**How is the state of a quantum mechanical system described**?**
**(1) At a fixed time t

Consider two kets, |Ψ(t

But |Φ> = a

Global phase factors and multipliers do not affect the physical predictions, but relative phases and multipliers are significant.

**How can we predict the results of a measurement?
**(2) Every measurable physical quantity is described by a Hermitian operator acting in
ℰ.
(We want the eigenvectors of the operator to form a basis for the vector space and its eigenvalues to be real.)

(3) The

(4) When a physical quantity described by the operator A is measured on a system in a normalized state |Ψ>, the probability of measuring the eigenvalue a

P(a

where {|u

This may be written in terms of the projector

P(a

(5) If a measurement of a system in the state |Ψ> gives the result a

|Ψ> --> P

(6) The Cartesian components of the observables

(b) Symmetrize the expression with respect to

**r**∙**p**--> ½(**r**∙**p + p**∙**r**) -->**R**∙**P + P**∙**R)**

**
How is the state of a quantum mechanical system evolve?**

(7) The time evolution of the state vector is governed by the **
Schroedinger equation**,

(iħ∂/∂t)|Ψ(t)> = H(t)|Ψ(t)>,

where H(t) is the observable associated with the total energy of the system.

For a conservative system, where all the forces can be derived by taking the gradient
of a scalar potential, the classical Hamiltonian of a particle is written as

H = T + U = p^{2}/(2m) + U(**r**,t).

This is an expression for the total energy
of the system.

The quantum mechanical operator is found by replacing p^{2} with P^{2}
= P_{x}^{2} + P_{y}^{2} + P_{z}^{2}
and U(**r**,t) with U(**R**,t). No product of non-commuting operators
is involved, so symmetrization is not required. The Schroedinger equation
becomes

iħ∂/∂t)|Ψ(t)> = [P^{2}/(2m) + U(**R**,t)]|Ψ(t)>.

The Schroedinger equation is first order in t. Given |Ψ(t_{0})>, |Ψ(t)> is
uniquely determined. The quantum state evolves in a perfectly deterministic way between
measurements.

**How can we know |Ψ(t _{0})> ?
**If, at t = 0, we measure a complete set of commuting observables, then |Ψ(t

Two commuting observables can be measured simultaneously, i.e. the measurement of one does not cause loss of information obtained in the measurement of the other. If we measure a C.S.C.O., then the state of the system after the measurement is one element of an unique eigenbasis. The results of the measurement specify the state completely.

**Example:**

- Take two commuting observables, A and B, [A,B] = 0.
There
exists a common eigenbasis

{|a_{n},b_{p},i(n,p)>}.

If both operators have degenerate eigenvalues, then |a_{n},b_{p},i> denotes the ith eigenvector with eigenvalues a_{n}and b_{p}.

We have

A|a_{n},b_{p},i(n,p)> = a_{n}|a_{n},b_{p},i(n,p)>,

B|a_{n},b_{p},i(n,p)> = b_{p}|a_{n},b_{p},i(n,p)>.

Assume that the system is in the state

Ψ(t_{0})> = ∑_{n,p,i}C_{n,p,i}(t_{0})|a_{n,}b_{p},i> with <Ψ(t_{0})|Ψ(t_{0})> = 1.

At t = 0 you measure A.

The result will be one of the eigenvalues a_{n}and after the measurement the system will be in the normalized eigenstate

P_{n}|Ψ(t_{0})>/(<Ψ(t_{0})|P_{n}|Ψ(t_{0})>)^{1/2}(postulate 5).

P_{n}|Ψ(t_{0})> = ∑_{p,i}|a_{n,}b_{p},i><a_{n,}b_{p},i|∑_{n',p',i'}C_{n',p',i'}(t_{0})|a_{n',}b_{p'},i'> = ∑_{p,i}C_{n,p,i}(t_{0})a_{n,}b_{p},i>.

The probability of obtaining the eigenvalue a_{n}is P(a_{n}) = ∑_{p,i}|C_{n,p,i}(t_{0})|^{2}(postulate 4).

An infinitesimal time later you measure B.

The result of this measurement will be one of the eigenvalues b_{p}, and after the measurement the system will be in the corresponding normalized eigenstate

P_{p}P_{n}|Ψ(t_{0})>/(<Ψ(t_{0})|P_{p}P_{n}|Ψ(t_{0})>)^{1/2}.

P_{p}P_{n}is the projector onto the subspace of eigenvectors with eigenvalues a_{n}and b_{p}

P_{p}P_{n}|Ψ(t_{0})> = ∑_{n',i'}|a_{n',}b_{p},i'><a_{n',}b_{p},i'| ∑_{p'',i''}C_{n,p'',i''}(t_{0})|a_{n,}b_{p''},i''> = ∑_{i}C_{n,p,i}(t_{0})|a_{n,}b_{p},i>.

[P_{p},P_{n}] = 0. This only holds because [A,B] = 0 and a common eigenbasis exist. If [A,B] ≠ 0 then P_{p}P_{n}is not a projector and [P_{p},P_{n}] ≠ 0.

P_{p}P_{n}|Ψ(t_{0})> = ∑_{i}C_{n,p,i}(t_{0})|a_{n,}b_{p},i>.

P_{an}(b_{p}) = probability of obtaining b_{p}after obtaining a_{n}= ∑_{i}|C_{n,p,i}(t_{0})|^{2}/(∑_{pi}|C_{n,p,i}(t_{0})|^{2}).

P(a_{n},b_{p}) = probability of the composite event = ∑_{i}|C_{n,p,i}(t_{0})|^{2}= <Ψ(t_{0})|P_{p}P_{n}|Ψ(t_{0})>.

The measurement of B does not cause loss of information obtained in the measurement of A (or vice versa), on the contrary, it adds to it. After the measurement of A we know the system is in an eigenstate of A, after the measurement of B we know the system is in an eigenstate of A and in an eigenstate of B.

In the Schroedinger picture of quantum mechanics the state vector |Ψ> of the system evolves in time. Its evolution is described by the Schroedinger equation (iħ∂/∂t)|Ψ(t)> = H(t)|Ψ(t)>. Results of measurements are predicted by letting operators, which do not evolve in time (which are time independent unless they contain time explicitly), operate on the evolving state vector. The predictions of quantum mechanics (the probabilities of obtaining given results if measurements are made) are expressed in terms of scalar products and matrix elements of operators.

Let {|a_{i}>} be a unique eigenbasis of the operator A, A|a_{i}>
= a_{i}|a_{i}>,
and let |Ψ> be a normalized state vector.
The
probability of obtaining a_{i} as a result of a measurement of A is P(a_{i})
= |<a_{i}|Ψ>|^{2} and the average value of the outcome of a
measurement of A is

<A> = ∑_{i}a_{i}P(a_{n}) = ∑_{i}a_{i}<Ψ|a_{i}><a_{i}|Ψ>
= ∑_{i}<Ψ|A|a_{i}><a_{i}||Ψ> = <Ψ|A|Ψ>.

**
These predictions of quantum mechanics do not depend on the representation, i.e. which
basis we choose to express our state vector.**