I asked a question on Physics Stack Exchange but no one answered the question and I didn't get enough views on it. I am asking it on QCSE because the question is related to experimental quantum computation realized through NMR.

For an ensemble of identical atoms in a superposition state $|\phi\rangle=\alpha|S_z^+\rangle+\beta|S_z^-\rangle$ where $|\alpha|^2$ and $|\beta|^2$ are not equal(although $|\alpha|^2 + |\beta|^2 = 1$) and gives the populations of $|S_z^+\rangle$ and $|S_z^-\rangle$ respectively, we can drive a transition between the two basis states and we call this phenomenon as population transfer(please correct me if I am wrong). Mathematically, the field induced transitions changes the values of $|\alpha|^2$ & $|\beta|^2$.

However, there can be one more phenomenon going on here called polarization transfer which is distinct from population transfer because polarization really counts the total spin magnetization of a state in the context of NMR and EPR and maybe Quantum Optics too.

Now the third phenomenon i.e. coherence transfer(no classical analogue) can't occur between two states but needs three level system say $|1\rangle$, $|2\rangle$ and $|3\rangle$ and the field driven transitions between states $|1\rangle$ & $|2\rangle$ and states $|1\rangle$ & $|3\rangle$ somehow creates transition of states $|2\rangle$ & $|3\rangle$ even though there is no driving field of frequency corresponding to $|2\rangle$ & $|3\rangle$ transition.

The last two phenomena written in bold are what I do not understand. Any insight into them will be very helpful and mathematics over them will be highly appreciated. This link also tries to explain the difference between population transfer, polarization transfer and coherence transfer through density matrices but I cannot see much physical explanation over the phenomenon.


Given a quantum system in a state defined by a density matrix $\rho$, it is an accepted terminology to use the term population for the diagonal matrix elements (not necessarily in the computational basis). Since a normalized vector corresponds to a pure state, thus we can define a population of the pure state $\psi$ by: $$P_{\psi} = \langle \psi | \rho | \psi \rangle$$ According to this definition, the sum of the populations of an orthonormal basis of pure states is a unity.

When the system evolves: $\rho = \rho(t)$, the populations also change in time. We can call a population transfer from a state $a$ to a state $b$ when the system starts with a high population of $a$ and low population of $b$ and ends with low population of $a$ and high population of $b$ with all other populations unchanged.

Usually, when the density matrix corresponds to photon polarization the same situation can be called a polarization transfer. When the density matrix corresponds to a spin system it can be called a spin transfer, and sometimes also a polarization transfer.

Sometimes (for example in spectroscopy) when the system is a tensor product of two subsystems for example two energy levels ($n$ and $n+1$) of a spinning electron. A suitable basis in this case can be chosen as: $\{| n \downarrow \rangle, | n \uparrow \rangle, | n+1 , \downarrow \rangle, | n+1 , \uparrow \rangle \}$, then the term population transfer can be reserved to the energy level reduced density matrix: $$\rho_E = \mathrm{tr}_{\downarrow, \uparrow} \rho$$ While the term spin transfer can be used for the spin reduced density matrix $$\rho_s = \mathrm{tr}_{n, n+1} \rho$$

Population transfers can be also defined for projected density matrices, for example spin transfer of the projected density matrix on the level $n$.

Coherences on the other hand are used to denote the off-diagonal elements of a density matrix $\langle b | \rho | a\rangle$ . The motivation for this definition is that a density matrix with solely diagonal terms describe a classical state and can be replaced by a probability distribution and the quantum correlations are achieved by means of the off-diagonal terms.

Again when the system evolves it can happen that the reduced density matrices of two subsytems switch from high coherence to low coherence and vice versa. This is the case of coherence transfer.

Clearly, when the system is two level, there is only one off-diagonal element $\rho_{12} = \rho_{21}$ and we cannot talk about a coherence transfer.

An example for spin coherence transfer in the above spectroscopy case is given by the following: (Tensor product notation is used where the left matrix lives in the energy level Hilbert space and the right one in the spin space) $$\rho_{\mathrm{init}} = \frac{1}{\sqrt{2}}\left(\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\otimes\begin{bmatrix}\frac{1}{2} & \frac{1}{2} \\\frac{1}{2} & \frac{1}{2} \end{bmatrix} + \begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}\otimes\begin{bmatrix}\frac{1}{2} & 0 \\0 & \frac{1}{2} \end{bmatrix} \right)$$

$$\rho_{\mathrm{fin}} = \frac{1}{\sqrt{2}}\left(\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\otimes\begin{bmatrix}\frac{1}{2} & 0 \\0 & \frac{1}{2} \end{bmatrix} + \begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}\otimes\begin{bmatrix}\frac{1}{2} & \frac{1}{2} \\\frac{1}{2} & \frac{1}{2} \end{bmatrix} \right)$$

Here, one can see a coherence transfer from the energy level $n$ to the energy level $n+1$, because the projected density matrices on the energy levels switched coherences.

Please see the following work by Aubourg and Viennot, where they define populations and coherences of a spin chain along the same lines (equations 5-7).

  • $\begingroup$ Your explanation of coherence transfer is very good. Thanks! But, I have a doubt with the explanation of polarization transfer($PT$). The gyromagnetic ratio of carbon($C$) is less then hydrogen($H$). If the NMR instrument is not sensitive enough to detect $C$ in NMR spectrum(we see peak buried in noise) and sensitive enough to detect $H$ then people suggest to transfer the polarization of $H$ to $C$ in that way peak of $C$ is seen clearly in the spectrum. I can relate your explanation to the case of $C$ and $H$ but do not understand that how it works in NMR. Can you add more details on $PT$. $\endgroup$ – Jitendra Sep 4 '18 at 12:40
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    $\begingroup$ In this case, you can define an atom type Hilbert space with the basis $|C\rangle$ and $|H\rangle$. The reduced density matrix in this Hilbert space must be diagonal because different atoms belong to different superselection sectors and cannot be mixed. Apart from this limitation you can repeat the above exercise with the energy level Hilbert space replaced by the atom type Hilbert space and the spin density matrices swapped from spin-up for $C$ and unpolarized for $H$ to the opposite. $\endgroup$ – David Bar Moshe Sep 4 '18 at 12:41

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