physical chemistry - Why don't equivalent hydrogens cause splitting in NMR?

When doing NMR spectroscopy, it is an observed fact that equivalent hydrogens do not split one another. Why don't equivalent hydrogens split each other's signals? For example, why is the NMR spectrum for ethane a singlet instead of a quartet or even a dodecuplet (due to the hydrogens on the same carbon)? What is so special about hydrogens being equivalent to one another that causes no splitting to be observed?

Answer

I will provide a full quantum mechanical explanation here.^[1] Warning: rather MathJax heavy. Hopefully, this lends some insight into how the diagrams that long and porphyrin posted come about.

Finding the states between which transitions occur

The Hamiltonian for two coupled spins is

$$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z} + \frac{2\pi J_{12}}{\hbar}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{1}$$

where $\omega_1$ and $\omega_2$ are the Larmor frequencies^[2] of the two nuclei and $J_{12}$ is the coupling constant (in Hz) between the two nuclei. (The factor of $2\pi/\hbar$ is simply there to bring it into energy units.) $\hat{I}_{\!i}$ is the operator for the spin angular momentum of nucleus $i$ and $\hat{I}_{\!iz}$ is the operator for its projection along the $z$-axis.

However, in NMR, it is customary to work in frequency units instead of energy units. Since $E = h\nu$, we simply need to divide through by $h$. Bearing in mind that $\omega = 2\pi\nu$ and $h = 2\pi\hbar$, we get:

$$\hat{H}_\text{freq} = \frac{\nu_1}{\hbar}\hat{I}_{\!1z} + \frac{\nu_2}{\hbar}\hat{I}_{\!2z} + \frac{J_{12}}{\hbar^2}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{2}$$

On top of that, to make the maths easier, it is also quite common to set $\hbar = 1$. Therefore, we have

$$\hat{H}_\text{freq} = \nu_1\hat{I}_{\!1z} + \nu_2\hat{I}_{\!2z} + J_{12}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{3}$$

We will deal with two spin-$1/2$ nuclei here, where $|\alpha\rangle$ and $|\beta\rangle$ represent the "up" and "down" spins (recall we set $\hbar = 1$ so it doesn't appear in the eigenvalues):

$$\begin{align} \hat{I}_{\!iz}|\alpha_i\rangle &= \frac{1}{2}|\alpha_i\rangle & \hat{I}_{\!iz}|\beta_i\rangle &= -\frac{1}{2}|\beta_i\rangle & (i = 1,2) \tag{4} \end{align}$$

Furthermore since we are dealing with equivalent nuclei we can simply set $\nu_1 = \nu_2 = \nu$ and drop the subscript in $J_{12}$ just to make it a bit cleaner:

$$\hat{H} = \nu(\hat{I}_{\!1z} + \hat{I}_{\!2z}) + J(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{5}$$

Now, we need to find the eigenstates and eigenvalues of $\hat{H}$. To do so, we will adopt the basis set of product functions $(|\alpha_1\alpha_2\rangle, |\alpha_1\beta_2\rangle, |\beta_1\alpha_2\rangle, |\beta_1\beta_2\rangle)$. From equation $(4)$ we have

$$\begin{align} \hat{I}_{\!1z}|\alpha_1\alpha_2\rangle &= \frac{1}{2}|\alpha_1\alpha_2\rangle & \hat{I}_{\!2z}|\alpha_1\alpha_2\rangle &= \frac{1}{2}|\alpha_1\alpha_2\rangle \tag{6} \\ \hat{I}_{\!1z}|\alpha_1\beta_2\rangle &= \frac{1}{2}|\alpha_1\beta_2\rangle & \hat{I}_{\!2z}|\alpha_1\beta_2\rangle &= -\frac{1}{2}|\alpha_1\beta_2\rangle \tag{7} \\ \hat{I}_{\!1z}|\beta_1\alpha_2\rangle &= -\frac{1}{2}|\beta_1\alpha_2\rangle & \hat{I}_{\!2z}|\beta_1\alpha_2\rangle &= \frac{1}{2}|\beta_1\alpha_2\rangle \tag{8} \\ \hat{I}_{\!1z}|\beta_1\beta_2\rangle &= -\frac{1}{2}|\beta_1\beta_2\rangle & \hat{I}_{\!2z}|\beta_1\beta_2\rangle &= -\frac{1}{2}|\beta_1\beta_2\rangle \tag{9} \end{align}$$

The action of the scalar product $\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}$ is more complicated. We need to introduce the shift operators (or ladder operators).

$$\begin{align} \hat{I}_{\!i+} &= \hat{I}_{\!ix} + \mathrm{i}\hat{I}_{\!iy} & \hat{I}_{\!i-} &= \hat{I}_{\!ix} - \mathrm{i}\hat{I}_{\!iy} \tag{10} \end{align}$$

from which we can obtain

$$\begin{align} \hat{I}_{\!ix} &= \frac{\hat{I}_{\!i+} + \hat{I}_{\!i-}}{2} & \hat{I}_{\!iy} &= \frac{\hat{I}_{\!i+} - \hat{I}_{\!i-}}{2\mathrm{i}} \tag{11} \end{align}$$

So, finally, we can write

$$\begin{align} \hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}} &= \hat{I}_{\!1x}\hat{I}_{\!2x} + \hat{I}_{\!1y}\hat{I}_{\!2y} + \hat{I}_{\!1z}\hat{I}_{\!2z} \tag{12} \\ &= \frac{\hat{I}_{\!1+}\hat{I}_{\!2-} + \hat{I}_{\!1-}\hat{I}_{\!2+}}{2} + \hat{I}_{\!1z}\hat{I}_{\!2z} \tag{13} \end{align}$$

where in going from $(12)$ to $(13)$ one simply substitutes in $(11)$ and does a fair bit of algebraic manipulation. The action of the shift operators are

$$\begin{align} \hat{I}_{\!i+}|\alpha_i\rangle &= 0 & \hat{I}_{\!i+}|\beta_i\rangle &= |\alpha_i\rangle \tag{14} \\ \hat{I}_{\!i-}|\alpha_i\rangle &= |\beta_i\rangle & \hat{I}_{\!i-}|\beta_i\rangle &= 0 \tag{14} \\ \end{align}$$

This allows you to work out the effect of $\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}$ on our basis states. The actual maths is left to the reader and I will simply quote the results:

$$\begin{align} \hat{H}|\alpha_1\alpha_2\rangle &= \left(\nu + \frac{J}{4}\right)|\alpha_1\alpha_2\rangle \tag{15} \\ \hat{H}|\alpha_1\beta_2\rangle &= -\frac{J}{4}|\alpha_1\beta_2\rangle + \frac{J}{2}|\beta_1\alpha_2\rangle \tag{16} \\ \hat{H}|\beta_1\alpha_2\rangle &= \frac{J}{2}|\alpha_1\beta_2\rangle - \frac{J}{4}|\beta_1\alpha_2\rangle \tag{17} \\ \hat{H}|\beta_1\beta_2\rangle &= \left(-\nu + \frac{J}{4}\right)|\beta_1\beta_2\rangle \tag{18} \\ \end{align}$$

Therefore in this basis the Hamiltonian matrix is

$$\mathbf{H} = \begin{pmatrix} \nu + J/4 & 0 & 0 & 0 \\ 0 & -J/4 & J/2 & 0 \\ 0 & J/2 & -J/4 & 0 \\ 0 & 0 & 0 & -\nu + J/4 \end{pmatrix} \tag{19}$$

Finding the eigenvectors and eigenvalues of this matrix is again left to the reader (it is not a difficult task) and they are (eigenvalues denoted $E_i$)

$$\begin{align} |1\rangle &= |\alpha_1\alpha_2\rangle & E_1 &= \nu + \frac{J}{4} \tag{20} \\ |2\rangle &= \frac{1}{\sqrt{2}}(|\alpha_1\beta_2\rangle + |\beta_1\alpha_2\rangle) & E_2 &= \frac{J}{4} \tag{21} \\ |3\rangle &= \frac{1}{\sqrt{2}}(|\alpha_1\beta_2\rangle - |\beta_1\alpha_2\rangle) & E_3 &= -\frac{3J}{4} \tag{22} \\ |4\rangle &= |\beta_1\beta_2\rangle & E_4 &= -\nu + \frac{J}{4} \tag{23} \\ \end{align}$$

The form of the eigenstates should be familiar: they are simply the triplet and singlet states of two spin-$1/2$ particles. These states arise from the coupling of two sources of angular momenta, $I_1$ and $I_2$, to form one overall angular momentum denoted $I$.

$$\vec{I} = \vec{I}_{\!1} + \vec{I}_{\!2} \tag{24}$$

The allowed values of $I$ are determined by the Clebsch-Gordan series:

$$I = I_1 + I_2, I_1 + I_2 - 1, \cdots, |I_1 - I_2| \tag{25}$$

Since $I_1 = I_2 = 1/2$, $I$ can take the values $1$ and $0$. The values of $M_I$, the projection of the total angular momentum along the $z$-axis, are as usual

$$M_I = I, I-1, \cdots, -I \tag{26}$$

so the states with $I = 1$ ("triplet") have $M_I = 1, 0, -1$ and the state with $I = 0$ ("singlet") has $M_I = 0$. One can use more quantum mechanics to work out which state is associated with which quantum numbers, but I will not do it here. They are:

$$\begin{array}{ccc} \hline \text{State} & I & M_I \\ \hline |1\rangle & 1 & 1 \\ |2\rangle & 1 & 0 \\ |3\rangle & 0 & 0 \\ |4\rangle & 1 & -1 \\ \hline \end{array}$$

Selection rules

We have four different states, which leads to ${4\choose 2} = 6$ different possible transitions. However, not all of these transitions are allowed.

The intensity of the transition is proportional to the square of the matrix element $\langle \psi_\mathrm{f} | \hat{H'} | \psi_\mathrm{i} \rangle$ (the so-called "transition dipole moment"), where $\hat{H'}$ is the Hamiltonian for the process that induces the transition. In the case of NMR transitions, the transition arises due to a magnetic field aligned along the $x$-axis.^[3] The corresponding Hamiltonian is therefore

$$\hat{H'} = \omega'(\hat{I}_{\!1x} + \hat{I}_{\!2x}) = \omega'\hat{I}_{\!x} \tag{27}$$

Exactly what $\omega'$ represents is not important here because we are only really concerned about whether the transition dipole moment is zero or not.^[4] Making use of the relations established in equations $(10)$ and $(11)$,^[5] one can find that the selection rules are

$$\Delta I = 0; \Delta M_I = \pm 1 \tag{28}$$

which means that the allowed transitions are $|4\rangle \leftrightarrow |2\rangle$ and $|2\rangle \leftrightarrow |1\rangle$. Transitions to and from the singlet state $|3\rangle$ are forbidden. The energies of the transitions are

$$\begin{align} E_{4\leftrightarrow2} &= \frac{J}{4} - \left(-\nu + \frac{J}{4}\right) = \nu \tag{29} \\ E_{2\leftrightarrow1} &= \left(\nu + \frac{J}{4}\right) - \frac{J}{4} = \nu \tag{30} \\ \end{align}$$

i.e. the two transitions are degenerate and only one line in the spectrum at frequency $\nu$ is observed. This is exactly what is depicted in the diagrams posted in the other answers.

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Notes and references

^[1] I am assuming the reader has some knowledge of the quantum mechanical treatment of angular momentum, which is a topic that is treated thoroughly in most quantum mechanics textbooks. See, for example, chapter 4 of Atkins's Molecular Quantum Mechanics (5th ed.).

^[2] The Larmor frequency is given by $\omega = -\gamma B_0$, where $\gamma$ is the magnetogyric ratio of the nucleus in question and $B_0$ is the strength of the external magnetic field. It represents the frequency with which a magnetic moment precesses about a magnetic field. See any textbook on magnetism for further details.

^[3] I am glossing over some details here. The so-called magnetic field in the $x$-axis is a component of the radiofrequency pulse applied in the $xy$-plane. If you are interested please consult a textbook on the vector model of NMR. In particular I recommend Keeler's Understanding NMR Spectroscopy (2nd ed.).

^[4] It is related to the strength of the magnetic field in the $x$-axis, $B_1$, by $w' = |\gamma|B_1$. The usual symbol is $\omega_1$, but I chose not to use this here to avoid potential confusion. Again, please consult a textbook on the vector model of NMR if you wish to find out more.

^[5] A full proof can be found in J. Chem. Educ. 1982, 59 (10), 819. There is also some discussion of the selection rules in Gunther's NMR Spectroscopy (3rd ed.), p 156 onwards.