Quantum State Vectors

An extended comparison of quantum state 'vectors', and the regular ol' physics vectors.

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Vectors

The state of a quantum system--all we can know about it--is represented by a "ket", written $$\ket{\psi}.\nonumber$$ That's the greek letter $\psi$, spelled as "psi" and pronounced as sigh.

Many physics quantities can be represented by vectors. $$\myv A\nonumber.$$

Completeness

Any possible quantum state (completeness) can be described as a linear superposition of basis states. The basis states of the $Z$ analyzer are $\ket +$, (spin up, or $S_z=+\hbar/2$) and $\ket -$ (spin down, or $S_z=-\hbar/2$). So any $\ket \psi$ of an electron (a spin-1/2 particle) can be written as $$\ket \psi = a\ket + + b\ket -.$$ where $a$ and $b$ are complex numbers.

There are only 2 possible outcomes for a measurement of of the $z$ component of the spin of an electron. But there are other fundamental particles that have 3 or more possible $S_z$ components, and this scheme can readily be generalized to describe state vectors in any old $n$-dimensional space.

Any vector (completeness) can be written as a linear sum of unit vectors in a particular coordinate system: $$\myv A=A_x\uv i+A_y\uv j,$$ where $A_x$ and $A_y$ are real numbers.

In addition to 2-d vectors, we routinely use 3-d vectors to describe positions, velocities, etc, in 3-d space. In special relativity we can talk about 4-d space-time vectors. And in principle this scheme can be generalized to any desired number of dimensions.

row vector: $$\ket \psi^\dagger\equiv \bra \psi\equiv \begincv a^* & b^*\endcv.$$

A vector $\myv A=A_x\uv i+A_y\uv j$ can be written more compactly as a column vector, containing just its components: $$\myv A \equiv \begincv A_x \\ A_y \endcv;\ \ \uv i\equiv \begincv 1\\0\endcv; \ \ \uv j\equiv \begincv 0\\1\endcv.$$ The Hermitian conjugate of the vector $\myv A$ is $\myv A^\dagger \equiv \begincv A_x^* & A_y^*\endcv$. But since the components of $\myv A$ are real, this is just... $$\myv A^\dagger\equiv \begincv A_x & A_y\endcv.$$

Dot product / "inner" product

Matrix multiplication: multiplying a row vector times a column vector results in a scalar--the inner product: $$\begincv a & b\endcv\begincv c\\d \endcv =ac+bd.$$

The inner product of two quantum states, $\innerp{\phi}{\psi}$ is a "bra"|"ket" (tee-hee: bracket!). The dot product of a quantum state with itself: $$\begineq \innerp{\psi}{\psi}&=\begincv a^* & b^*\endcv \begincv a\\b \endcv\\ &=a^*a+b^*b=|a^2|+|b^2|.\endeq$$ Norm of a complex number:... Let's write out the complex number $a$ as $a=x+yi$, where $x$ and $y$ are real numbers, such that $x$ is the real part of $a$ and $y$ is the imaginary part.

Viewing $a$ as a vector in the 'complex plane', the length of this vector is $\sqrt{x^2+y^2}$. You can calculate the (length)^2 by multiplying a complex number by its complex conjugate: $$\begineq aa^*\equiv|a|^2&=(x+yi)(x-yi)=\\ &=x^2+xyi-xyi-y^2\times i^2\\ &=x^2-y^2\times (-1)=x^2+y^2. \endeq$$

The inner product, or dot product of two vectors: $$\myv A\cdot\myv B=\begincv A_x&A_y\endcv\begincv B_x\\B_y\endcv=A_xB_x+A_yB_y.$$ The dot product of a vector with itself is $$\myv A\cdot\myv A=A_x^2+A_y^2$$.

The length (or "norm") of a vector is a scalar, written as $|\myv A|$ or $A$.

Using Pythagoras' theorem, we know how to calculate the length from the Cartesian components, and we can write it in terms of a dot product... $$A = \sqrt{A_x^2+A_y^2} =\sqrt{\myv A\cdot\myv A}.$$

Unit vectors

Your text points out that one difference between regular vectors and quantum state vectors is that...

All physical quantum states should be normalized to 1. That is: $$\innerp{\psi}{\psi}=1.$$

Therefore, if you are given some quantum state, $\psi_u=a\ket + + b\ket -$, which might be "unnormalized", you can normalize it by dividing by the 'length' of the un-normalized vector. So, this state vector is always guaranteed to be normalized: $$\ket \psi = \frac{\ket{\psi_u}}{\sqrt{\innerp{\psi_u}{\psi_u}}} = \frac{\ket{\psi_u}}{\sqrt{a^*a+b^*b}}.$$

Given a vector $\myv A$, you can construct a unit vector, $\uv A$, that points in the same direction, but has a length of 1, like this: $$\uv A = \frac{\myv A}{A} = \frac{\myv A}{\sqrt{\myv A\cdot\myv A}}.$$

Apparently, if a quantum state is already normalized to 1, $\innerp{\psi}{\psi}=1$, we must have: $$a^*a+b^*b =|a|^2+|b|^2 = 1.$$

Homework

Chapt 1: Problems 1ac, 3

Examples 1.1 and 1.2 also show normalization.