# Work definition does not make any sense for coherent states? TPM (two projective measurement) scheme

I'm talking about the relation that if we are in a isolated, quantum system that only allows for work exchange with the surrounding system we know from the first law of thermodynamics that $$\Delta E = W$$ where $$E$$ is the systems internal energy.

There is the so called TPM (two projective measurement) scheme in which one considers an initially equilibrium system $$\rho$$. This system gets measured at the beginning giving us some initial energy $$E_i = Tr(\rho H)$$ after which it will get evolved according to some unitary $$U$$. Now after the evolution another measurement is taken of the system $$\rho_f$$ which gives us the final energy $$E_f = Tr(\rho_f H)$$

We know (can be proven) that as long as the state $$\rho$$ has been classical (only diagonal entries) the relation of the first law of thermodynamic holds for $$\Delta E = E_i - E_f = W$$ where W is the work that we used to drive the system with the unitary $$U$$, holds.

But apparently this is not the case if $$\rho$$ shows coherence (has off-diag.) entries.

But the problem is I calculated $$E_i$$ and $$E_f$$ for such a coherent $$\rho$$ but it seems as if these off diagonal terms have no effect on the systems energy.

Bonus: it is implied that the first measurement destroys any coherence in the system making it impossible to calc. $$W$$ for systems that carry coherence.

• "first measurement destroys any coherence" - yes that is what measurement does by interaction with macroscopic measurement device. I can expand on that if you say which bit you don't understand. – Bruce Greetham Dec 16 '18 at 17:41
• >"In such open systems the Schrodinger evolution mixes energy eigenstates (think how atomic energy level transitions can occur). In this case coherences will make a difference." Could you maybe refer me to an exact (simple) example that showcases this exact behavior? – Benjamin Jabl Dec 16 '18 at 21:47

I will calculate throughout in the energy basis of the free Hamiltonian of a 2 state system.

Notation : $$H_0 \lvert 1 \rangle = \hbar \omega_1 \lvert 1 \rangle , H_0 \lvert 2 \rangle = \hbar \omega_2 \lvert 2 \rangle , \omega = \omega_2 - \omega_1, H_0 = \begin{bmatrix} \hbar \omega_1 & 0 \\ 0 & \hbar \omega_2\end{bmatrix}$$

Case 1: Free Time Evolution Here I will show there is no mixing of the energy eigenstates. As a result the average energy is constant and coherence terms in the density matrix have no effect: you can't distinguish between a pure state and a mixture.

$$\lvert \psi (0) \rangle = c_1 \lvert 1 \rangle + c_2 \lvert 2 \rangle$$

$$\rho (0) = \begin{bmatrix} p & \alpha \\ \alpha^* & 1-p \end{bmatrix}$$

Do time evolution under free Schrodinger equation:

$$\lvert \psi (t) \rangle = c_1 e^{i \omega_1 t} \lvert 1 \rangle + c_2 e^{i \omega_2 t} \lvert 2 \rangle$$

$$\rho (t) = \begin{bmatrix} p & \alpha e^{-i \omega t} \\ \alpha^* e^{i \omega t} & 1-p \end{bmatrix}$$

So $$\langle H_0 (t)\rangle = Tr ( \rho(t) H_0 ) = p \hbar \omega_1 + (1 - p) \hbar \omega_2$$ , a constant independent of the coherence term $$\alpha$$.

Case 2: Time Dependent Perturbation causing Rabi Oscillation Here I will show that if a unitary transformation is applied to the two state system so as to mix the free energy eigenstates then the coherence terms in the density matrix become important. I will defer the derivation of the Schrodinger evolution but note for now that a suitable choice of time-dependent interaction Hamiltonian between measurements can apply a unitary transformation to the state vector. Let the Rabi oscillation frequency be $$\Omega$$.

$$\psi (t) \rangle =\cos {(\Omega t)} e^{i \omega_1 t} \lvert 1 \rangle + \sin {(\Omega t)} e^{i \omega_2 t} \lvert 2 \rangle$$

$$\rho (t) = \begin{bmatrix} \cos^2(\Omega t) & \sin (\Omega t) \cos (\Omega t) e^{-i \omega t} \\ \sin (\Omega t) \cos (\Omega t) e^{i \omega t} & \sin^2(\Omega t) \end{bmatrix}$$

So $$\langle H_0 (t)\rangle = Tr ( \rho(t) H_0 ) = \cos^2(\Omega t) \hbar \omega_1 + \sin^2(\Omega t) \hbar \omega_2$$

Note the average energy calculation still does not explicitly depend on the coherence terms. However the coherence terms play a role in the time evolution. This means the result will depend on whether a measurement is performed midway through the evolution as we will now demonstrate.

We assume initial state is pure ground state.

$$\rho (0) = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$$

Time evolve through 45-degree oscillation:

$$\rho (45^o) = \begin{bmatrix} \frac {1}{2} & \frac{1}{2} e^{-i \omega t} \\ \frac{1}{2} e^{i \omega t} & \frac{1}{2}\end{bmatrix}$$

Case 2a. Repeat 45-degree pulse without intermediate measurement

$$\rho (90^o) = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$$

Average energy = $$\hbar \omega_2$$

Case 2b. Repeat 45-degree pulse after intermediate measurement

The energy measurement after the first pulse results in decoherence in the energy basis resulting in the pure state being replaced by a mixed state (following Born's rule). We can discuss this step in more detail later.

$$\rho (45^o) = \begin{bmatrix} \frac {1}{2} & 0 \\ 0 & \frac{1}{2}\end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix} + \frac{1}{2} \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$$

$$\rho (90^o) = \frac{1}{2} \begin{bmatrix} \frac {1}{2} & \frac{1}{2} e^{-i \omega t} \\ \frac{1}{2} e^{i \omega t} & \frac{1}{2}\end{bmatrix} + \frac{1}{2} \begin{bmatrix} \frac {1}{2} & \frac{1}{2} e^{-i \omega t} \\ \frac{1}{2} e^{i \omega t} & \frac{1}{2}\end{bmatrix} = \begin{bmatrix} \frac {1}{2} & \frac{1}{2} e^{-i \omega t} \\ \frac{1}{2} e^{i \omega t} & \frac{1}{2}\end{bmatrix}$$

So average energy = $$\frac {1}{2} (\hbar \omega_1 + \hbar \omega_2 )$$.

• 1. Aha! That was so helpful thank you very very much! Especially for all the effort you've put in I really do appreciate it. What I basically learned is that after we've measured a coherent state the evolution afterwards will behave very differently to the evolution before. The measurement destroys coherence and thus changes the dynamics of the whole evolution. Thank you so much ! Finally I understand this issue. One minor question. They way I've learned this that going from one state $\mid 1 \rangle$ to another state $\mid 2 \rangle$ is usually done via unitary transformations. – Benjamin Jabl Dec 27 '18 at 14:56
• 2. Yet you managed to tackle this whole protocl without ever mentioning unitaries. Are they somewhere implied or are we not using them at all? You know $\rho^f \rightarrow U \rho U^{\dagger}$ is what I am talking about. It should be similar to your example but I can't seem to find the connection. Also if you have any time left I have another quite tricky question which hasn't been answered yet. I you feel like it I'd welcome any help. physics.stackexchange.com/questions/450442/… – Benjamin Jabl Dec 27 '18 at 14:57
• Yes they are related - this took me a long time to piece together so happy to share it. I may add details to answer later. Here is key points: 1. a closed unmeasured system changes by unitary transformation (ensuring conservation of probabilities). A pure state changes $\psi \to U\psi$ which gives rise to $\rho \to U \rho U^{\dagger}$ ... – Bruce Greetham Dec 27 '18 at 15:06
• ... but then think more fundamentally what causes that unitary transformation : it must be the dynamics of the system. So that is what Schodingers equation is: describing the unitary state evolution due to time translation by H. – Bruce Greetham Dec 27 '18 at 15:10
• Now to actually implement an interesting unitary change for a two level system like at electron in an atom (ground + one excited state) requires studying quite a lot of quantum optics! If an atom is truly isolated then each energy eigenstate just evolves on their own - there is no transition between energy levels. – Bruce Greetham Dec 27 '18 at 15:15

As an interesting example consider Rabi oscillation.Reference Sakurai, Modern Quantum Mechanics p.320.

We have a two state system with two energy levels $$E_1$$ and $$E_2$$.

$$H_0 = E_1 \lvert 1 \rangle \langle 1 \rvert + E_2 \lvert 2 \rangle \langle 2 \rvert$$

We apply a time dependent perturbation at the resonant frequency $$\omega = (E_2 -E_1)/\hbar$$

$$V(t) = e^{i \omega t} \lvert 1 \rangle \langle 2 \rvert + e^{-i \omega t} \lvert 2 \rangle \langle 1 \rvert$$

It can be shown that a pure state oscillates between the two energy eigenstates at frequency $$\omega/2$$.

So the mean energy of the system oscillates between $$E_1$$ and $$E_2$$.

$$\langle E \rangle = E_1 \sin^2 \frac{\omega t}{2} + E_2 \cos^2 \frac{\omega t}{2}$$

Note at $$45 ^o$$ through a cycle the pure state will be in a equal coherent superposition of the two energy eigenstates.

Now suppose at this point an energy measurement is performed and the measurement result discarded. Now the state should be described as a equal statistical mixture of each of the two eigenstates.

Evolving the mixture from the point of measurement each eigenstate independently performs Rabi oscillation, but out of phase with each other by $$90^o$$.

So the average energy of the mixed state is

$$\langle E \rangle = \frac{1}{2} ( E_1 \sin^2 \frac{\omega t}{2} + E_2 \cos^2 \frac{\omega t}{2} )+ \frac{1}{2} ( E_1 \cos^2 \frac{\omega t}{2} + E_2 \sin^2 \frac{\omega t}{2} )$$

$$= \frac{1}{2} (E_1 + E_2)$$

So we see in the pure state the perturbation is causing a transfer of energy to and from the system (absorbtion and emission). In the mixed state there is no net energy transfer.

• 1. Thank you very much ! I understood parts of it and it was immensely helpful. But I tried to re-do your calculation step by step and sadly was not able to follow it 100%. 1. Let's say we have a 2D system that starts in the pure groundstate $\mid 1 \rangle$. How does your unitary turn this pure state into the 'correct' superposition which should be... $\mid \psi \rangle = sin(\omega t/2)\mid 1 \rangle + cos(\omega t/2)\mid 2\rangle$ ? Cont. – Benjamin Jabl Dec 25 '18 at 19:03
• 2. Assuming this is the correct superposition for the pure state the corresponding density matrix should go like this \begin{equation} \begin{split} \rho = sin^2(\omega t/2)\mid 1 \rangle \langle 1 \mid + sin(\omega t/2)cos(\omega t/2)\mid 1 \rangle \langle 2 \mid + sin(\omega t/2)cos(\omega t/2)\mid 2 \rangle \langle 1 \mid + cos^2(\omega t/2)\mid 2 \rangle \langle 2 \mid \end{split} \end{equation} Now I'm quite unsure how this superposition density matrix turns into the mixed density matrix you've proposed. Could you maybe explain a little bit more? Thank you sooo much ! – Benjamin Jabl Dec 25 '18 at 19:08
• BTW, I would like to correct my comment earlier about "open system". The example I have given is not an open system, but it is influenced by a time-dependent Hamiltonian. The point I was trying to make was that a time-independent Hamiltonian does not work. We can return to this point when we've done the explicit calculation. – Bruce Greetham Dec 25 '18 at 20:10
• Ah I see. Ok I look forward to it (: – Benjamin Jabl Dec 25 '18 at 20:19
• I have restructured in a separate answer. Later I will address your questions as notes to the new answer. Please check calculation so far. – Bruce Greetham Dec 26 '18 at 18:30