Introduction to Quantum Mechanics Unit 0. Review of Linear Algebra
A. Linear Space 1. Definition: A linear space is a collection S of vectors |a>, |b>, |c>, …. (usually infinite number of them) on which vector addition + is defined so that (i) (S,+) forms an Abelien group, i,e,, S is closed under + + is associate: (|a>+|b>)+|c> = |a>+(|b>+|c>) + is commutative: |a>+|b>=|b>+|a> Existence of identity: |a> + |0> = |a> for all |a> Existence of inverse: |a> + (|-a>) = |0> and also scalar (usually complex number) * is defined so that (ii) S is closed under * (iii) * and complex number multiplication is associative: α(β*|a>)=(αβ)*|a> (iv) 1|a>=|a> (and hence 0|a>=|0> and |a>=-|a>) (v) * and + is distributive: (α+β)|a>=α|a>+β|a> and α(|a>+|b>)=α|a>+α|b>
(i) Definition of a basis: A basis is a collection of vectors B={|a1>, |a2>,|a3>,….|aN>} such that any vector in S can be written as a linear combination of a1>, |a2>,|a3>,….|aN>:
a = α1 a1 + α 2 a 2 + α3 a 3 +Lα N a N where α1 ,α 2 ,Kα N are complex numbers Furthermore, ={|a1>, |a2>,|a3>,….|aN> are independent of each other, i.e.
0 = α1 a1 + α 2 a 2 + α 3 a 3 +Lα N a N if and only if α1 = α 2 = K = α N = 0 (ii) Choice of a basis is not unique. (iii) However, all basises have the same number of elements. This number (say, N) depends only on the linear space S, and it is known as the dimension of the linear space S. (iv) If the basis has too few elements (
3. Coordinate and column matrix (i) For a particular basis B={|a1>, |a2>,|a3>,….|aN>}, we can express any vector as a linear combination of the basis vectors:
a = α1 a1 + α 2 a 2 + α3 a 3 +Lα N a N where α1 ,α 2 ,Kα N are complex numbers
α1, α2,α3,….αN are called the coordinates of the vector |a>. (ii) Note that α1, α2,α3,….αN are complex numbers.
(iii) Not an N-dimensional basis, each vector within the space can be expressed as N coordinates. (iv) We can write the coordinates as a column matrix: ⎛ α ⎞ ⎜ 1 ⎟ ⎜ α 2 ⎟ a = ⎜ α ⎟ ⎜ 3 ⎟ ⎜ M ⎟ ⎜ ⎟ ⎝α N ⎠ (v) A column matrix corresponds to a contravariant rank 1 tensor. (vi) Above notation is kind of confusing, because the coordinates of a vector depend on the choice of the basis. In other words, a column matrix of complex number has no meaning unless we know the basis corresponds to it. (vii) If necessary, we will write the name of the basis next to the column like this:
⎛ α ⎞ ⎜ 1 ⎟ ⎜α 2 ⎟ ⎜ ⎟ a = B α 3 ⎜ ⎟ ⎜ M ⎟ ⎜ ⎟ ⎝α N ⎠
We say the vector |a> is presented under “B-represntation”. (viii) If we use another basis C, the column matrix will be different for the same vector |a>.
B. Linear Transformation 1. Definition (i) A linear transformation T is a mapping of vectors from linear space S1 to another linear space S2 (T: S1→ S2) such that if T(|a>)= |b> where |a>∈ S1 and |b>∈S2 then T(α1|a1>+α2|a>)=α1(T|a1>)+α2(T|a>) (ii) Note that in general S1 and S2 can be two completely different spaces of different dimensions. (iii) We do not need to define a linear transformation over all vectors within the space. Once we know how the vectors in a particular basis of S1 are transformed, the linear transformation is defined. For example, if B1={|a1>, |a2>,|a3>,….|aN>}is a particular basis of S1 and we know what are T(|a1>),T (|a2>), …T (|a1>), then for any vector |a> in S1:
a = α1 a1 + α 2 a 2 + α 3 a 3 +Lα N a N
⇒ T( a ) = α1T( a1 ) + α 2T( a 2 ) + α 3T( a 3 ) +Lα N T( a N )
(iv) T can be written in matrix form, with respect to certain basises in S1 and S2. Let B1={|a1>, |a2>,|a3>,….|aN>}be a basis in S1 and B2={|c1>, |c2>,|c3>,….|cM>}be a basis in S2. If T(|a1>)=α11| c1>+α21| c2>+α31| c3>+…….+αM1| cM> T(|a2>)=α12| c1>+α22| c2>+α32| c3>+…….+αM2| cM> ………… T(|aN>)=α1N| c1>+α2N| c2>+α3N| c3>+…….+αMN| cM> We can write the linear transformation as a matrix: ⎛ α α α α ⎞ ⎜ 11 12 13 K 1N ⎟ ⎜ α 21 α 22 α 23 L α 2N ⎟ T = ⎜ α α α α ⎟ ⎜ 31 31 31 K 31 ⎟ ⎜ M M M M M ⎟ ⎜ ⎟ ⎝α M1 α M2 α M3 L α MN ⎠ (v) Above notation is kind of confusing, because the matrix representation of T depends on the choice of the two basises in S1 and S2. We may want to write the names of the basises at the side of the matrix as a reminder. However, we must be more careful this time because matrix has two “directions”- up/down and left/right. In the above matrix, we note that the horizontal index (the second subscript 1 to N) runs through the basis B1 of S1 and the vertical index (the second subscript 1 to M) runs through the basis B2 of S2. Hence, we should write it this way:
B 1 ⎛ α α α α ⎞ ⎜ 11 12 13 K 1N ⎟
⎜ α 21 α 22 α 23 L α 2N ⎟
T = B ⎜ α α α α ⎟ 2 ⎜ 31 31 31 K 31 ⎟ ⎜ M M M M M ⎟ ⎜ ⎟ ⎝α M1 α M2 α M3 L α MN ⎠
We will discuss about this procedure more carefully when we discuss NxN matrix.
(vi) Now the linear transformation of any other vector can be calculated easily with matrix multiplication.
B1 ⎛ β ⎞ ⎛ α α α α ⎞ ⎛ β ⎞ ⎛ γ ⎞ ⎜ 1 ⎟ ⎜ 11 12 13 K 1N ⎟ ⎜ 1 ⎟ ⎜ 1 ⎟ ⎜ β 2 ⎟ ⎜ α 21 α 22 α 23 L α 2N ⎟ ⎜ β 2 ⎟ ⎜ γ 2 ⎟ If a = B ⎜ β ⎟ Then T( a ) = B ⎜ α α α α ⎟ B ⎜ β ⎟ = B ⎜ γ ⎟ 1 ⎜ 3 ⎟ 2 ⎜ 31 31 31 K 31 ⎟ 1 ⎜ 3 ⎟ 2 ⎜ 3 ⎟ ⎜ M ⎟ ⎜ M M M M M ⎟ ⎜ M ⎟ ⎜ M ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ β N ⎠ ⎝α M1 α M2 α M3 L α MN ⎠ ⎝ β N ⎠ ⎝γ N ⎠
Note how the name of B1 disappears during the matrix multiplication. We start with a S1-vector in B1 representation, and end up in an S2-vector in B2 representation.
Rule: In matrix multiplication, for it to be meaningful, the multiplying row and column must have the same basis and that name will disappear after the multiplication. This corresponds to Einstein notation in tensor multiplication.
2. Linear functional – one important kind of linear transformation
(i) Consider linear transformation T mapping vectors in linear space S to complex number T:S → C.
(ii) Define addition and scalar multiplication on these transformations according to the operations in the linear space S: (T1+ T2) |a> = T1(|a>)+ T2 (|a>) (αT)|a> = T(α|a>) for all vectors |a> in S. In this way, we can write linear functional as linear combinations of the others.
(iii) All linear functionals form a linear space S*, with the same dimension (N) as the original linear space S (only if the dimension is finite). S* is called the dual space of S. Example: Reciprocal space is the dual space of real space. First rank covariant tensor is dual to the first rank contravariant tensor.
(iv) We will use notation
(v) Since S and S* have the same dimension, we can easily set up a one-to-one relationship between the elements in this two space. For example, we can pick up A={|a1>, |a2>,|a3>,….|aN>} from S and B={|b1>, |b2>,|b3>,….|bN>} from B* and set up a correspondence like:
a1 a2 a3 L aN b b b b
b1 b2 b3 L bN We said a metric has been defined for the vector space S. Obviously there are infinite numbers of ways to set up this relationship.
(vi) In the above relationship, we said
(vii) Because of the complex conjugate in the above expression, we have the nice properties
(vii) The metric relationship allows us to use the same name for a vector in S and its correspond dual vector in S*. For example, if (viii) Metric definition is important because it define how we calculate the inner products between two vectors |x> and |y> in S. The inner product between |y> and |x> = (ix) A basis A={|a1>, |a2>,|a3>,….|aN>}is said to be orthonormal if (ix) Because of the complex conjugate in the above expression, we have the nice properties (x) Consider two basises A={|a1>,|a2>,|a3>,…..|aN>}and B={|b1>,|b2>,|b3>,…..|bN>}, provided that A is an orthonormal basis. B is also an orthonormal basis if (xi) Only in terms of orthonormal basis B, we can write dual vector in matrix form: Related by the metric key ⎛ α1 ⎞ ⎜ ⎟ α ⎜ 2 ⎟ B a = ⎜ α ⎟ B ⎜ 3 ⎟ * * * * Dual with a = ()α1 α 2 α 3 L α N ⎜ M ⎟ ⎜α ⎟ ⎝ N ⎠ We can then calculate the inner product like matrix multiplication: ⎛ β1 ⎞ B ⎜ ⎟ ⎜ β 2 ⎟ < a | b >= (α * ,α * ,α * , α * ) ⎜ β ⎟ B = α * β + α * β + α * β + + α * β 1 2 3 L N ⎜ 3 ⎟ 1 1 2 2 3 3 L N N ⎜ M ⎟ ⎜ ⎟ ⎝ β N ⎠ Note the complex conjugate and how the names of the basises disappear during the multiplication process. We can only do this if B is orthonormal. This matrix multiplication does not take care of 3. Mapping between spaces of the same dimension – another important kind of linear transformation (i) We now consider a linear transformation from one space S to another S’ of the same dimension. The metric key we discussed above is one of the examples. T: S → S’ (S and S’ have the same dimension) There are many important transformation, like the Hermitian transformation, and unitary transformation belong to this class of transformation. (ii) As discussed before, we just need to know how vectors in one basis in S are mapped into S’ (in terms of another basis in S’). The mapping of all other vectors in S will then also be known. (iii) Here is how things work: Let B= {|a1>, |a2>,|a3>,….|aN>}be a basis in S, and B’= B={|c1>, |c2>,|c3>,….|cN>}be a basis in S’. If under linear transformation T we have the following mapping: T(|a1>)=α11| c1>+α21| c2>+α31| c3>+…….+αN1| cM> T(|a2>)=α12| c1>+α22| c2>+α32| c3>+…….+αN2| cM> ………… T(|aN>)=α1N| c1>+α2N| c2>+α3N| c3>+…….+αNN| cM> We can drop away the base vectors and write the coefficient as coordinates in B’ representation: ⎛ α ⎞ ⎛ α ⎞ ⎛ α ⎞ ⎜ 11 ⎟ ⎜ 12 ⎟ ⎜ 1N ⎟ ⎜α 21 ⎟ ⎜α 22 ⎟ ⎜α 2N ⎟ T a = ⎜α ⎟ B' T a = ⎜α ⎟ B' ……… T a = ⎜α ⎟ B' 1 ⎜ 31 ⎟ 2 ⎜ 32 ⎟ N ⎜ 3N ⎟ ⎜ M ⎟ ⎜ M ⎟ ⎜ M ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝α N1 ⎠ ⎝α N2 ⎠ ⎝α NN ⎠ We know that the matrix form of T can be obtained by stacking these numbers together. If we stacked these column matrices vertically in column, we will have a matrix like this: B ⎛ ⎞ B ⎜ ⎟ B' T = B' ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ If we stacked these column matrices horizontally in row, we will have a matrix like this: B' B’ ⎛ ⎞ ⎜ ⎟ B T = B ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ In here we pay more attention to the vertical and horizontal representation of the matrix. We know that we are going to multiply it (from the left) with an S-vector in B representation, so the matrix must be in B- representation horizontally and hence the column matrices must be stacked vertically in column like the first case. In the future we may pay more attention to the representation than the actual content of the matrix! (iv) Some definition of matrices, Inverse matrix: T’ is an inverse of T if TT’=T’T=I where I is the identity matrix. We will write T’ as T-1. Note that the inverse interchanges its vertical and horizontal representations with that of T (because inverse is a reverse mappint). For example, if we write T as B’ B T Then T-1 has to be in this form: B B’ T-1 For this reason, an identity matrix should always have the same representation both vertically and horizontally. If that is not the case, the identity matrix simply does not represent an identity transformation (it represents a change of basis). Transpose: Transpose of a matrix is the interchange of its column and row T indices, i.e. (T )ij=Tji. The superscript T represent the transpose. Again, the transpose interchanges its vertical and horizontal representations with that of T. If T is in the above form, TT has to be in this form (like the inverse): B B’ TT Adjoint: The adjoint of a matrix is its transpose with all elements in complex + conjugate. i.e., (T )ij=(Tji)* . The superscript + represent the transpose. Like the transpose, the adjoint interchanges its vertical and horizontal representations with that of T. (iv) The linear transformation can be a mapping within the same space, i.e. T: S → S Only matrix of this kind of linear transformation can have the same representation horizontally and vertically. If we use B representation for both cases, T can be written as: B B T Of course, we have the freedom to use two different representations vertically and horizontally. This type of transformation is often called “operator” in quantum mechanics. 4. Change of basis (i) Change of basis is an important operation, which can be considered as T: S → S. We choose a basis (say B) for the first S, and another basis (say B’) for the second S. Change of basis from B to B’ can be accomplished by establishing a mapping (can be considered as a linear transformation) between these two basises. I reserve the symbol ∆ (or sometimes U) for the change of basis operation. B’ B ∆ coverts a vector from B representation to B’ representation. The way to construct ∆ is exactly the same as the way we used before, except that in changing basis, it is more customary to write vectors in B’ in terms of vectors in the original basis B, and hence there is a switch in the row and column representation. (ii) If the two basises B and B’ are legitimate basises, and the mapping defined by ∆ is one to one, then ∆ must have an inverse. This inverse is the change of basis from B’ back to B. (iii) One important difference between ∆ and other type of linear transformation is that it does not really map a vector to another one. It just changes the representation, but the vector |a> is still the same object in the space: B B’| |a> = B’ ∆-1B| |a> For this reason, ∆ is actually an identity mapping, i.e. B B ≡ B’ ∆ B’ I However, to avoid unnecessary confusion, I will strictly confine the use of “I” to mean identity transformation under the same representation (hence the matrix is identity also). (iv) Now consider a linear transformation T, under a representation we call A: A A| |a> = A T A| |a> or simply |a>=T|a>. If we now perform a change of basis from A to B, the vector part can be easily done as follow: A B| |a> = B ∆-1 A| |a> Note that we are forced to use ∆-1 instead of ∆ to ensure the representation workout properly. How about the matrix T? Note that T has two representations, one horizontal and one vertical. Hence there should be changes of basis, one takes care of the horizontal representation and the other takes care for the vertical representation: B A A B -1 B T = B ∆ A T A ∆ (v) For this reason, ∆-1A∆ is called the similar transformation of matrix A. Similar transformation is needed to apply to a matrix under basis transformation. C. Eigenvalue Problem 1. Eigenvalue, eigenvector, and diagonalization (i) |a> is an eigenvector of operator A if A |a> = λ |a> (representation independent) λ (complex) an eigenvalue of A. Note that above a vectorial equation, independent of the representation, or the equation is representation invariant. The eigenvector itself as an object in the vector space, and the value of the eigenvalues, are independent of the choice of basis. (ii) In an N-dimensional space (hence A can be represented as an NxN matrix), there will be N eigenvalues and at most N independent eigenvectors. Some of these eigenvalues can be equal. (iii) When two or several eigenvectors have the same eigenvalue, their linear combination is also an eigenvector with the same eigenvalue. (iv) Since there are N eigenvectors of an operator, they can be used as a basis for the linear space. (v) Procedure: To find eigenvalues and eigenvectors: (a) Start with a certain representation (say, O-representation), write operator A as a matrix (A). (b) A |a> = λ |a> ⇒ (A-Iλ)|A>=0 ⇒ det(A-Iλ)=0. (c) By solving det(A-Iλ)=0 we will get N roots. These are the N eigenvalues of the operator A. (d) For every of the eigenvalue, plug it back into A |a> = λ |a>, this will allow us to solve for the components of under O-representation. (e) If needed, we can normalize all the eigenvectors with their lengths. All eigenvectors form a basis, let us call this basis N. To diagonalize matrix A: (f) Note that the given matrix A is in representation O. It is a diagonal matrix with eigenvalues along the diagonal if it is in representation N. A change of basis is needed if we want to diagonalized A. (g) Pack the eigenvectors in column to form an NxN matrix. This will be the transformation matrix: O −1 N ∆ Perform similar transformation: N O O N N T = N ∆-1 O T O ∆ (vi) An operator always have N eigenvalues, Case 1. If all N eigenvalues have different value, then there must be N independent eigenvectors which can be used as a basis. The matrix is diagonalizable. Case 2. If m of these N eigenvalues have the same value, then Case 2(a) These m multiple eigenvalues still give rise to m linearly independent eigenvectors, then the situation will be similar to case 1 and the matrix is still diagonalizable. Any linear combination of eigenvectors with the same eigenvalue is again an eigenvector of the same eigenvalue. Case 2(b) These m multiple eigenvalues give rise to less than m linearly independent eigenvectors, the matriz is not diagonizable. (vii) Rule 1. A matrix A is diagonalizable if [A, A+]= AA+- A+A=0. (ix) Rule 2. Consider two operators A and B, they can be diagonalized simultaneously with the same set of eigenvectors (but different eigenvalues) if and only if they commute, [A,B] = AB-BA =0. 2. Hermitian operator (i) An operator (i.e. linear transformation T: S →S) T is Hermitian if T+ = T (representation independent) (ii) Hermitian operator is important in quantum mechanics because any observable physical quantity, like energy and momentum, can be represented as a Hermitian operator. (iii) Rule 3. A Hermitian matrix is always diagonalizable and its eigenvalues are always real. (iv) Rule 4. Eigenvectors of a Hermitian operator with different eigenvalues are orthogonal to each other. For eigenvectors with the same eigenvector, they may not be orthogonal, but we can always make them orthogonal. 3. Unitary transformation (i) A matrix U is known as unitary if UU+= U+U = I (representation independent) (ii) U is reserved to denote unitary transformation. (iii) Unitary matrix is important in quantum mechanics because it is used for changing basis (representation). (iv) Rule 5. Unitary transformation refers to change between orthonormal bases. If any one of this basis is not orthonormal, the transformation matrix will not be unitary (so U is a particular case of ∆). In other words, unitary transformation “preserves” orthonormality. 4. Completeness and orthonormal (i) A set is said to be orthonormal if (ii) Assuming the original basis O={ ai, (The notes is incomplete!)