# More Quantum Mechanics: Wavefunctions and Operators

In Some Basics of Quantum Mechanics, we explained the role of vector spaces (which we first discussed in Vector Spaces, Modules, and Linear Algebra) in quantum mechanics. Linear transformations, which are functions between vector spaces, would naturally be expected to also play an important role in quantum mechanics. In particular, we would like to focus on the linear transformations from a vector space to itself. In this context, they are also referred to as linear operators.

But first, we explore a little bit more the role of infinite-dimensional vector spaces in quantum mechanics. In Some Basics of Quantum Mechanics, we limited our discussion to “two-state” systems, which are also referred to as “qubits”. We can imagine a system with more “classical states”. For example, consider a row of seats in a movie theater. One can sit in the leftmost chair, the second chair from the left, the third chair from the left, and so on. But if it was a quantum system, one can sit in all chairs simultaneously, at least until one is “measured”, in which case one will be found sitting in one seat only, and the probability of being found in a certain seat is the “absolute square” of the probability amplitude, which is the coefficient of the component of the “state vector” corresponding to that seat.

The number of “classical states” of the system previously discussed is the number of chairs in the row. But if we consider, for example, just “space”, and a system composed of a single particle in this space, and whose classical state is specified by the position of the particle, the number of states of the system is infinite, even if we only consider one dimension. It can be here, there, a meter from here, $0.1$ meters from here, and so on. Even if the particle is constrained on, say, a one meter interval, there is still an infinite number of positions it could be in, since there are an infinite number of numbers between $0$ to $1$. Hence the need for infinite-dimensional vector spaces.

As we have explained in Eigenvalues and Eigenvectors, sets of functions can provide us with an example of an infinite-dimensional vector space. We elaborate more on why functions would do well to describe a quantum system like the one we have described above. Let’s say for example that the particle is constrained to be on an interval from $0$ to $1$. For every point on the interval, there is a corresponding value of the probability amplitude. This is exactly the definition of a function from the interval $[0,1]$ to the set of complex numbers $\mathbb{C}$. We would also have to normalize later on, although the definition of normalization for infinite-dimensional vector spaces is kind of different, involving the notion of an integral (see An Intuitive Introduction to Calculus). For that matter, the square of the probability amplitude is not the probability, but the probability density.

The function that we have described is called the wave function. It is also a vector, an element of an infinite-dimensional vector space. It is most often written using the symbol $\psi(x)$, reflecting its nature as a function. However, since it is also a vector, we can also still use Dirac notation and write it as $|\psi\rangle$. The wave function is responsible for the so-called wave-particle duality of quantum mechanics, as demonstrated in the famous double-slit experiment.

We have noted in My Favorite Equation in Physics that in classical mechanics the state of a one-particle system is given by the position and momentum of that particle, whlie in quantum mechanics the wave function is enough. How can this be, since the wave function only contains information about the position? Well, actually the wave function also contains information about the momentum – this is because of the so-called de Broglie relations, which relates the momentum of a particle in quantum mechanics to its wavelength as a wave.

Actually, the wave function is a function, and does not always have to look like what we normally think of as a wave. But whatever the shape of the wave function, even if it does not look like a wave, it is always a combination of different waves. This statement is part of the branch of mathematics called Fourier analysis. The wavelengths of the different waves are related to the momentum of the corresponding particle, and we should note that like the position, they are also in quantum superposition.

There is one thing to note about this. Suppose our wave function is really a wave (technically we mean a sinusoidal wave). This wave gives us information about where we are likely to find the particle if we make a measurement – it is near the “peaks” and the “troughs” of the wave. But there are many “peaks” and “troughs” in the wave, and so it is difficult to determine where the particle will be when we measure it. On the other hand, since the wave function is composed of only one wave, we can easily determine what the momentum is.

We can also put several different waves together, resulting in a function that is “peaked” only at one place. This means there is only one place where the particle is most likely to be. But since we have combined different waves together, there will not be a single wavelength, hence, the momentum cannot be determined easily! To summarize, if we know more about the position, we know less about the momentum – and if we know more about the momentum, we know less about the position. This observation leads to the very famous Heisenberg uncertainty principle.

The many technicalities of the wave function we leave to the references for now, and proceed to the role of linear transformations, or linear operators, in quantum mechanics. We have already encountered one special role of certain kinds of linear transformations in Eigenvalues and Eigenvectors. Observables are represented by self-adjoint operators. A self-adjoint operator $A$ is a linear operator that satisfies the condition $\displaystyle \langle Au|v\rangle=\langle u|Av\rangle$.

for a vector $|v\rangle$ and linear functional $\langle u|$ corresponding to the vector $|u\rangle$. The notation $|Av\rangle$ refers to the image of the vector $|v\rangle$ under the linear transformation $A$, while $\langle Au|$ refers to the linear functional corresponding to the vector $|Au\rangle$, which is the image of $|u\rangle$ under $A$. The role of linear functionals in quantum mechanics was discussed in Some Basics of Quantum Mechanics.

There is, for example, an operator corresponding to the position, another corresponding to the momentum, another corresponding to the energy, and so on. If we measure any of these observables for a certain quantum system in the state $|\psi\rangle$, we are certain to obtain one of the eigenvalues of that observable, with the probability of obtaining the eigenvalue $\lambda_{n}$ given by $\displaystyle |\langle \psi_{n}|\psi\rangle|^{2}$

where $\langle \psi_{n}|$ is the linear functional corresponding to the vector $|\psi_{n}\rangle$, which is the eigenvector corresponding to the eigenvalue $\lambda_{n}$. For systems like our particle in space, whose states form an infinite-dimensional vector space, the quantity above gives the probability density instead of the probability. After measurement, the state of the system “collapses” to the state given by the vector $|\psi_{n}\rangle$.

Another very important kind of linear operator in quantum mechanics is a unitary operator. Unitary operators are kind of like the orthogonal matrices that represent rotations (see Rotating and Reflecting Vectors Using Matrices); in fact an orthogonal matrix is a special kind of unitary operator. We note that the orthogonal matrices had the special property that they preserved the “magnitude” of vectors; unitary operators are the same, except that they are more general, since the coefficients of vectors (the scalars) in this context are complex.

More technically, a unitary operator is a linear operator $U$ that satisfies the following condition: $\displaystyle \langle u|v\rangle=\langle Uu|Uv\rangle$

with the same conventions as earlier. What this means is that the probability of finding the system in the state given by the vector $|u\rangle$ after measurement, given that it was in the state $|v\rangle$ before measurement remains the same if we rotate the system – or perform other “operations” represented by unitary operators such as letting time pass (time evolution), or “translating” the system to a different location.

So now we know that in quantum mechanics observables correspond to self-adjoint operators, and the “operations” of rotation, translation, and time evolution correspond to unitary operators. We might as well give a passing mention to one of the most beautiful laws of physics, Noether’s theorem, which states that the familiar “conservation laws” of physics (conservation of linear momentum, conservation of angular momentum, and conservation of energy) arise because the laws of physics do not change with translation, rotation, or time evolution. So Noether’s theorem in some way connects some of our “observables” and our “operations”.

We now revisit one of the “guiding questions” of physics, which we stated in My Favorite Equation in Physics:

“Given the state of a system at a particular time, in what state will it be at some other time?”

For classical mechanics, we can obtain the answer by solving the differential equation $F=ma$ (Newton’s second law of motion). In quantum mechanics, we have instead the Schrodinger equation, which is the “ $F=ma$” of the quantum realm. The Schrodinger equation can be written in the form $\displaystyle i\hbar\frac{d}{dt}|\psi(t)\rangle=H|\psi(t)\rangle$

where $i=\sqrt{-1}$ as usual, $\hbar$ is a constant called the reduced Planck’s constant (its value is around $1.054571800\times 10^{-34}$ Joule-seconds), and $H$ is a linear operator called the Hamiltonian. The Hamiltonian is a self-adjoint operator and in many cases corresponds to the energy observable. In the case where the Hamiltonian is time-independent, this differential equation can be solved directly to obtain the equation $\displaystyle |\psi(t)\rangle=e^{-\frac{i}{\hbar}Ht}|\psi(0)\rangle$.

Since $H$ is a linear operator, $e^{-\frac{i}{\hbar}Ht}$ is also a linear operator (actually a unitary operator) and is the explicit form of the time evolution operator. For a Hamiltonian with time-dependence, one must use other methods to obtain the time evolution operator, such as making use of the so-called interaction picture or Dirac picture. But in any case, it is the Schrodinger equation, and the time evolution operator we can obtain from it, that provides us with the answer to the “guiding question” we asked above.

References:

Wave Function on Wikipedia

Matter Wave on Wikipedia

Uncertainty Principle on Wikipedia

Unitary Operator on Wikipedia

Noether’s Theorem on Wikipedia

Schrodinger Equation on Wikipedia

Introduction to Quantum Mechanics by David J. Griffiths

Modern Quantum Mechanics by Jun John Sakurai

Quantum Mechanics by Eugen Merzbacher