{\partial x}\right\rangle_\psi$$
but taking the expectation value also eliminates the position dependence of the r.h.s. You cannot speak about the integral of force against displacement in quantum mechanics - heuristically because "displacement" is ill-defined to begin with. Quantum mechanical work must simply be defined by the change in energy, i.e. the r.h.s.
However, consider that at least for conservative forces, your "work-energy theorem" is really nothing but energy conservation since then the l.h.s. becomes the change in potential energy. And $\Delta V = \Delta E$ for $\Delta E$ the difference between the expectation values of kinetic energies and $\Delta V$ the different between the expectation values of the potential energies is again true by Ehrenfest's theorem.
A: As at @ACuriousMind points out, in a quantum mechanical context it is generally impossible to integrate a "force" against a "displacement" since the latter cannot be defined.
And yet there exists a class of situations that come as close as possible to admitting a "quantum work-energy theorem": systems whose time-independent Hamiltonian $H(\{\vec{R}_\alpha\})$ and energy spectrum $E(\{\vec{R}_\alpha\})$ depend on a set of parameters $\{\vec{R}_\alpha\}$ that happen to represent position vectors. 
Prime examples: electronic clamped nucleus Hamiltonians in the Born-Oppenheimer approximation for molecular systems, where the $\{\vec{R}_\alpha\}$ are simply the clamped positions of the atomic nuclei.
For such systems the Hellmann-Feynman theorem relates the force on nucleus $\sigma$, in the combined field of all electrons, to the variation of the electronic energy $E(\{\vec{R}_\alpha\})$ with the displacement of nucleus $\sigma$. That is, 
$$
\vec{F}_\sigma(\{\vec{R}_\alpha\}) = - \nabla_{\vec{R}_\sigma}E(\{\vec{R}_\alpha\}) = \langle \Psi(\{\vec{R}_\alpha\}) \;|\; -\nabla_{\vec{R}_\sigma}H(\{\vec{R}_\alpha\}) \;|\;\Psi(\{\vec{R}_\alpha\}) \rangle
$$
where $\Psi(\{\vec{R}_\alpha\})$ is the electronic eigenfunction as parametrized by the clamped positions $\{\vec{R}_\alpha\}$ of the nuclei.
In this case integrating the force $\vec{F}_\sigma(\{\vec{R}_\alpha\})$ along (parametric) displacements of the nuclei does make sense, and the resulting "work-energy theorem" relates the "work done to displace the nuclei in the field of the electrons" to the (adiabatic) change in electronic energy that such a displacement induces.
But you can also try your hand on a much simpler textbook example: Take a particle in the ground state of an infinite 1-D box of width $L$. If one of the walls were allowed to slide outwards for a small distance $\Delta L$, what would be the work done by the particle while displacing the wall?   
