Quantum Mechanics

Quantum mechanics is a theory that describes the interactions of all particles and systems. It underlies all physical phenomena, including scattering.

Wavefunction

A quantum system is completely specified by its Wave Function:

Integral Notation	Dirac Notation
$\psi (x)$	$\langle x|\psi \rangle$

The wavefunction is typically normalized:

Integral Notation	Dirac Notation
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int |\psi (x)|^{2}\mathrm {d} x=1}	$\langle \psi |\psi \rangle =1$

The distribution of the particle described by $\psi (x)$ is given by:

Integral Notation	Dirac Notation
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \Pr(x)\mathrm {d} x=|\psi (x)|^{2}\mathrm {d} x}	Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |\langle x|\psi \rangle |^{2}}

In the Copenhagen Interpretation, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \Pr(x)} is the probability of finding the particle at location $x$. In Universal Wave Function interpretations (e.g. MWI), Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \Pr(x)} can be thought of as the spatial distribution of the particle. The wavefunction contains all the information one can know about a system. It can thus be thought of as 'being' the particle/system in question. However, the wavefunction can be described in an infinite number of different ways. That is, there is not a unique basis for describing the wavefunction. So, for instance, one can describe the wavefunction using position-space or momentum-space:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \psi (x)\longleftrightarrow {\tilde {\psi }}(k)}

These representations can be inter-related (c.f. Fourier transform):

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \psi (x)={\frac {1}{\sqrt {2\pi }}}\int {\tilde {\psi }}(k)e^{ikx}\mathrm {d} k}

${\tilde {\psi }}(k)={\frac {1}{\sqrt {2\pi }}}\int {\psi }(x)e^{-ikx}\mathrm {d} x$

State

Note that the wavefunction describes the state of the system; there are various choices of basis one can use as an expansion.

$\psi =\sum _{n}c_{n}\psi _{n}$

A basis should be orthonormal:

Integral Notation	Dirac Notation
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int |\psi _{n}(x)|^{2}\mathrm {d} x=1}	Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \langle \psi _{n}|\psi _{n}\rangle =1}	normalized
$\int \psi _{m}(x)^{*}\psi _{n}(x)\mathrm {d} x=0$	Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \langle \psi _{m}|\psi _{n}\rangle =0}	orthogonal

An operator defines a particular convenient basis: one can always expand $\psi$ using the basis defined by an operator, in which case the Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \psi _{n}} above are the eigenvectors (or eigenstates) of that basis. This can also be viewed as a vector in the Hilbert space. The Dirac notation (bra-ket notation) is useful in this regard. A particular state is a (column) vector:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |\psi \rangle ={\begin{bmatrix}c_{1}\\c_{2}\\\vdots \\c_{n}\end{bmatrix}}}

Which is a 'ket'. We define a 'bra' (the 'final state') as a (row) vector:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \langle \psi |={\begin{bmatrix}c_{1}^{*}&c_{2}^{*}&\dots &c_{n}^{*}\end{bmatrix}}}

And note that the 'bra' is the conjuagte transpose of the 'ket':

$\langle \psi |^{\dagger }=|\psi \rangle$

Wave packet

TBD

Heisenberg Indeterminacy Relations

(Also known as Heisenberg Uncertainty Principle.)

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \Delta _{x}\Delta _{p}\geq {\frac {\hbar }{2}}}

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \Delta _{E}\Delta _{t}\geq {\frac {\hbar }{2}}}

Superposition

If Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \psi _{1}(x)} and $\psi _{2}(x)$ are both allowed states for a given system, then the following state is also allowed:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \psi (x)=\alpha \psi _{1}(x)+\beta \psi _{2}(x)}

This leads to a notable consequence:

${\begin{alignedat}{2}\Pr(x)&=|\alpha \psi _{1}(x)+\beta \psi _{2}(x)|^{2}\\&=(\alpha \psi _{1}+\beta \psi _{2})(\alpha \psi _{1}+\beta \psi _{2})^{*}\\&=|\alpha |^{2}|\psi _{1}|^{2}+|\beta |^{2}\psi _{2}^{2}+\alpha \beta ^{*}\psi _{1}\psi _{2}^{*}+\alpha ^{*}\beta \psi _{1}^{*}\psi _{2}\\&=\mathrm {Pr} _{1}(x)+\mathrm {Pr} _{2}(x)+\mathrm {interference} \\\end{alignedat}}$

Notice that the final terms represent 'interference' between the two constituent states. This interference has no classical analogue; it is a quantum effect. Thus a superposition is not merely a 'joining' of the two states (e.g. "the particle can be in state 1 or state 2"), but a truly coherent interference between the two states. The superposition may be more generally written as:

Integral Notation	Dirac Notation
$\int |\psi (x)|^{2}\mathrm {d} x=1$	$\langle \psi |\psi \rangle =1$

The distribution of the particle described by $\psi (x)$ is given by:

Integral Notation	Dirac Notation
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \psi (x)=\sum _{n}c_{n}\psi _{n}}	Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |\psi \rangle =c_{1}|1\rangle +c_{2}|2\rangle +c_{3}|3\rangle +\cdots }

Operators

Observables in QM appears as operators (${\hat {O}}$).

Examples: TBD.

Measurement

The transition of the wavefunction $\psi$ into state Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \phi } can be thought of as:

$\int \phi ^{*}\psi \mathrm {d} x$

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \langle \phi |\psi \rangle =a_{1}^{*}c_{1}+a_{2}^{*}c_{2}+a_{3}^{*}c_{3}+\cdots }

When acting on a wavefunction with operator ${\hat {O}}$ the probability that the wavefunction ends up in state Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \phi _{n}} is given by:

$\Pr(O_{n})=|c_{n}|^{2}$

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \Pr(O_{n})=|\langle n|\psi \rangle |^{2}=|c_{n}|^{2}}

The solutions take the form of an eigenvalue problem:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {O}}\phi _{n}=o_{n}\phi _{n}}

The allowed solutions of the equation, for operator ${\hat {O}}$, involve an eigenstate Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \phi _{n}} with associated eigenvalue Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle o_{n}} . A measurement on a quantum system can be thought of as driving the wavefunction into an eigenstate defined by the operator; the value of the associated observable is then fixed to be the corresponding eigenvalue. (As noted above, the probability of ending up in a particular eigenstate is regulated by the coefficient of that eigenstate in the original wavefunction decomposition.)

Expectation value

A given operator, e.g. Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {A}}} , implies an expectation value of:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \langle A\rangle _{\psi }=\int \psi ^{*}{\hat {A}}\psi \mathrm {d} x}

$\langle A\rangle _{\psi }=\langle \psi |{\hat {A}}|\psi \rangle$

If the system is in an eigenstate of the operator:

$\psi =\sum _{n}c_{n}\psi _{n}=\psi _{n}$

We know that:

${\hat {A}}\psi _{n}=a_{n}\psi _{n}$

And so:

${\begin{alignedat}{2}\langle A\rangle &=\int \psi _{n}^{*}{\hat {A}}\psi _{n}\mathrm {d} x\\&=\int \psi ^{*}a_{n}\psi \mathrm {d} x\\&=a_{n}\int \psi ^{*}\psi \mathrm {d} x\\&=a_{n}\\\end{alignedat}}$

In other words, the expectation value of an eigenstate is simply the eigenvalue.

Schrödinger Equation

TBD

Entanglement

TBD

Decoherence

TBD

See Also

• Wikipedia: Quantum Mechanics