Nuclear magnetic resonance (NMR) spectroscopy uses the intrinsic magnetic moment that arises from the spin angular momentum of a spin-active nucleus.[1] If the element of interest has a nuclear spin that is not zero,[1] the nucleus may exist in different spin angular momentum states, where the energy of these states can be affected by an external magnetic field. For a spin I = ½ nucleus two energy levels may be considered: spin up and spin down, depending on how the spin aligns with the external magnetic field.[2] It is important to remember that, in the presence of an external magnetic field, individual nuclei may have random orientations other than up and down. However, the sample's bulk magnetization, that is, the sum of the total magnetic moments will determine the strength of the NMR signal.[3] In addition, the energy of the applied radio frequency used in NMR must be consistent with the energy difference between the spin states.[3]
The Hamiltonian operator represents the energy operator. The Spin Hamiltonian for a nuclear spin under an applied magnetic field (B0) is,[3]
Ĥone spin = -γB0ÎZ
Where γ is the gyro-magnetic ratio and ÎZ is the z-component of the nuclear spin angular momentum.
The energy of the nuclear spin level is given by this Hamiltonian operator, since we know the eigenvalue for ψ. We will first determine the energy of states and subsequently convert it to frequency units since in NMR, energy expressed in frequency units is more common.
The equation of the Hamiltonian contains an angular momentum operator. So it will be easy if we find the eigenvalues of the angular momentum operator first and then substitute it into the Hamiltonian. For a spin half nucleus there are two eigenfunctions for ÎZ.[3]
Let m = +1/2 and m = -1/2 and eigenfunctions are,
ÎZ ψm = mħψm
Applying the equation of nuclear spin angular momentum (ÎZ ψm) to one spin Hamiltonian (Ĥone spin) will give,[3]
Ĥone spin ψm = -mħγB0ψm
From this, eigenvalue is,
Em = -mħγB0
In frequency units,
Em = -mγB0/2πHz
Introducing Larmor frequency (v0), Em = mv0Hz
Hence the Hamiltonian in frequency units, Ĥone spin = v0ÎZ
If there are two spin states, then we have to change the Hamiltonian in such a way that it accommodates both the spin states.[3]
Ĥtwo spins, no coupling = v0,1Î1Z + v0,2Î2Z
v0,1 is the Larmor frequency of first spin and v0,2 is the Larmor frequency of second spin. Similarly Î1Z is the z-component of angular momentum operator of first spin and Î2Z is the z-component of angular momentum operator of first spin. Here in this case coupling is not considered. Here while considering the wave function we have to look into both spin states of both spin 1 and 2. The spin up state is represented by α and spin down is β. The wave functions hence will have four combinations as below. ψα,1 ψα,2 = αα ψα,1 ψβ,2 = αβ ψβ,1 ψα,2 = βα ψβ,1 ψβ,2 = ββ Applying these combinations into the two spin Hamiltonian above will give the eigenvalue which is the energy state. This is tabulated below.
| Spin states | Eigenfunction | Eigenvalue (energy) |
| αα | ψα,1 ψα,2 | +(1/2)v0,1 + (1/2)v0,2 |
| αβ | ψα,1 ψβ,2 | +(1/2)v0,1 - (1/2)v0,2 |
| βα | ψβ,1 ψα,2 | -(1/2)v0,1 + (1/2)v0,2 |
| ββ | ψβ,1 ψβ,2 | -(1/2)v0,1 - (1/2)v0,2 |
In general, the energy level (eigenvalue) can be written as;
Em = m1v0,1 + m2v0,2
To consider coupling of spin 1 and 2 a coupling constant (J) and corresponding coupling term is introduced in the Hamiltonian:[3]
Ĥtwo spins = v0,1Î1Z + v0,2Î2Z + J12Î1ZÎ2Z
Applying the wave functions in this Hamiltonian gives the eigenvalues as tabulated below.
| Number | Spin states | Eigenfunction | Eigenvalue (energy) |
| 1 | αα | ψα,1 ψα,2 | +(1/2)v0,1 + (1/2)v0,2 + (1/4)J12 |
| 2 | αβ | ψα,1 ψβ,2 | +(1/2)v0,1 - (1/2)v0,2 - (1/4)J12 |
| 3 | βα | ψβ,1 ψα,2 | -(1/2)v0,1 + (1/2)v0,2 - (1/4)J12 |
| 4 | ββ | ψβ,1 ψβ,2 | -(1/2)v0,1 - (1/2)v0,2 + (1/4)J12 |
When two spins couple each other, the Hamiltonian operator will be,[3]
Ĥtwo spins = v0,1Î1Z + v0,2Î2Z + J12Î1ZÎ2Z
The eigenvalue,
Em = m1v0,1 + m2v0,2 + m1m2J12
The selection rule for allowed transition is + or -1.[1] Here we are considering homonuclear protons. Thus their αβ and βα states will have the same energy. The transition energy can be calculated by reducing the energy (eigenvalue) of the upper state from the lower state. The transition energy in frequency units is tabulated below.
| Transitions | Spin states | Frequency |
| 1 to 2 | αα to αβ | -v0,2 - (1/2)J12 |
| 3 to 4 | βα to ββ | -v0,2 + (1/2)J12 |
| 1 to 3 | αα to βα | -v0,1 - (1/2)J12 |
| 2 to 4 | αβ to ββ | -v0,1 + (1/2)J12 |
The transitions given in the above table is represented in the figure below.
