In this video, we'll discuss rotation of spin one-half system. The lowest dimensions where you can actually perform rotation operation is two-dimensional space and spin one-half system is described by a two-dimensional vector space spanned by the two independent orthogonal basis kets spin up and spin down. If we choose the eigenkets of the z component of spin S of z operator, so z plus and z minus corresponding to the spin up and spin down states along the z direction. Then you can express these x, y, z component of spin operators as shown here. If you represent them in a matrix form, then you can get this. These are all the results that we have already discussed in previous lectures. Now, using either this expression or this matrix representation, you can show that these spin operators x, y, z component of spins do satisfy these commutation relation that we have derived for the general angular momentum vector. Now, because spin operators satisfy this commutation relation, we can describe rotational motion of kets using these spin operators. They do generate rotational motion, in other words. Let's consider the rotation about z-axis by an angle phi. This operator D of R of z phi represents a rotational operation on about Zx by angle phi acting on a ket alpha. We're going to denote the rotated ket by a subscript r here. Now this operator of course is defined by the usual exponential function of these angular momentum operator S. Because we're talking about the rotation about z-axis, we just simply select the z component of the spin angular momentum here. Now, consider the expectation value of the x component of spin. Use the rotated ket and rotated bra to evaluate the expectation value of s of x from the definition of these rotated ket denoted by subscript r, we get this. The operator here that we are concerned with is the x component of spin operator Sx transformed by the similarity transformation with these rotational operators. We need to evaluate this rotated operator of S of x. Now, we can do that by just substituting these exponential expression for these rotation operators. This here is the dagger, so Hermitian adjoint. We take the complex conjugate and then S of z operator is Hermitian so it doesn't change. Then on this side we retain the negative sign here. Express S of x in terms of the basis kets of S of z from the equations given in the first slide. Then we can act these S of z operator on these ket z plus and z minus. Because z plus and z minus are the eigenkets of S of z, that simply results in the same ket with its Eigenvalue multiply to it plus H-bar over 2 for z plus minus H-bar over 2 for z minus. Then same thing, this operator here will act on a bra from the right. Once again, z plus and z minus are the eigenbras of these Sz operator. We get either plus H-bar omega 2 or negative H-bar omega 2 as their eigenvalues. If you do that operation, we take out these H-bar over 2 out in front and you're left with this. Here you multiply these two exponential terms and then you use these expression for S of x and S of y given in the first slide. Then you can rearrange these terms in a compact form as this. You can see that the expectation value of the x component of spin after rotation about z axis by angle Phi is rotated by this, which is what you would expect for a rotation of a normal vector in three-dimensional space about z axis by an angle Phi. Similarly, you can show that the z expectation value doesn't change and y expectation value is rotated by this angle once again, by angle Phi. The results of these expectation values shows that the expectation value gets rotated just like a regular vector in three-dimensional space. This is true for any angular momentum operator, J. If you're dealing with a test space that is a larger dimension, then you're going to have to deal with these larger dimensional ket space. But the angular momentum vector, the three components of the angular momentum vector, J_x, J_y, and J_z will rotate in exactly the same way as shown here for the spin angular momentum vector. Now, we consider the rotation of an arbitrary ket Alpha, and we express this ket Alpha in terms of z plus and z minus. The eigenkets of S_z operator. The coefficients are given by these inner product here. As usual, now apply the rotation operator. Rotation about z axis by angle Phi. Then once again, because z plus and z minus are the eigenkets of S_z operator, we get this. We notice that if we rotate by 2Pi, 360 degrees about z axis, it doesn't come back to itself. It comes back to itself with a negative sign. That is because in this exponential term here, we get Phi divided by 2 instead of just Phi. So if you substitute 2Pi for angle Phi here, you get Pi in the exponential factor here, which gives you this negative sign. After rotation by 360 degree by 2Pi angle, these ket specifying a spin state doesn't come back to itself. It comes back to itself with a negative sign. You have to rotate by 2Pi. You have to make two complete turns to bring the ket to the same ket as the original ket, which seems unphysical. Now, this minus sign here, after rotation by two Pi, disappears when calculating the expectation value. Why? Because there are two exponential factors multiplied together as shown in the previous slide. They disappear when you actually evaluate the expectation value. But when you're dealing with just ket itself, this negative sign persist. Now the question is, does this negative sign manifests itself in real experiments as on observable? Now, to answer that question, we consider the spin precession problem that we have dealt with before. See the Hamiltonian for a spin in a magnetic field is given by this dot-product. So if your B field is along the z direction, so this simply is given by this. Omega here is defined as these constants proportional to the magnitude or amplitude of your applied magnetic field. Now, time evolution operator here in this case, which is defined as the exponential function of the Hamiltonian operator, turns out to be this which contains the spin angular momentum operator, simply because your H, Hamiltonian is linearly proportional to your S_z operator. Time evolution operator simply is a rotation operator by an angle Omega t. Time evolution, in this case, is equivalent to rotation and as we have seen before, actually the spin vector rotates about z-axis with an angular frequency Omega and this is a direct consequence that your Hamiltonian operator is linearly proportional to the angular momentum operator that generates rotation. Now to be precise the expectation value of S of x and S of y rotate with the omega angular frequency Omega, as we would normally expect. However, the state kets rotate with angular frequency Omega divided by 2 and this can be shown by expressing an arbitrary ket Alpha in terms of z plus and z minus the basis kets, the eigenstates of SZ operator and these are the constants. The rotation results in these two terms, exponential terms, which these 1 over 2 factor of 1 over 2 comes from the fact that the eigenvalue of z plus and z minus are H bar over 2 plus n minus H bar over 2 which is the direct result that the spin state has a spin of one-half. We're dealing with a spin one-half system. You can see that these ket Alpha rotate with an angular frequency Omega over 2 instead of Omega. Now can we see this actually, so we consider a neutron interferometry experiment where you send in a neutron beam and you split into two and these neutrons going through path 1, simply pass through and go through another updates here or some sort of a lens system and bring these neutrons back to a detector but the other neutrons sent to path 2 are subject to a constant B field. The neutrons going through this path 2 are subject to a motion that's similar to the spin precession of electrons that we just discussed. Neutron is also a spin one-half system so the angular frequency here is defined exactly the same way as the electron case, except that now we have to use the neutron mass instead of the electron mass and this g factor here should be chosen to be an appropriate g factor for neutron that may be different from the electron case. The amplitude of neutrons arriving via path 2 is the amplitudes for the case of no B field times this exponential factor representing the rotation of spin. Of course, the neutrons arriving via path 2 the amplitude should be independent of B field because they are not subject to the magnetic field. Now, if you combine them and observe the intensity due to the interference of the state kets due to the interference of the wave functions if you will, although the spin state cannot be represented by a spatial function, it's something that we're familiar with so if it is a perfect on analog of the interference patterns in the case of the double-slit experiments but we're doing it in the spin caspase. The interference pattern produces this interference term, which varies with a frequency Omega over 2 instead of Omega. This interference pattern with Omega over 2 is a direct consequence of the rotational properties of spin caspase, which rotate with an angular frequency of Omega over 2 instead of Omega.