# Quantum Mechanics Study Notes 1: basic rules

/************ Definitions ************/

Math definition 1: A linear operator X has a number of eigenvalues x1, x2, …, xn, and their corresponding eigenstates, |x1>, |x2>, …, |xn>, which satisfies:

X|x1> = x1|x1>
X|x2> = x2|x2>

X|xn> = xn|xn>

Those eigenvalues are complex numbers.

Math definition 2: each vector |x> comes with a dual vector <x|, and a dot product system, such that

1. <x|y> = (<y|x>)*
2. <x|(c|y>)=(c<x|)y> where c is a complex number.
3. <x|(|y>+|z>)=<x|y>+<x|z>

Because of 1, we know <x|x> is always a real number. Define it as the norm of |x>.

Example of such systems: |x> is a linear vector, <x| is the complex conjugate of its transpose, multiplied by any constant c. Any c will work for QM — the absolute value of <x|y>does not really matter; we just care about their relative scales. In other words, we just care about removing a dimension of freedom.

For convenience, from now on, we would go without saying that all vectors are normalized.

Math definition 3: <a|X is such a thing that, for any |b>, (<a|X)|b> = <a|(X|b>).

Math definition 4: For an operator X, X* is such a thing that, for any |a> and |b>, <a|X|b> = (<b|X*|a>)*.

Math definition 5: operator X is called “Hermitian” if X = X*.

/************ Theorems ************/

Math theorem 1: if |y> = a|x>, then <y|=a*<x|.

Proof: take any vector |z>, we have a*<x|z> = a*(<z|x>)*=(a<z|x>)*=(<z|y>)*=<y|z>.

Math theorem 2: any vector |a> can be expressed as sum of eigenvectors: |a> = a1|x1>+a2|x2>+…+an|xn>.

Proof: (missing… )

Math theorem 3: Hermitian operator only has real eigenvalues.

Proof: <a|X|a> = <a|X*|a> = (<a|X|a>)*, therefore <a|X|a> is a real number for any |a>.
Now if |a> is a eigenvector, then <a|X|a>= <a|(a|a>)=a<a|a>, and <a|a> is a real number, therefore a is a real number.

Math theorem 4: for a hermitian X and its eigenvector |a>, <a|X = a*<a|.

Proof: for any |b>, <a|X|b> = (<b|X*|a>)*=(<b|X|a>)*=(a<b|a>)*=a*<a|b>.

Math theorem 5: if two eigenstates have different eigenvalues, then their dot product is 0.

Proof: <xi|X|xj> = <xi|(xj|xj>) = xj<xi|xj>

Also, <xi|X|xj> = (<xi|X)|xj> = xi<xi|xj>.

Therefore, xi = xj or <xi|xj> = 0.

/************ QM Rules ************/

QM Rule 1: An observable is represented by a hermitian operator X.

QM Rule 2: Eigenvalues {x1, … , xn} are all the possible outcomes of measurement X. It could be discrete, e.g. spin of an electron, or infinte and continuous, e.g. position of a particle.

QM Rule 3: A state is represented by a vector in that same linear space, e.g. |a>.

QM Rule 4: expand|a> = a1|x1>+a2|x2>+…+an|xn>, where ai are complex numbers, and|xi> are eigenvectors with norm 1, then the probability of measuring X on |a> and get xi outcome is ai*ai.

Note on QM Rule 4: Sum ai*ai = 1 is guaranteed.

This is required o make ai*ai a plausible probability distribution. Luckily this is guaranteed by the normalization rules above: 1=<a|a> = Sum_i_j ai*aj<xi|xj> = Sum_i ai*ai <xi|xj> = Sum ai*ai.

However, this normalization rule is by no means required by QM. If we don’t require normalization, then the probability of getting xi in measurement would simply become ai*ai<xi|xi>/<a|a>.

QM Rule 4 (continuous space): expand|a> = S f(x)|x>dx, where f(x) are complex numbers, and the norm of |x> are all 1, then the probability of measuring X on |a> and get x in range of u, v is S f(x)*f(x)dx.

Random thoughts of a random guy. I know that I know nothing.

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Random thoughts of a random guy. I know that I know nothing.