Matemática para Mecânica Quântica

Como já estudei o básico das ferramentas matemáticas necessárias, por enquanto quero priorizar a compreensão física. Assim, aprofundarei a matemática até o nível necessário para compreender, aplicar e questionar criticamente a estrutura física em estudo. Demonstrações e desenvolvimentos puramente matemáticos podem ser deixados para quando se tornarem necessários ou quando eu decidir aprofundar os fundamentos.

Por exemplo, não é necessário, neste momento, aprofundar a demonstração do teorema espectral para matrizes Hermitianas, pois sua prova é predominantemente matemática e não acrescenta, por si só, uma nova interpretação física ao problema. Em contrapartida, vale aprofundar resultados matemáticos que estabelecem relações diretamente relevantes para a teoria física, como demonstrar que dois operadores que possuem uma mesma base completa de autovetores comutam, pois essa relação permite compreender matematicamente a compatibilidade entre observáveis.

Porém estarei compartilhando aqui uma revisão puramente matemática, uma reprodução quase na íntegra deste material: Delft University of Technology (2026).

Álgebra Linear¶

Aritmética Matricial Básica¶

Operação de Matrizes¶

Very often, an $m\times n$-matrix $A$ with real entries represents a linear map from the space $\mathbb{R}^{n}$ of vectors of length $n$ to the space $\mathbb{R}^{m}$ of vectors of length $m$. Similarly, a matrix with complex entries represents a linear map from $\mathbb{C}^{n}$ to $\mathbb{C}^{m}$. This map is defined by the matrix-vector product.

Definition: Given a real (resp. complex) $m\times n$-matrix $A$, the linear map associated with $A$ is the map that sends each vector $\boldsymbol{v}\in\mathbb{R}^{n}$ to the vector $A\boldsymbol{v}\in\mathbb{R}^{m}$ (resp. each vector $\boldsymbol{v}\in\mathbb{C}^{n}$ to the vector $A\boldsymbol{v}\in\mathbb{C}^{m}$). :::

Matrix multiplication is defined so that the product $AB$ represents the linear map gotten by first applying $B$ and then $A$. In other words, we have

Note that on the left we have one matrix multiplication and one matrix-vector multiplication, while on the right we just have two matrix-vector multiplications. Commutators play an extremely important role in quantum physics.

Definition: Given two $n\times n$-matrices $A$ and $B$, the commutator of $A$ and $B$ is the $n\times n$-matrix $[A,B]=AB-BA.$

Definition: The transpose of an $m\times n$-matrix $A$ is the $n\times m$-matrix $A^{\top}$ defined by $(A^{\top})_{i,j}=A_{j,i}.$

Definition: The adjoint (conjugate transpose or Hermitian transpose) of an $m\times n$-matrix $A$ is the $n\times m$-matrix $A^{\dagger}$ obtained by applying the complex conjugate and the transpose operation: $(A^{\dagger})_{i,j}=\overline{A_{j,i}}.$ The adjoint of $A$ is also known as the ** or ** of $A$.

Definition: The trace of an $n\times n$-matrix $A$ is the sum of the diagonal entries of $A$: $\text{Tr}\left(A\right)=\sum^{n}_{i=1}A_{i,i}.$

The determinant of an $n\times n$-matrix $A$ can be defined in various ways. We will make use of permutations. Using the Einstein summation convention^[1] and the Levi-Civita symbol^[2] commonly used in physics, we can also write this as

Properties of the determinant :

1. A square matrix $A$ is invertible if and only if $\det(A)$ is non-zero.

2. Given two $n\times n$-matrices $A$ and $B$, we have $\det(AB)=(\det A)(\det B).$

3. The determinant of a diagonal matrix (and more generally of an upper or lower triangular matrix) is the product of the diagonal entries.

4. 4. If $B$ is obtained from $A$ by scaling a row by a scalar $c$, then $\det B=c\det A$.

5. 5. If $B$ is obtained from $A$ by swapping two rows, then $\det B=-\det A$.

6. 6. If $B$ is obtained from $A$ by adding a multiple of a row to another row, then $\det B=\det A$.

Properties 4-6 also hold for columns instead of rows.

Finally, we define the characteristic polynomial of a square matrix; this will be used in diagonalisation.

Definition: The characteristic polynomial of an $n\times n$-matrix $A$ is the polynomial of degree $n$ in a variable $t$ defined by $\chi_{A}(t)=\det(x\mathbb{I}-A).$^[3]

Basis transformations¶

Until now, we have been working with one specific coordinate system in which vectors are identified with their coordinate vectors with respect to the standard basis vectors:

In other words, a vector $\boldsymbol{v}=\begin{pmatrix}v_{1}\\ v_{2}\\ \vdots\\ v_{n} \end{pmatrix}$ means the same as the corresponding linear combination of the standard basis vectors:

In many situations, it is useful to express vectors in a different coordinate system. For this we need the general concept of a basis.

Definition: A basis of $\mathbb{R}^{n}$ consists of $n$ vectors $(\boldsymbol{b}_{1},\ldots,\boldsymbol{b}_{n})$ (or $\left\{ \boldsymbol{b}_{i}\right\} ^{n}_{i=1}$) such that every vector $\boldsymbol{v}\in\mathbb{R}^{n}$ can be written in exactly one way as a linear combination:$\boldsymbol{v}=c_{1}\boldsymbol{b}_{1}+\cdots+c_{n}\boldsymbol{b}_{n}$, with scalars $c_{1},\ldots,c_{n}$. These scalars are called the coordinates of $\boldsymbol{v}$ relative to $\left\{ \boldsymbol{b}_{i}\right\} ^{n}_{i=1}$. The vector $\left(\begin{array}{ccc} c_{1} & \dots & c_{n}\end{array}\right)^{T}$ is called the coordinate vector of $\boldsymbol{v}$ relative to $\left\{ \boldsymbol{b}_{i}\right\} ^{n}_{i=1}$.

Alternatively, a basis consists of $n$ vectors $\left\{ \boldsymbol{b}_{i}\right\} ^{n}_{i=1}$ with the following two properties:

• they span the space $\mathbb{R}^{n}$, i.e. every vector is a linear combination of $\left\{ \boldsymbol{b}_{i}\right\} ^{n}_{i=1}$;

• they are linearly independent, i.e. the only way to write the zero vector as a linear combination of $\left\{ \boldsymbol{b}_{i}\right\} ^{n}_{i=1}$is as $0\boldsymbol{b}_{1}+\cdots+0\boldsymbol{b}_{n}$.

A priori, we could have defined a basis as a collection of some number (say $m$) of vectors that span the space and are linearly independent. However, it can be shown that in fact any two bases contain the same number of vectors. Since the standard basis of $\mathbb{R}^{n}$ consists of $n$ vectors, the same therefore holds for any basis.

Suppose we have two bases of $\mathbb{R}^{n}$, say

Definition: The basis transformation matrix (or change-of-basis matrix) from $B$ to $C$ is the matrix $P=P_{B\to C}$ such that the $j$-th column of $P$ contains the coordinates of $\boldsymbol{b}_{j}$ relative to the basis $(\boldsymbol{c}_{1},\ldots,\boldsymbol{c}_{n})$.

The definition of $P$ means:

Given a vector $\boldsymbol{x}$, we write $\boldsymbol{x}^{(B)}=(x^{(B)}_{1},\ldots,x^{(B)}_{n})$ for the coordinate vector of $\boldsymbol{x}$ with respect to the basis $\left\{ \boldsymbol{b}_{i}\right\} ^{n}_{i=1}$, and we define $\boldsymbol{x}^{(C)}$ similarly. Then we calculate

which shows that

Similarly, given a linear map $A$, we write $A^{(B)}$ for the matrix of $A$ with respect to the basis $\left\{ \boldsymbol{b}_{i}\right\} ^{n}_{i=1}$, and likewise for $A^{(C)}$. Then $A^{(B)}$ and $A^{(C)}$ are related by

Namely, consider what happens when we multiply this matrix to a coordinate vector of the form $\boldsymbol{x}^{\left(C\right)}$. First, applying $P^{-1}$^[4] to $\boldsymbol{x}^{\left(C\right)}$ gives $\boldsymbol{x}^{\left(B\right)}$. Next, applying $A^{(B)}$ gives $(A\boldsymbol{x})^{(B)}$. Finally, applying $P$ gives $(A\boldsymbol{x})^{(C)}$, which is what we wanted to show.

The trace and the determinant are invariant under basis transformation: if $A$ and $P$ are square matrices of the same size with $P$ invertible, then we have

Because the characteristic polynomial is also defined as a determinant, it too is invariant under basis transformation.

Diagonalisation¶

Hilbert Spaces¶

Operators on Hilbert spaces¶

Tensor products¶