Quantum mechanics is probabilistic. We cannot predict the results of an experiment precisely, only the probability that a certain result is measured.
Spin measurements are quantized. Only discrete results are measured.
Quantum measurements disturb the system. Measuring one physical observable destroys existing information about other observable.
The state of a quantum mechanical system, including all the information you can know about it, is represented mathematically by a normalized ket $\ket{\psi}$.
A physical observable is represented mathematically by an operator $A$ that acts on kets.
The only possible result of a measurement of an observable $A$ is one of its eigenvalues $a_n$.
The probability of obtaining of obtaining the eigenvalue $a_n$ in a measurement of an observable $A$ on the system in the state $\ket{\psi}$ is $$P_{a_n}=|\braket{a_n}{\psi}|^2$$.
After a measurement $A$ with eigenvalue $a_n$, the new state is the normalized projection of the original state onto the measurement's ket(s). $$\ket{\psi'}=\frac{P_n\ket{\psi}}{\sqrt{\braketo{\psi}{P_n}{\psi}}}$$.
The time evolution of a quantum system is determined by the Hamiltonian (total energy operator), $H(t)$, through the Schrödinger equation:
$$i\hbar\frac{d}{dt}\ket{\psi(t)}=H(t)\ket{\psi(t)}$$
Postulate 1
The state of a quantum mechanical system, including all the information you can know about it, is represented mathematically by a normalized ket $\ket{\psi}$.
In 2D-space, two kets, $\ket{\pm}$ form a basis. The basis kets have 3 properties:
Completeness: $\ket{\psi}=a\ket{+}+b\ket{-}$, $a, b$ are complex scalars.
The general quantum state vector $\ket{\psi}$ can be written as a linear combination of the two basis kets.
This is referred to as a superposition of states.
The complex conjugate vector of the general state vector is:
$$\bra{\psi}=a^*\bra{+}+b^*\bra{-}$$
By convention, state vectors are written in the $S_z$ basis, and $\ket{\pm}$ represent the measurements along the $z$-axis.
Ex 1. Determine $\braket{+}{\psi}$.
$$\begin{align*}
\braket{+}{\psi}&=\bra{+}(a\ket{+}+b\ket{-})\\
&=\braketo{+}{a}{+} + \braketo{+}{b}{-}\\
&= a\braket{a}{a} + b\braket{+}{-}\\
&= a
\end{align*}$$
* Note that $\braket{-}{\psi}=b$
We can also show that $\braket{\phi}{\psi}=\braket{\psi}{\phi}^*$.
From Postulate 1, we require all state vectors to be normalized. Given a state vector, here's how we normalize it:
$$\begin{align*}
\braket{\psi}{\psi}&=(a^*\bra{+}+b^*\bra{-})(a\ket{+}+b\ket{-})\\
1&=a^*a\braket{+}{+} + a^*b\braket{+}{-} + b^*a\braket{-}{+} + b^*b\braket{-}{-}\\
1&= |a|^2+ |b|^2\quad\text{OR}\\
1&= |\braket{+}{\psi}|^2 + |\braket{-}{\psi}|^2
\end{align*}$$
Each term in the above sum represents the probability that the general quantum state is measured in the corresponding basis state.
$$P_{S_z=+\hbar/2}=|\braket{+}{\psi}|^2\quad\quad P_{S_z=-\hbar/2}=|\braket{-}{\psi}|^2$$
Postulate 4
The probability of obtaining of obtaining the eigenvalue $a_n$ in a measurement of an observable $A$ on the system in the state $\ket{\psi}$ is
$$P_{a_n}=|\braket{a_n}{\psi}|^2$$
The inner product, $\braket{+}{\psi}$ is called the probability amplitude. The probability amplitude, when written in Bra-Ket notation, represents the input state on the right and the output state on the left. (i.e. $\braket{+}{\psi}$ represents measuring an input general state vector in the $+z$ direction)
If we have a state vector in the $S_z$ basis and measure it along the x-axis, we get the following probabilities:
$$P_{+,+x}=|_x\braket{+}{+}|^2=\frac{1}{2},\quad P_{+,-x}=|_x\braket{-}{+}|^2=\frac{1}{2}\\ P_{-,+x}=|_x\braket{+}{-}|^2=\frac{1}{2},\quad
P_{-,-x}=|_x\braket{-}{-}|^2=\frac{1}{2}$$
By the completeness property, each ket written in the $S_x$ basis can be written as a superposition of basis kets. Plugging them in to the equations above, we get complex coefficients which we can write as:
$$\ket{+}_x=\frac{1}{\sqrt{2}}(\ket{+}+e^{i\alpha}\ket{-}),\quad\ket{-}_x=\frac{1}{\sqrt{2}}(\ket{+}+e^{i\beta}\ket{-})$$
We can choose one of the coefficients to be real and positive because the overall phase is irrelevant.
Further applying the orthonormality property onto the $S_x$ kets, we get that $e^{i\alpha}=-e^{i\beta}$. Choosing $\alpha=0$, we get
$$\ket{+}_x=\frac{1}{\sqrt{2}}(\ket{+}+\ket{-}),\quad\ket{-}_x=\frac{1}{\sqrt{2}}(\ket{+}-\ket{-})$$
In a similar way, together with the additional requirement that $|_y\braket{+}{+}_x|^2=\frac{1}{2}$, we get that
$$\ket{+}_y=\frac{1}{\sqrt{2}}(\ket{+}+i\ket{-}),\quad\ket{-}_y=\frac{1}{\sqrt{2}}(\ket{+}-i\ket{-})$$
State vectors can be represented as matrices. By convention, they are written in the $S_z$ basis. Ket vectors are represented as column vectors. The basis kets are represented as:
$$\ket{+}\doteq\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right),\quad \ket{-}\doteq\left(\begin{smallmatrix}0\\1\end{smallmatrix}\right)$$
Basis kets are unit vectors when written in their own basis.
The top element represents the amplitude in the spin-up state and the bottom element represents that in the spin-down state. The $S_x$ kets can be written as:
$$\ket{+}_x\doteq\frac{1}{\sqrt{2}}\left(\begin{smallmatrix}1\\1\end{smallmatrix}\right),\quad \ket{-}_x\doteq\frac{1}{\sqrt{2}}\left(\begin{smallmatrix}1\\-1\end{smallmatrix}\right)$$
Thus, we get:
The general state vector is represented as:
$$\ket{\psi}\doteq\left(\begin{smallmatrix}\braket{+}{\psi}\\\braket{-}{\psi}\end{smallmatrix}\right)$$
i.e. if $\ket{\psi}=a\ket{+}+b\ket{-}$, then $\ket{\psi}\doteq\left(\begin{smallmatrix}a\\b\end{smallmatrix}\right)$.
Conversely, bra vectors are represented as row vectors. So, $\bra{\psi}=a^*\bra{+}+b^*\bra{-}$, then $\bra{\psi}\doteq\left(\begin{smallmatrix}a^*&b^*\end{smallmatrix}\right)$.
Writing the state vectors in matrix notation is convenient because we can use the rules of linear algebra. For example,
$$\begin{align*}
\braket{\psi}{\psi}&=\left(\begin{smallmatrix}a^*&b^*\end{smallmatrix}\right)\left(\begin{smallmatrix}a\\b\end{smallmatrix}\right)\\
&=|a|^2+|b|^2
\end{align*}$$
A bra is represented by a row vector that is the complex conjugate and transpose of the column vector representing its corresponding ket. Tip: To remember whether a ket/bra is a column or row vector, follow the orientation of the starting symbol of the ket/bra
Ket: starts with a vertical bar ( $|$ ) == column vector
Bra: starts with a left angle braket ( $\langle$ ) == row vector
The properties of state vectors in a spin-$\frac{1}{2}$ system can be generalized to other quantum systems. Consider a system with an observable $A$ that has measurements $a_n$ for some finite $n$. We have the basis kets as $\ket{a_i}$ for $1\leq i\leq n$.
$$\braket{a_i}{a_j}=\delta_{ij}\quad\text{orthonormality}$$
$$\ket{\psi}=\sum_i\braket{a_i}{\psi}\ket{a_i}\quad\text{completeness}$$
An operator is a mathematical object that acts on a ket and transforms it into a new ket.
Postulate 2
A physical observable is represented mathematically by an operator $A$ that acts on kets.
There are special kets called eigenvectors that are not changed by a particular operator $A$, except for a multiplicative constant (eigenvalues). i.e. $A\ket{\psi}=a\ket{\psi}$.
Postulate 3
The only possible result of a measurement of an observable is one of the eigenvalues $a_n$ of the corresponding operator $A$.
For example, with a physical observable $S_z$, we get the following eigenvalue equations:
$$S_z\ket{+}=+\frac{\hbar}{2}\ket{+}\quad S_z\ket{-}=-\frac{\hbar}{2}\ket{-}$$
Matrix Representation of Operators
In matrix notation, operators are represented by an $n \times n$ matrix. For example, $S_z$ is represented by the following $2 \times 2$ matrix:
$$S_z\doteq\frac{\hbar}{2}\left(\begin{matrix}1&0\\0&-1\end{matrix}\right)$$
An operator is always diagonal in its own basis, with its diagonal elements as its eigenvalues.
For an observable $A$ in a spin-$\frac{1}{2}$ system, we have
$$A\doteq\left(\begin{matrix}\braketo{+}{A}{+}&\braketo{+}{A}{-}\\\braketo{-}{A}{+}&\braketo{-}{A}{-}\end{matrix}\right)$$
In general, $A_{ij}=\braketo{i}{A}{j}$, where $i$ and $j$ are the rows and columns respectively.
A general direction $\hat{n}$, a unit vector, is represented in spherical coordinates as:
$$\hat{n}=\hat{i}\sin\theta\cos\phi +\hat{j}\sin\theta\sin\phi + \hat{k}\cos\theta$$
The spin component along $\hat{n}$ is obtained by projecting the spin vector $S$ onto it:
$$\begin{align*}
S_n&=S\cdot\hat{n}\\
&=S_x\sin\theta\cos\phi + S_y\sin\theta\sin\phi + S_z\cos\theta\\
S_n&\doteq\frac{\hbar}{2}\left(\begin{matrix}\cos\theta&\sin\theta e^{-i\phi}\\\sin\theta e^{i\phi}&-\cos\theta\end{matrix}\right)
\end{align*}$$
This method is used to find the eigenvalues and eigenvaectors given an operator. We start with the eigenvalue equation: $A\ket{a_n}=a_n\ket{a_n}$
Rewrite the eigenvalue equation as $(A - a_nI)\ket{a_n}=0$ and express it in matrix notation.
$$\left(\begin{matrix}A_{11}-a_n&A_{12}\\A_{21}&A_{22}-a_n\end{matrix}\right)\left(\begin{matrix}c_1\\c_2\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$
We get two sets of homogenous equations, which yield non-trivial solutions if the determinant of the coefficient matrix is zero. Letting $\lambda=a_n$, we get the characteristic equation from $\det(A-\lambda I)=0$.
Solve the characteristic equation for $\lambda$. We get two roots which are the eigenvalues $a_1$ and $a_2$.
Substitute it back into the eigenvalue equation and solve for the eigenvectors corresponding to each eigenvalue.
Ex 2. Find the eigenvalues and eigenvectors for $S_y$.
Rewrite the eigenvalue equation as $(S_y - \lambda I)\ket{\lambda}=0$
Solve for the characteristic equation.
$$\begin{align*}
\det\left(\begin{matrix}-\lambda&-i\frac{\hbar}{2}\\i\frac{\hbar}{2}&-\lambda\end{matrix}\right)&=0\\
\lambda^2-\bigg(\frac{\hbar}{2}\bigg)^2&=0
\end{align*}$$
Solve the characteristic equation for the eigenvalues:
$\lambda=\pm\frac{\hbar}{2}$
Substitute each eigenvalue back into the eigenvalue equation and solve for the corresponding eigenvector.
For $+\frac{\hbar}{2}$,
$$\begin{align*}
S_y\ket{+}_y&=+\frac{\hbar}{2}\ket{+}_y\\
\frac{\hbar}{2}\left(\begin{matrix}0&-i\\i&0\end{matrix}\right)\left(\begin{matrix}a\\b \end{matrix}\right)&=+\frac{\hbar}{2}\left(\begin{matrix}a\\b \end{matrix}\right)\\
\left(\begin{matrix}-ib\\ia \end{matrix}\right)&=\left(\begin{matrix}a\\b \end{matrix}\right)
\end{align*}$$
Solving this, we get: $a=\frac{1}{\sqrt{2}}$ and $b=i\frac{1}{\sqrt{2}}$. Thus, the eigenvector corresponding to the $+\frac{\hbar}{2}$ eigenvalue is $\ket{+}_y\doteq\frac{1}{\sqrt{2}}\left(\begin{matrix}1\\i \end{matrix}\right)$.
Likewise, we can find that $\ket{-}_y\doteq\frac{1}{\sqrt{2}}\left(\begin{matrix}1\\-i \end{matrix}\right)$.
Diagonalizing the operator $S_n$ gives us:
$$\ket{+}_n\doteq\cos\frac{\theta}{2}\ket{+}+\sin\frac{\theta}{2}e^{i\phi}\ket{-},\quad\ket{-}_n\doteq\sin\frac{\theta}{2}\ket{+}-\cos\frac{\theta}{2}e^{i\phi}\ket{-}$$
Similar to how operators act on the left of a bra, operators act on the right of kets (i.e. $\bra{\zeta} = \bra{\psi}A$ ). Note that $\bra{\zeta}$ does not necessarily equal $\ket{\phi}=A\ket{\psi}$. In order to find that corresponding $\bra{\phi}$, we use the hermitian adjoint of the operator $A$, $\ud{A}$, such that $\bra{\phi} = \bra{\psi}\ud{A}$.
The Hermitian adjoint, $\ud{A}$ is found by transposing and taking the complex conjugate of $A$.
An operator is Hermitian if $A=\ud{A}$.
In QM, all physical observables are Hermitian operators.
Hermitian operators have real eigenvalues. This ensures that the results of measurements are always real.
The eigenvectors of a Hermitian matrix comprise a set of complete basis states. This ensures that we can use the eigenvectors of any observable as a valid basis.
A ket in the $S_z$ basis can be written as
$$\begin{align*}
\ket{\psi}&=a\ket{+}+b\ket{-}\\
&=(\braket{+}{\psi})\ket{+}+(\braket{-}{\psi})\ket{-}\\
&=(\ketbra{+}{+} + \ketbra{-}{-})\ket{\psi}
\end{align*}$$
Completeness Relation
$$\quad\ketbra{+}{+} + \ketbra{-}{-}=\mathbb{I}$$
The above relation is true for any basis. Each term in the relation, known as an outer product, is a projection operator. For spin-$\frac{1}{2}$ systems, the $S_z$ projection operators are
$$\mathbb{P}_+=\ketbra{+}{+}\doteq\left(\begin{matrix}1&0\\0&0\end{matrix}\right), \quad \mathbb{P}_-=\ketbra{-}{-}\doteq\left(\begin{matrix}0&0\\0&1\end{matrix}\right)$$
Note that a projector acting on its corresponding eigenstate results in the eigenstate, whereas if it acts on an orthogonal eigenstate results in zero:
$$\mathbb{P}_+\ket{+}=\ketbra{+}{+}\ket{+}=\ket{+}, \quad \mathbb{P}_-\ket{-}=\ketbra{-}{-}\ket{-}=\ket{-}$$
Spectral Representation
In general, any operatorcan be written as a linear combination of its projection operators:
$$A=\sum_n a_n\ketbra{\psi_n}{\psi_n}$$
where $a_n$ is the corresponding eigenvalue.
Ex 3. Write $S_x$ in its spectral representation and show that it equals its matrix representation.
Another use of projection operators is in calculating probabilities.
$$\begin{align*}
P_n&=|{}_n\braket{+}{\psi}|^2\\
&= {}_n\braket{+}{\psi}^*_n\braket{+}{\psi}\\
&=\braket{\psi}{+}_{nn}\braket{+}{\psi}\\
&=\braketo{\psi}{\mathbb{P}_n}{\psi}
\end{align*}$$
The probability of a measurement of a particular state $n$ from an input state $\ket{\psi}$ can be calculated using the matrix element of the projection operator $P_n$. For example, the probability of the measurement $S_z=\frac{\hbar}{2}$:
$$\begin{align*}
P_+&=|\braket{+}{\psi}|^2\\
&=\braket{\psi}{+}\braket{+}{\psi}\\
&=\braketo{\psi}{\mathbb{P}_+}{\psi}
\end{align*}$$
Projection Postulate
We can also view this representation of probability as the projection operator projecting the input ket onto the direction of the projection, with a multiplicative constant of the probability amplitude. Normalizing by dividing by the probability amplitude (square root of probability), we can represent the new state as
$$\ket{\psi'}=\frac{\mathbb{P}_n\ket{\psi}}{\sqrt{\braketo{\psi}{\mathbb{P}_n}{\psi}}}$$
Postulate 5
After a measurement of $A$ that yields the result $a_n$, the system is in a new state: the normalized projection of the input state onto the ket corresponding to the measurement.
$$\ket{\psi'}=\frac{\mathbb{P}_n\ket{\psi}}{\sqrt{\braketo{\psi}{\mathbb{P}_n}{\psi}}}$$
Note that the expectation value is not the expected result of a single experiment. Rather, it is the mean result of a large number of experiments on identically-prepared systems.
Ex 4. Calculate the expectation value of $S_z$ for a system prepared in the state $\ket{+}_x$.
$$\begin{align*}
\angle{S_z}&={}_x\braketo{+}{S_z}{+}_x\\
&\doteq\frac{1}{\sqrt{2}}\left(\begin{matrix}1&1\end{matrix}\right)\frac{\hbar}{2}\left(\begin{matrix}1&0\\0&-1\end{matrix}\right)\frac{1}{\sqrt{2}}\left(\begin{matrix}1\\1\end{matrix}\right)\\
&=\frac{\hbar}{4}\left(\begin{matrix}1&1\end{matrix}\right)\left(\begin{matrix}1\\-1\end{matrix}\right)\\
&=0
\end{align*}$$
As a sanity check, we can check that the expectation value is real and is between $-\frac{\hbar}{2}$ and $\frac{\hbar}{2}$.
Standard Deviation
$$\Delta A=\sqrt{\angle{A^2}-\angle{A}^2}$$
where $\angle{A^2}=\braketo{\psi}{A^2}{\psi}$.
Ex 5. Calculate the standard deviation of $S_x$ for an input ket $\ket{+}$.
Two observables commute if their commutator is equal to zero. If two operators commute, they have a common set of eigenstates.
Commuting observables share common eigenstates.
Given two commuting observables $A$ and $B$, if we take a measurement of $A$ to get a state $\ket{a}$ and then take a subsequent measurement of $B$, the second measurement does not change the state of $\ket{a}$. We say that the 2 observables can be measured simultaneously. Thus, we can measure one observable without "erasing" our previous knowledge of the other observable.
If two observables do not commute, they are incompatible and cannot be measured simultaneously. For example, $S_x, S_y, S_z$ do not commute and hence measuring one observable and then measuring another disturbs the original state.
Pure quantum states are states that are prepared through a projection measurement (i.e. prepared by a ket $\ket{\psi}$). They cannot be expressed as a mixture of other states.
Mixed quantum states are a probabilistic mixture of pure states (i.e. a probabilistic mixture of states $\ket{\psi_i}$).
Mixed states are not the same as a coherent superposition.
State is one of the states but we don't know which one
Expectation Value
Consider an experiment that produces states $\ket{\psi_n}$ with probabilities $p_i$ such that $\sum_i p_i=1$. Note that the states $\ket{\psi_i}$ need not be orthogonal. How do we calculate the expectation value of $S_n$, $[S_n]$, of this mixed state?
$$\begin{align*}
[S_n]&=\sum_i p_i\braketo{\psi_i}{S_n}{\psi_i}\\
&=\sum_i p_i\braketo{\psi_i}{(\frac{\hbar}{2}\ket{+}_{nn}\bra{+}-\frac{\hbar}{2}\ket{-}_{nn}\bra{-})}{\psi_i}\\
&=\sum_i p_i\braketo{\psi_i}{(S_n\ket{+}_{nn}\bra{+}-S_n\ket{-}_{nn}\bra{-})}{\psi_i}\\
&=\sum_i p_i[\braketo{\psi_i}{S_n}{+}_{nn}\braket{+}{\psi_i} + \braketo{\psi_i}{S_n}{-}_{nn}\braket{-}{\psi_i}]\\
&=\sum_i p_i[{}_n\braket{+}{\psi_i}\braketo{\psi_i}{S_n}{+}_n+ {}_n\braket{-}{\psi_i}\braketo{\psi_i}{S_n}{-}_n]\\ &\quad*\braket{-}{\psi_i}\text{ is a complex number so we can bring it in front}\\
&={}_n\braketo{+}{\sum_i p_i}{\psi_i}\braketo{\psi_i}{S_n}{+}_n+ {}_n\braketo{-}{\sum_i p_i}{\psi_i}\braketo{\psi_i}{S_n}{-}_n
\end{align*}$$
Note that this is the sum of the diagonal elements of a matrix represented by $\sum_i p_i\ketbra{\psi_i}{\psi_i}S_n$ written in the $S_n$ basis.
Density Operator
$$\rho=\sum_i p_i\ketbra{\psi_i}{\psi_i}$$
The density operator (or density matrix) is useful for characterizing mixed states. Note that two different mixed states can have the same density matrix
Therefore, $[S_n]=Tr(\rho S_n)$. This formula for the expectation value for a mixed state can be applied to any observable in any dimension: $[A]=Tr(\rho A)$.