# Phase on Hilbert Spaces

Why, in somes cases, such as spin 1/2 systems like in the Stern-Gerlach experiment, the representation of some kets has a term with an complex exponential? For example: $\vert S_{x}; + \rangle = \dfrac{1}{\sqrt{2}} \vert + \rangle + \dfrac{1}{\sqrt{2}} \exp [i \delta] \vert - \rangle$

Your notation is misleading. Certainly, the eigenvector of the $S_x$ operator:

$$S_x = \frac{\hbar}{2}\begin{pmatrix}0 & 1 \cr 1& 0\end{pmatrix}$$

is:

$$\mid S_x,+\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\cr 1\end{pmatrix}$$

With an arbitrary choice of global phase. Ok, so this is what I guess you meant by $\mid S_x,+\rangle$. Now, the most general state for this system is given by two numbers (the two original complex numbers need four real numbers, but one you remove by the normalization requirement and the second by the freedom of choosing a global phase). These are usually represented by the Bloch sphere. Your state is just a set of states in a circle on this sphere, but not necessarily an eigenvector of $S_x$.

Edit (01/27/2010): In view of the claim that I'm not answering the question, let's try to fully understand the origin of this phase. Let's say you want to measure the spin in a direction given by the unit vector:

$$\vec{n} = (\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)$$

It is a common exercise in QM to show that the eigenvalues of $\vec{S}\cdot\vec{n}$ are $\pm 1$ and that the eigenvector associated to the eigenvalue $+1$ is:

$$\mid\vec{S}\cdot\vec{n},+\rangle = \cos(\theta /2)\mid +\rangle + \sin(\theta /2)e^{i\varphi}\mid -\rangle$$

where the basis are eigenvectors of $S_z$ and with the state written up to an arbitrary global phase, as I explained before. If you have have trouble finding this answer, look it up in Sakurai's book, chapter 3. He derives this result over and over again. Therefore, by looking at the state in the question, we immediately identify $\theta=\pi/2$ and $\varphi$ as the phase you left undetermined $\varphi = \delta$.

So, this state is a spin 1/2 system which, if you measure $\cos\varphi\; S_x + \sin\varphi\; S_y$ (meaning, if you align your Stern-Gerlach device in the direction $\vec{n}=(\cos\varphi,\sin\varphi,0)$), you will get always the result $+1$.

• Dear Rafael, you could have identified the state. It's an eigenstate with respect to an axis in the $xy$ plane that is separated from $x$ by angle $\delta$... Moreover, I don't think you're answering the question. – Luboš Motl Jan 27 '11 at 13:20
• well, I "stole" the notation that I used from the JJ Sakurai "Modern Quantum Mechanics" Revised Edition, page 26. What I can't understand is: sometimes that exponential term don't appear, so could I choose arbitrarily $\delta = 0$ in this cases? – Rodrigo Thomas Jan 27 '11 at 14:48
• Oh, I see. The question there is that he hasn't yet determined which directon is the $x$ direction. He started with an experimental observation that if you have particles polarized perpendicularly to your device, the beam splits in two with equal "intensity" (probability). Let's, as Sakurai did, call the direction of the SG device the $z$ direction. Saying that is perpendicular means only that your are in a plane. This phase is exactly the freedom of choosing the $x$ direction any direction in this plane. This more geometric derivation is in chapter 3, keep on reading. – Rafael Jan 27 '11 at 15:02
• By the way, in this part where you are reading, he assumes yet another perpendicular direction to both the SG device and the previous polarization (call it the $y$ direction :) ). And he shows that the coefficients have to necessarily be complex, more or less the aspect of your question that Lubos explored in his answer. – Rafael Jan 27 '11 at 15:04

Dear Rodrigo, all amplitudes - the coefficients in front of the states such as $|+\rangle$ and $|-\rangle$ you mentioned - are complex numbers in quantum mechanics. See e.g.

http://motls.blogspot.com/2010/08/why-complex-numbers-are-fundamental-in.html

Why complex numbers are fundamental in physics

In particular, coefficients $1$ and $\exp(i\delta)$ for any real $\delta$ are physically equivalent if one places them in front of the whole wave function. It's because all predicted probabilities only depend on $\psi(x)\psi^*(y)$ where the phase cancels: note that $\exp(i\delta)^* = \exp(-i\delta)$ which is also the inverse to $\exp(i\delta)$.

The imaginary unit is necessary because the commutator of $[x,p]$ has to be $i\hbar$, to guarantee the uncertainty principle - one may show that the commutator of two Hermitian ("real") operators is antihermitian ("pure imaginary"). Also, Schrödinger's equation, Heisenberg equations, and Feynman path integrals have to contain $i$ in their basic form, too. That's because wave functions have to oscillate in the complex plane.

The relative phases between two basis vectors are very important and physical, however. They may be measured. It's because we may measure the probability that a given state vector has another property. That's given by the square of the absolute value of the inner product of this state vector with a particular eigenvector of the property. All phases influence the result because they influence the inner product.

The relative phases between component waves coming from the two slits in a double-slit experiment decide about the location of the interference maxima and minima. Phases decide about many other important things. Wave functions with energy $E$ depend as $\exp(i\delta)$ on time where $\delta = Et/i\hbar$; the dependence on space is $\exp(i\delta)$ where $\delta = px/\hbar$ where $p$ is the momentum. And I could go on and on and on (angular momentum $j_z$ and many other things).

The particular vector you wrote is the eigenstate of a component of the spin; it is the "up" eigenstate with respect to an axis in the $xy$ plane whose relative angle from the axis $x$ is $\delta$. It can be seen by writing the vector $(1,1)/\sqrt{2}$ which is the "up" eigenstate with respect to the $x$-axis. Then you may rotate this vector around the axis $z$ by angle $\delta$. This changes the two components differently because the $+$ and $-$ states have the opposite $j_z$ angular momentum. You get $(\exp(-i\delta/2),\exp(+i\delta/2))$ where the factor $1/2$ comes from the fact that we deal with a spin-1/2 system. After you multiply this vector by an overall phase $\exp(i\delta/2)$ that is physically irrelevant, as explained at the beginning, we obtain the vector you wrote. You should have perhaps used $x'$ in the description of the vector because it's not the normal $x$-axis but an axis whose direction depends on $\delta$.

You'll know the Pauli matrices $\sigma_1=\left(\begin{matrix}0 & 1\\1 & 0\end{matrix}\right)$, $\sigma_2=\left(\begin{matrix}0 & -i\\i & 0\end{matrix}\right)$, and $\sigma_3=\left(\begin{matrix}1 & 0\\0 & -1\end{matrix}\right)$. Take a spinor such as your $v=\frac{1}{\sqrt{2}}\left(\matrix{1 \\e^{i\delta}}\right)$, taking your basis vectors $\left|+\right>,\left|-\right>$ to be $\left(\matrix{1 \\0}\right)$ and $\left(\matrix{0 \\1}\right)$, and construct the 3-vector $(v^\dagger\!\sigma_1\! v, v^\dagger\!\sigma_2\! v,v^\dagger\!\sigma_3\! v)$, taking both the transpose of $v$ and the complex conjugate, which is called the Hermitian conjugate; you'll find that it's $(\cos(\delta),\sin(\delta),0)$. This is only saying what Rafael and Lubos already said.

• Update: There's a significant addition to this, pushing a little beyond the level of your question, which is that as well as the 3-vector $(v^\dagger\!\sigma_1\! v, v^\dagger\!\sigma_2\! v,v^\dagger\!\sigma_3\! v)$, we can also construct the null 4-vector $(v^\dagger v, v^\dagger\!\sigma_1\! v, v^\dagger\!\sigma_2\! v,v^\dagger\!\sigma_3\! v)$, where $(v^\dagger v)^2$ is the same as the sum of the squares of the other three components. This is important when we come to relativistic systems. The relative phase you asked about still corresponds to a rotation of the 3-vector part of this null 4-vector, but now there's also a possibility of Lorentz boosts introducing a relative phase, as well as rotations.

If you change the complex phase $\delta$, the new spinor corresponds to a different 3-vector, rotated through the difference between the angles. If you consider a spinor such as $\left(\matrix{\cos(\alpha) \\ \sin(\alpha)e^{i\delta}}\right)$ you get a vector somewhere between the $x-y$ plane and the $z$ axis. This turns out to be very useful and significantly different from other ways of constructing vectors. On the other hand, if you just multiply the spinor by $e^{i\alpha}$, you'll get the same 3-vector. In Physics we just throw away the overall complex phase of a spinor, but in Mathematics we can consider something like the real and complex parts of the complex 3-vector $(v^T\sigma_2\sigma_i v)$, which turn out to be two more 3-vectors that are always orthogonal to $(v^\dagger\sigma_i v)$ and the same length. Nice! The mapping $v^\dagger\mapsto v^T\sigma_2$ is known as the "charge conjugate" in Physics, $v\mapsto v^c$; it looks a little screwy in this notation, but it can be constructed more naturally.

The interpretation of spinors is a minefield, however. A spinor can as well represent rotations, or a triad of vectors, as a single vector, and for Dirac spinors there has at times been a whole industry devoted to trying to figure out ways to think about them, largely because there is not as simple a relationship between Dirac spinors and 4-vectors and rotations in Minkowski space. Perhaps best to think of them as just happening to be the right number of degrees of freedom, and being a representation of the Lorentz group in a useful way, when we're describing experimental results.