Lecture 02 · Measurement

Measurement & Quantum Logic

Why measurement is invasive, the spin-½ particle, correlations between measurements, the failure of classical logic, the six spin eigenstates, the Born rule, and the non-commutativity of observables.

Part I — Classical Measurement Is Arbitrarily Gentle

In classical physics, there is no fundamental lower bound on how much a measurement disturbs a system. To observe a particle's position, you shine light on it — but you can always use dimmer light, lower-energy photons, weaker probes. In principle, the disturbance can be made arbitrarily small.

This is the classical ideal: measurement is passive. The state of the system is a pre-existing fact, and the measurement merely reads it off without changing it. A good experimenter is one who perturbs the system as little as possible — and in principle, the perturbation can be driven to zero.

Quantum mechanics demolishes this idea entirely.

In quantum mechanics, measurement is not a passive act of reading. It is an active physical interaction that irreversibly disturbs the system. This is not an engineering limitation — it is a logical impossibility. There is no quantum measurement procedure that reads the value of one observable without potentially disturbing incompatible observables. The disturbance is built into the structure of the theory.

Part II — The Spin-½ Particle

The simplest quantum system is the spin-½ particle — a particle (like an electron) with a magnetic moment that a Stern-Gerlach device can measure. The first and most important experimental shock:

Shock #1: No matter how the apparatus is oriented, it always reads either $+1$ or $-1$. Never any value in between.

This is deeply non-classical. A classical spinning top oriented at angle $\theta$ from the apparatus axis would register $\cos\theta$ — a continuous value between $-1$ and $+1$. The quantum spin refuses: it always snaps to one of the two extreme values.

We label the three spin observables $\sigma_z$, $\sigma_x$, $\sigma_y$. Each takes values only in $\{+1, -1\}$.

State Preparation and Reproducibility

State preparation: selecting an outcome prepares the state. If we measure $\sigma_z$ and discard all electrons except those giving $+1$, the remaining electrons are in a definite state — the spin-up eigenstate along $z$.

Reproducibility: measuring the same observable twice in quick succession always gives the same answer. If $\sigma_z = +1$ now, then measuring $\sigma_z$ immediately again gives $+1$ with certainty. The first measurement has prepared the state.

Part III — Correlations Between Measurements

Prepare a spin along direction $\hat{n}$ (so the state is the $+1$ eigenstate of $\sigma_{\hat{n}}$), then measure along a different direction $\hat{m}$. The individual results are always $\pm 1$, but their average over many repetitions is:

Spin Correlation Formula
$$\langle\sigma_{\hat{m}}\rangle = \hat{n} \cdot \hat{m} = \cos\theta$$

where $\theta$ is the angle between $\hat{n}$ and $\hat{m}$. This is a remarkable result: the average of the quantum measurement equals the classical dot product. Quantum mechanics recovers classical behavior at the level of expectations — but the individual measurements are always discretely $\pm 1$, never the intermediate classical value $\cos\theta$.

Part IV — The Failure of Classical Logic

In classical logic, the statement "A or B" is true if either A or B is true, regardless of the order in which you check them. Checking A first, then B, gives the same result as checking B first, then A. This order-independence is so obvious in classical reasoning that it's never even stated as an assumption.

Quantum mechanics violates it.

The Experiment

Experiment A: Prepare $\sigma_z = +1$. Ask: "Is $\sigma_z = +1$?" Answer: Yes, with probability 1.

Experiment B: Prepare $\sigma_z = +1$. First measure $\sigma_x$ — this gives $\pm 1$ randomly, and collapses the state to $|r\rangle$ or $|l\rangle$. Then ask: "Is $\sigma_z = +1$?" Now the answer is random — probability $\frac{1}{2}$ for $+1$.

The measurement of $\sigma_x$ has disturbed $\sigma_z$. The question "Is $\sigma_z = +1$?" depends on whether you checked $\sigma_x$ first. Classical logic does not allow for this.

The Mathematical Reason: Projectors Don't Commute

Let $P_z$ be the projector onto the $\sigma_z = +1$ eigenstate, and $P_x$ onto the $\sigma_x = +1$ eigenstate. In quantum mechanics:

$$P_z P_x \neq P_x P_z$$

The operators do not commute, and neither do the physical operations they represent. The structure of quantum logic — the lattice of projectors — is fundamentally non-distributive.

Part V — The Six Spin Eigenstates

There are three spin observables ($\sigma_z$, $\sigma_x$, $\sigma_y$) and each has two eigenstates, giving six fundamental states in total:

DirectionStateEigenvalue
Up along $z$$|u\rangle$$\sigma_z = +1$
Down along $z$$|d\rangle$$\sigma_z = -1$
Right along $x$$|r\rangle$$\sigma_x = +1$
Left along $x$$|l\rangle$$\sigma_x = -1$
In along $y$$|i_s\rangle$$\sigma_y = +1$
Out along $y$$|o\rangle$$\sigma_y = -1$

In the $\{|u\rangle, |d\rangle\}$ basis, these states are represented as column vectors:

$$|u\rangle = \begin{pmatrix}1\\0\end{pmatrix}, \quad |d\rangle = \begin{pmatrix}0\\1\end{pmatrix}$$ $$|r\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}, \quad |l\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$$ $$|i_s\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}, \quad |o\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix}$$

Why Complex Numbers Are Essential

The states $|i_s\rangle$ and $|o\rangle$ involve the imaginary unit $i$. This is not a cosmetic choice. In a real 2D vector space, you cannot find two orthonormal vectors that simultaneously satisfy all the required probability constraints for the $y$-direction measurements. The third axis demands complex phases. A real vector space is simply insufficient to represent all three spin directions consistently. Complex numbers are not a convenience — they are a necessity imposed by the physics of spin.

Part VI — Inner Product and Orthogonality

For two states $|\phi\rangle = (a, b)^T$ and $|\psi\rangle = (c, d)^T$ in $\mathbb{C}^2$, the inner product uses complex conjugation on the bra:

Complex Inner Product
$$\langle\phi|\psi\rangle = a^*c + b^*d$$

In Dirac notation: the ket $|\psi\rangle$ is a column vector; the bra $\langle\phi|$ is the row vector of complex conjugates of $|\phi\rangle$; the bracket $\langle\phi|\psi\rangle$ is their product.

Orthogonality = physical distinguishability. Two states are orthogonal if and only if each is invisible to the other — measuring one gives zero probability for the other. The three pairs of eigenstates are mutually orthogonal:

$$\langle u|d\rangle = (1,0)\begin{pmatrix}0\\1\end{pmatrix} = 0 \checkmark$$ $$\langle r|l\rangle = \frac{1}{2}(1,1)\begin{pmatrix}1\\-1\end{pmatrix} = \frac{1}{2}(1-1) = 0 \checkmark$$ $$\langle i_s|o\rangle = \frac{1}{2}(1,-i)\begin{pmatrix}1\\-i\end{pmatrix} = \frac{1}{2}(1 + (-i)(-i)) = \frac{1}{2}(1 + i^2) = \frac{1}{2}(1-1) = 0 \checkmark$$

Important subtlety: $|u\rangle$ and $|d\rangle$ are antiparallel as physical directions (180° apart in space) but orthogonal as vectors in $\mathbb{C}^2$. Conversely, $|u\rangle$ and $|r\rangle$ are 90° apart in physical space but are not orthogonal as vectors — their inner product is $1/\sqrt{2} \neq 0$. The geometry of state space is not the geometry of physical space.

Part VII — The Born Rule

Given a system in state $|\psi\rangle$, the probability of finding outcome $|\phi\rangle$ upon measurement is:

Born Rule
$$P(\phi\,|\,\psi) = |\langle\phi|\psi\rangle|^2$$

Verification with our known states:

$$P(r\,|\,u) = |\langle r|u\rangle|^2 = \left|\frac{1}{\sqrt{2}}(1,1)\begin{pmatrix}1\\0\end{pmatrix}\right|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} \checkmark$$

This confirms: a spin prepared along $z$ has a 50% chance of registering spin-right along $x$. The general probability formula for spin measurements is:

$$P(+1\,|\,\hat{n}, \hat{m}) = \frac{1 + \hat{n}\cdot\hat{m}}{2} = \frac{1 + \cos\theta}{2} = \cos^2\!\frac{\theta}{2}$$

This is consistent with the correlation formula $\langle\sigma_{\hat{m}}\rangle = \cos\theta$: if $P(+1) = \cos^2(\theta/2)$ and $P(-1) = \sin^2(\theta/2)$, then $\langle\sigma\rangle = \cos^2(\theta/2) - \sin^2(\theta/2) = \cos\theta$. ✓

Part VIII — Non-Commutativity of Observables

The three Pauli matrices represent the three spin components as $2\times 2$ matrices in the $\{|u\rangle, |d\rangle\}$ basis:

$$\sigma_z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad \sigma_x = \begin{pmatrix}0&1\\1&0\end{pmatrix}, \qquad \sigma_y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$$

These matrices do not commute. For example:

$$\sigma_x\sigma_z = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$$ $$\sigma_z\sigma_x = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}0&1\\-1&0\end{pmatrix}$$
Commutation Relation
$$[\sigma_x, \sigma_z] = \sigma_x\sigma_z - \sigma_z\sigma_x = \begin{pmatrix}0&-2\\2&0\end{pmatrix} = 2i\sigma_y \neq 0$$

The general commutation relation for all three Pauli matrices is:

$$[\sigma_i, \sigma_j] = 2i\,\varepsilon_{ijk}\,\sigma_k$$

where $\varepsilon_{ijk}$ is the Levi-Civita symbol ($+1$ for cyclic permutations, $-1$ for anticyclic). This algebraic structure — the Pauli algebra — is the mathematical backbone of quantum spin and reflects the non-commutativity of rotations in 3D space.

Exercises — Lecture 2

Eight exercises on measurement, spin states, orthogonality, the Born rule, and quantum logic. Click any exercise to reveal the full worked solution.

1
Reproducibility of Measurements
EasyCollapse

The spin is measured along $z$ and gives $+1$. It is immediately measured along $z$ again.

(a) What result do you always get? Why?

(b) If instead you first measure $\sigma_x$ (getting some result), then measure $\sigma_z$, is the result determined?

Complete Solution
a

Always $+1$. The first measurement collapsed the state to the eigenstate $|u\rangle$. Mathematically:

$$P(\sigma_z = +1\,|\,|u\rangle) = |\langle u|u\rangle|^2 = |1|^2 = 1$$

An eigenstate of an observable always returns the corresponding eigenvalue with certainty. The state has been "prepared" by the first measurement.

b

No — the result is not determined. Measuring $\sigma_x$ has collapsed the state to either $|r\rangle$ or $|l\rangle$ (each with probability $\frac{1}{2}$). In either case:

$$P(\sigma_z=+1\,|\,|r\rangle) = |\langle u|r\rangle|^2 = \frac{1}{2}, \qquad P(\sigma_z=-1\,|\,|r\rangle) = \frac{1}{2}$$
Key Insight
Measuring $\sigma_x$ destroys the definite value of $\sigma_z$. The intervening measurement between two $\sigma_z$ readings — even if discarded — randomizes the second $\sigma_z$ outcome.
2
Orthogonality of $|i_s\rangle$ and $|o\rangle$
EasyInner Product

Verify that $\langle i_s|o\rangle = 0$, showing that the spin-in and spin-out states along $y$ are orthogonal.

Use $|i_s\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}$ and $|o\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix}$.

Complete Solution
1

Form the bra $\langle i_s|$ by taking the complex conjugates of $|i_s\rangle$'s components and transposing:

$$|i_s\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix} \implies \langle i_s| = \frac{1}{\sqrt{2}}\begin{pmatrix}1^* & i^*\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & -i\end{pmatrix}$$
2

Compute the bracket:

$$\langle i_s|o\rangle = \frac{1}{\sqrt{2}}(1,\,-i)\cdot\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix} = \frac{1}{2}\bigl(1\cdot 1 + (-i)\cdot(-i)\bigr)$$
$$= \frac{1}{2}(1 + i^2) = \frac{1}{2}(1 - 1) = 0 \checkmark$$
3
Conclusion
$\langle i_s|o\rangle = 0$. The spin-in and spin-out states are orthogonal — they are physically distinguishable. Measuring $\sigma_y = +1$ gives zero probability for the outcome $\sigma_y = -1$.
3
Born Rule: Probability for $\sigma_y$
MediumBorn Rule

A spin is prepared in state $|r\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$. What is $P(\sigma_y = +1)$?

Recall the $\sigma_y = +1$ eigenstate is $|i_s\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}$.

Complete Solution
1

Compute the inner product $\langle i_s|r\rangle$. The bra of $|i_s\rangle$ is $\frac{1}{\sqrt{2}}(1,-i)$:

$$\langle i_s|r\rangle = \frac{1}{\sqrt{2}}(1,\,-i)\cdot\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = \frac{1}{2}(1\cdot 1 + (-i)\cdot 1) = \frac{1}{2}(1 - i)$$
2

Apply the Born rule. The modulus squared of a complex number $a+bi$ is $a^2+b^2$:

$$P(\sigma_y=+1) = |\langle i_s|r\rangle|^2 = \left|\frac{1-i}{2}\right|^2 = \frac{|1-i|^2}{4} = \frac{1^2 + (-1)^2}{4} = \frac{2}{4} = \frac{1}{2}$$
3
Answer
$P(\sigma_y = +1\,|\,|r\rangle) = \dfrac{1}{2}$, and by symmetry $P(\sigma_y = -1) = \dfrac{1}{2}$. Starting from the $x$-right eigenstate, measurement along $y$ is completely random — the state $|r\rangle$ has no preference along $y$.
4
General Probability Formula
MediumBorn Rule

A spin is prepared along $\hat{z}$ (in state $|u\rangle$). It is measured along a direction at angle $\theta = 60°$ from $\hat{z}$.

Use the formula $P(+1) = \cos^2(\theta/2)$ to find the probability of each outcome, and verify consistency with $\langle\sigma_\theta\rangle = \cos\theta$.

Complete Solution
1

At $\theta = 60°$, the half-angle is $30°$:

$$P(+1) = \cos^2(30°) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$$
$$P(-1) = \sin^2(30°) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$
2

Check: probabilities sum to 1: $\frac{3}{4} + \frac{1}{4} = 1$. ✓

Compute the expectation value:

$$\langle\sigma_\theta\rangle = (+1)\cdot\frac{3}{4} + (-1)\cdot\frac{1}{4} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$$
3
Verification
$\langle\sigma_\theta\rangle = \frac{1}{2} = \cos(60°)$ ✓. The average of $\pm 1$ quantum outcomes equals the classical dot product — even though no individual outcome equals $\frac{1}{2}$.
5
Quantum Logic Failure: Order Dependence
MediumConceptual

A spin is prepared so $\sigma_z = +1$. We want to evaluate the statement "$\sigma_z = +1$ OR $\sigma_x = +1$".

Method A: Check $\sigma_z$ first. What is the probability this is true?

Method B: Check $\sigma_x$ first (getting $+1$), then check $\sigma_z$. What is $P(\sigma_z = +1)$ after obtaining $\sigma_x = +1$? Why does the order matter?

Complete Solution
A

The spin is in $|u\rangle$ (eigenstate of $\sigma_z$ with eigenvalue $+1$). Measuring $\sigma_z$:

$$P(\sigma_z = +1) = |\langle u|u\rangle|^2 = 1$$

The statement is true with certainty. No additional measurement needed.

B

Start in $|u\rangle$. Measure $\sigma_x$ — we assume we get $+1$, so the state collapses to $|r\rangle = \frac{1}{\sqrt{2}}(1,1)^T$. Now measure $\sigma_z$ from $|r\rangle$:

$$P(\sigma_z = +1\,|\,|r\rangle) = |\langle u|r\rangle|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$

The probability of $\sigma_z = +1$ has dropped from 1 to $\frac{1}{2}$!

Key
Why Classical Logic Fails
In classical logic, the order of checking a disjunction doesn't matter. Here it does: checking $\sigma_x$ first physically changes the system. The measurement is invasive. Classical logic assumes facts are pre-existing; quantum mechanics says the act of asking changes the answer.
6
Commutator $[\sigma_x, \sigma_y]$
MediumPauli Matrices

Compute $[\sigma_x, \sigma_y]$ directly using matrix multiplication.

$\sigma_x = \begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\quad\sigma_y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$

Complete Solution
1

Compute $\sigma_x\sigma_y$:

$$\sigma_x\sigma_y = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix} = \begin{pmatrix}0\cdot0+1\cdot i & 0\cdot(-i)+1\cdot0 \\ 1\cdot0+0\cdot i & 1\cdot(-i)+0\cdot0\end{pmatrix} = \begin{pmatrix}i&0\\0&-i\end{pmatrix}$$
2

Compute $\sigma_y\sigma_x$:

$$\sigma_y\sigma_x = \begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}0\cdot0+(-i)\cdot1 & 0\cdot1+(-i)\cdot0 \\ i\cdot0+0\cdot1 & i\cdot1+0\cdot0\end{pmatrix} = \begin{pmatrix}-i&0\\0&i\end{pmatrix}$$
3

The commutator:

$$[\sigma_x,\sigma_y] = \sigma_x\sigma_y - \sigma_y\sigma_x = \begin{pmatrix}i&0\\0&-i\end{pmatrix} - \begin{pmatrix}-i&0\\0&i\end{pmatrix} = \begin{pmatrix}2i&0\\0&-2i\end{pmatrix} = 2i\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$
Result
$[\sigma_x,\sigma_y] = 2i\sigma_z$. ✓ This is the cyclic permutation $x\to y\to z\to x$ with $\varepsilon_{xyz} = +1$, consistent with $[\sigma_i,\sigma_j] = 2i\varepsilon_{ijk}\sigma_k$.
7
Why Real Numbers Fail
HardComplex Numbers

In a real 2D vector space, can you find unit vectors $\mathbf{v}_1, \mathbf{v}_2 \in \mathbb{R}^2$ with $\mathbf{v}_1 \perp \mathbf{v}_2$ such that each has a 50% probability of giving spin-up along $z$ (i.e., $|\mathbf{v}_k \cdot \hat{e}_1|^2 = \frac{1}{2}$, where $\hat{e}_1 = (1,0)$) and a 50% probability of giving spin-up along $x$ and along $y$?

Then explain why $|i_s\rangle$ and $|o\rangle$ succeed in a complex space.

Complete Solution
Setup

For a real unit vector $\mathbf{v}_1 = (a,b)$ with $a^2+b^2=1$ and $|\mathbf{v}_1\cdot(1,0)|^2 = a^2 = \frac{1}{2}$, we need $a = \pm\frac{1}{\sqrt{2}}$. The four options are $(\pm\frac{1}{\sqrt{2}}, \pm\frac{1}{\sqrt{2}})$.

Orthogonal pair

Take $\mathbf{v}_1 = \frac{1}{\sqrt{2}}(1,1)$ and $\mathbf{v}_2 = \frac{1}{\sqrt{2}}(1,-1)$. These are real, unit, and orthogonal. Also $|\mathbf{v}_2\cdot(1,0)|^2 = \frac{1}{2}$. So far, these work for the $z$ and $x$ conditions.

But notice: $\mathbf{v}_1 = |r\rangle$ and $\mathbf{v}_2 = |l\rangle$. The eigenstates of $\sigma_x$ are real. Now the problem: to represent the $y$-axis eigenstates, we need a vector with $|\langle\mathbf{v}|(1,0)^T|^2 = \frac{1}{2}$ and $|\langle\mathbf{v}|(1,1)^T/\sqrt{2}|^2 = \frac{1}{2}$ and orthogonal to another vector satisfying the same. In $\mathbb{R}^2$, we only have two orthogonal directions — already used by $\mathbf{v}_1$ and $\mathbf{v}_2$.

$$\text{Any real unit vector in }\mathbb{R}^2 = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 \text{ with } c_1,c_2\in\mathbb{R}$$

For a $y$-eigenstate, we need: $|\langle u|\psi_y\rangle|^2 = \frac{1}{2}$, $|\langle r|\psi_y\rangle|^2 = \frac{1}{2}$, and the two $y$-eigenstates mutually orthogonal. There is no real solution satisfying all three conditions simultaneously.

Complex solution

In $\mathbb{C}^2$, $|i_s\rangle = \frac{1}{\sqrt{2}}(1,i)^T$ and $|o\rangle = \frac{1}{\sqrt{2}}(1,-i)^T$ satisfy all conditions:

$$|\langle u|i_s\rangle|^2 = \frac{1}{2}, \quad |\langle d|i_s\rangle|^2 = \frac{1}{2}, \quad |\langle r|i_s\rangle|^2 = \frac{1}{2}, \quad |\langle l|i_s\rangle|^2 = \frac{1}{2}$$
Conclusion
The imaginary unit $i$ in $|i_s\rangle = \frac{1}{\sqrt{2}}(1,i)^T$ provides the extra "direction" in $\mathbb{C}^2$ that is unavailable in $\mathbb{R}^2$. Quantum mechanics of spin requires $\mathbb{C}^2$ — not as a mathematical luxury, but as a physical necessity.
8
Hermitian Symmetry of the Inner Product
HardProof

Prove that for any two quantum states $|\phi\rangle$ and $|\psi\rangle$ in $\mathbb{C}^2$:

$$\langle\phi|\psi\rangle = \langle\psi|\phi\rangle^*$$

(This property is called Hermitian symmetry, or conjugate symmetry.) Then use it to show that $\langle\psi|\psi\rangle$ is always real and non-negative.

Complete Solution
Setup

Let $|\phi\rangle = (a,b)^T$ and $|\psi\rangle = (c,d)^T$ with $a,b,c,d\in\mathbb{C}$. By definition:

$$\langle\phi|\psi\rangle = a^*c + b^*d$$
Prove Hermitian symmetry

Compute $\langle\psi|\phi\rangle$:

$$\langle\psi|\phi\rangle = c^*a + d^*b$$

Take its complex conjugate, using $(z_1 + z_2)^* = z_1^* + z_2^*$ and $(z_1 z_2)^* = z_1^* z_2^*$:

$$\langle\psi|\phi\rangle^* = (c^*a)^* + (d^*b)^* = (c^{**})(a^*) + (d^{**})(b^*) = ca^* + db^* = a^*c + b^*d = \langle\phi|\psi\rangle \checkmark$$
Norm is real

Set $|\phi\rangle = |\psi\rangle$ in the Hermitian symmetry relation:

$$\langle\psi|\psi\rangle = \langle\psi|\psi\rangle^*$$

A complex number equal to its own conjugate must be real. Explicitly:

$$\langle\psi|\psi\rangle = c^*c + d^*d = |c|^2 + |d|^2 \geq 0$$
Conclusion
$\langle\psi|\psi\rangle$ is always real and $\geq 0$. It equals zero only if $|\psi\rangle = 0$. This makes $\sqrt{\langle\psi|\psi\rangle}$ a valid norm. Physical states are normalized so $\langle\psi|\psi\rangle = 1$.