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+---
+title: Operators quantum mechanics
+---
+
+# Operators in quantum mechanics
+
+The lecture on operators in quantum mechanics consists of different parts, each with their own video:
+
+- [Definition of operators](#definition-of-operators)
+- [Properties of operators](#properties-of-operators)
+- [Manipulating operators](#manipulating-operators)
+
+**Total video length: xxx **
+
+
+In the previous lecture we presented the mathematical language to describe the *quantum states* of a physical system. We saw that the state of a quantum system is described by its *wave function*, an element of a special vector space called the *Hilbert space}. We also presented the Dirac notation and discussed that we can assign a *probabilistic interpretation* to the wave function.
+
+Now we need to introduce the mathematical language required to *extract information} about physical properties of the system from its state vector, which we will denote by *observables*.
+
+We emphasize that this distinction between the *state of a quantum system} (given by the wave function) and the *observables*that we can extract from it is new in quantum mechanics, since it is absent in classical physics.
+
+With this motivation, to represent fundamental physical quantities of a quantum system that we can measure such as position, momentum, or energy, we need to introduce a special mathematical entity known as an *operator*.
+
+## Definition of operators
+
+Operators in quantum mechanics are mathematical entities used to represent physical processes that result in the change of the state of the system, such as the evolution of these states with time.
+
+These operators can also represent physical properties of a system that can be experimentally measured (position, momentum, and energy), the *observables* associated to this quantum system.
+
+An operator is a mathematical object that *acts*on the wave function
+a system and produces another state vector. To be precise, if we denote an operator by $\hat{A}$ and $|\psi\rangle$ is an element of the Hilbert space of the system, then we will have that
+$$\hat{A} |\psi\rangle = |\phi\rangle \, ,$$
+where the state vector $|\phi\rangle$ *also*belongs to the same Hilbert space.
+
+There are many types of important operators in quantum mechanics.
+In this lecture we will present some of these, such as the
+*unitary operators* that determine the time evolution of a quantum system
+and the *Hermitian operators* which can be associated to
+physically observable properties of a system, such as momentum or energy.
+
+## Properties of operators
+
+In this course we are interested in the so-called *linear operators},
+which are those operators $\hat{A}$ such that for any arbitrary pair of state vectors
+$|\psi_1\rangle$ and $|\psi_2\rangle$ and for any complex numbers $c_1$ and $c_2$ one has
+$$\hat{A}[c_1|\psi_1\rangle+c_2|\psi_2\rangle]=c_1\hat{A}|\psi_1\rangle+c_2\hat{A}|\psi_2\rangle \, .$$
+
+Linearity of operators has an important consequence. Recall that in the previous lecture we discussed that any state vector $|\psi\rangle can be expressed as a linear combination of a complete set of basis states $\{|\phi_i\rangle,i=1,2,3,...,n\}$ associated to this Hilbert space:
+$$|\psi\rangle=\sum_{i=1}^nc_i|\phi_i\rangle \, , \quad c_i = \braket{\phi_i}{\psi} \, ,$$
+where the values of the coefficients $c_i$ can be fixed thanks to
+the orthogonality properties of the basis, $\braket{\phi_i}{\phi_j}=\delta_{ij} $.
+
+Then one can see that for linear operators
+ one has
+$$
+ \hat{A}|\psi\rangle= \hat{A}\sum_{i=1}^nc_i|\phi_i\rangle
+ = \sum_{i=1}^nc_i ( \hat{A}|\phi_i\rangle ) \, .
+$$
+This results tells us that if we know the effects of the operator
+$\hat{A}$ for each of the elements of the basis $|\phi_i\rangle$,
+we can easily determine its effects for a *general state vector}
+$|\psi\rangle$ belonging to the same Hilbert space.
+
+Another important properties of operators can be stated
+ as follows.
+%
+ If two operators $\hat{A}$ and $\hat{B}$ are such that
+ $$
+ \hat{A}|\psi\rangle=\hat{B}|\psi\rangle
+ $$
+ for all state vectors $|\psi\rangle$ belonging to the Hilbert
+ space of the system, then two operators must be identical''
+ $$
+ \hat{A}=\hat{B} \, .
+ $$
+ Note that this is true only if the action of two operators
+ is identical for all elements of the Hilbert space.
+
+ As in general vector spaces, in Hilbert spaces
+ we also have the identity (or unit) and zero (or null)
+operators defined as
+ \begin {enumerate}
+ The *unit operator} $\hat{I}$ is the operator that
+ $$
+ \hat{I}|\psi\rangle=|\psi\rangle
+ $$
+ The *zero operator*$\hat{0}$ is the operator that
+ satisfies
+ $$
+ \hat{0}|\psi\rangle=0
+ $$
+ \end{enumerate}
+ In both cases, these relations hold for all state vectors $|\psi\rangle$
+ of the Hilbert space.
+
+
+## Manipulating operators
+
+We can combine and manipulate operators in various ways.
+%
+In doing so, we should be careful since manipulations
+of operators can be quite different as compared
+to manipulations with scalar of complex numbers.
+%
+For example, if you have two complex numbers, the result
+of their multiplication does not depend on the order in which
+you multiply them, but for operators it does!
+
+ As mentioned above, in these lectures we will be focusing
+ only on linear operators.
+
+
+*Addition of operators*:
+ the sum of two operators $\hat{A}$ and $\hat{B}$ is defined by
+ $$
+ (\hat{A}+\hat{B})|\psi\rangle=\hat{A}|\psi\rangle+\hat{B}|\psi\rangle \, ,
+ $$
+ for all state vectors $|\psi\rangle$.
+ The sum of two operators defines another operator, $\hat{C}$:
+ $$
+ \hat{C}|\psi\rangle=(\hat{A}+\hat{B})|\psi\rangle=\hat{A}|\psi\rangle+\hat{B}|\psi\rangle
+ $$
+ for all states $|\psi\rangle$, and so we can write
+ $$
+\hat{C} = \hat{A} + \hat{B} \, .
+ $$
+
+*Multiplication of an operator by a complex number*:
+ if we have an operator that acts on a state vector as
+ $$
+ \hat{A}|\psi\rangle=|\phi\rangle \, ,
+ $$
+ then we can define the operator $\hat{C}=\lambda \hat{A}$, where $\lambda$ is a complex number
+ $$
+ \hat{C} |\psi\rangle= (\lambda \hat{A})|\psi\rangle=\lambda(\hat{A}|\psi\rangle)=\lambda |\phi\rangle \, .
+ $$
+
+ *Multiplication of operators}:
+ assume that an operator $\hat{A}$ acting on a ket vector $|\psi\rangle$ maps it into another ket vector $|\phi\rangle$, and that the operator $\hat{B}$ acting on $|\phi\rangle$ results into
+ a third ket vector $|\rho\rangle$:
+$$
+ \hat{B} \left( \hat{A}|\psi\rangle \right)=\hat{B}|\phi\rangle=|\rho\rangle \, .
+$$
+One can then define the product of the two operators as a new operator,
+$\hat{C}=\hat{B}\hat{A}$, such that its action on to the initial ket vector $|\psi\rangle$
+is defined as
+$$
+ \hat{C}|\psi\rangle=\hat{B}\hat{A}|\psi\rangle = |\rho\rangle \, .
+$$
+Note that in general the multiplication of two operators is *non commutative}:
+the order in which we multiply $\hat{A}$ and $\hat{B}$ is important.
+
+*Commutator of two operators}: the difference between the product
+ of operators $\hat{B}\hat{A}$ and the same but in the opposite order,
+ $\hat{B}\hat{A}$, is defined as the commutator of these two operators:
+$$
+ [\hat{A},\hat{B}]\equiv \hat{A}\hat{B}-\hat{B}\hat{A} \, .
+$$
+The commutator plays a fundamental role in the physical interpretation of quantum
+mechanics.
+%
+In a nutshell,
+it tells us whether or not two observable properties of a system
+can be determined simultaneously with arbitrary precision.
+
+Using the properties of the addition and multiplication of operators
+that we just discussed, one can check that the commutator satisfies the following
+properties:
+
+
+- $[\hat{A},\hat{B}]= -[\hat{B},\hat{A}] $
+- $[\hat{A},\alpha\hat{B}+\beta\hat{C}]=\alpha[\hat{A},\hat{B}]+\beta[\hat{A},\hat{C}] $
+- $[\hat{A}\hat{B},\hat{C}]=\hat{A}[\hat{B},\hat{C}] +[\hat{A},\hat{C}] \hat{B}$
+- $[\hat{A},[\hat{B},\hat{C}]]+[\hat{C},[\hat{A},\hat{B}]]+[\hat{B},[\hat{C},\hat{A}]]=0$
+
+## Projection operators
+
+
+An operator $\hat{A}$ that has the following property
+$$
+ \hat{A}^2= \hat{A}\hat{A}= \hat{A}
+$$
+is said to be a *projection operator}.
+
+
+Let us give an explicit example of such operator.
+ %
+ Assume we have an $n$-dimensional Hilbert space with a basis given by $\{|\phi_i\rangle\}$.
+ %
+ We can define the operator $\hat{B}_i$ as follows
+ $$
+ \hat{B}_i|\phi_j \rangle=\delta_{ij}|\phi_j \rangle \, .
+ $$
+ Note that as demonstrated below, once we indicate the behaviour of an operator
+ for the basis vectors, we automatically know how it will act for any general
+ state vector of the Hilbert space.
+
+Let's demonstrate that this operator is a projection operator.
+$$
+ \left( \hat{B}_i\right)^2|\phi_j \rangle=\hat{B}_i\hat{B_i}|\phi_j \rangle=\delta_{ij}\hat{B_i}|\phi_j \rangle=\delta_{ij}^2|\phi_j \rangle = \delta_{ij}|\phi_j \rangle = \hat{B}_i|\phi_j \rangle .
+$$
+so from this we have that $\hat{B_i}^2=\hat{B_i}$, as requested
+for a projection operator.
+
+This projection operator has one important property.
+ %
+ Let's act with $\hat{B}_i$ on an arbitrary vector $|\psi\rangle$,
+ expanded in terms of the basis vectors $\{|\phi_j \rangle\}$.
+ %
+ We have that
+$$
+ \hat{B}_i|\psi\rangle=\hat{B}_i\sum_j|\phi_j \rangle\braket{\phi_j}{\psi}=\sum_j\left(\hat{B_i}|\phi_j \rangle\right) \braket{\phi_j}{\psi}
+$$
+$$
+ \hat{B}_i|\psi\rangle=\sum_j\delta_{ij}|\phi_j \rangle\braket{\phi_j}{\psi}=|\phi_i\rangle\braket{\phi_i}{\psi} \, .
+$$
+In other words, we see that the operator $ \hat{B}_i$
+*projects*the state vector $|\psi\rangle$ into the direction
+given by the basis vector $|\phi_i\rangle$.
+%
+We clearly see in this case why these operators are called
+*projection operators}.
+
+
+## The Hermitian adjoint
+
+In general for an operator
+ $\hat{A}|\psi\rangle=|\phi\rangle$ one has that $\bra{\psi}\hat{A}\neq \bra{\phi}$.
+ %
+ We can introduce another operator related to $\hat{A}$ and written as $\hat{A}^\dag$
+ which has the following defining property
+ $$
+ \hat{A}|\psi\rangle=|\phi\rangle\; \text{then}\; \bra{\psi}\hat{A^\dag}= \bra{\phi}
+ $$
+ The operator $\hat{A^\dag}$ is known as the *Hermitian adjoint*of $\hat{A}$.
+
+ What is then the action of this Hermitian adjoint operator on a ket vector?
+ We can consider the following product
+ $$
+ \bra{\rho}\hat{A}|\psi\rangle=\bra{\rho}(\hat{A}|\psi\rangle)=\bra{\rho}|\phi\rangle
+ $$
+ since $\hat{A}|\psi\rangle=|\phi\rangle$.
+ Then one take the complex conjugate of the previous expression to find
+ $$
+ \bra{\rho}\hat{A}|\psi\rangle^*=\bra{\rho}|\phi\rangle^*=\bra{\phi}|\rho\rangle
+ $$
+ if $\bra{\psi}\hat{A^\dag}= \bra{\phi}$ then
+ $$
+ \bra{\rho}\hat{A}|\psi\rangle^*=\bra{\phi}|\rho\rangle=(\bra{\psi}\hat{A^\dag})|\rho\rangle=\bra{\psi}\hat{A^\dag}|\rho\rangle \, .
+ $$
+ As we will see next, in quantum mechanics we are interested in operators
+ for which $\hat{A}=hat{A}^\dag$, that is, where the operator coincides
+ with its Hermitian adjoint.
+
+
+