q3-qubit-dirac-notation

Question 3: Quantum Information (Qubit String and Dirac’s Notation)

Source: Chapter 02, slides 3-4, 7-9, 19, 21, 27, 32-33

Why questions (Q&A)

Q1. Why does a system of $n$ qubits need a state vector of size $2^n$ rather than just $n$ numbers, unlike $n$ classical bits?

Step 1, the classical case. $n$ classical bits each hold one independent value (0 or 1), and nothing more. Listing those $n$ values fully describes the whole register: $n$ numbers, no more needed.

Step 2, the quantum case. $n$ qubits, in contrast, can be in a superposition over every joint assignment of the $n$ bits at once. Each of the $2^n$ possible basis strings

$$|x_1{x}_2\cdots x_n⟩,x_i\in \{0,1\}$$

can carry its own complex amplitude, independently of every other basis string. Describing the general state requires listing one complex amplitude per basis string, that is $2^n$ complex numbers in total.

Step 3, why this is not just “n qubits, n numbers”. The register is not a list of $n$ separately-described qubit states; it is a single joint vector living in a $2^n$-dimensional space. This is the same $2^n$ dimension counted from a different angle in Question 1, Math Q1, where it was used to count physical degrees of freedom.

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Q2. Why is Dirac (bra-ket) notation useful compared to writing out full column vectors every time?

Reason 1, direct labeling. $|01⟩$ names the basis vector by its bit pattern directly. The reader immediately sees which classical string an amplitude belongs to, with no need to cross-reference a numeric position index in a long column vector.

Reason 2, algebraic manipulations stay compact. Bra-ket notation lets the inner product $⟨\varphi |\psi ⟩$, the tensor product $|a⟩\otimes |b⟩$, and the adjoint of an operator all be written and manipulated as short algebraic expressions on labels, rather than as index-by-index sums over long vectors or matrices.

Reason 3, this scales. This compactness matters increasingly as the qubit count grows: at just 4 qubits a column vector already has $2^4=16$ entries, and at 10 qubits it has $2^{10}=1024$ entries. Dirac notation stays exactly as readable at any qubit count, because it never requires writing out the full vector.

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Q3. Why does the bra $⟨\psi |$ correspond to the conjugate transpose of $|\psi ⟩$, rather than a plain transpose?

Step 1, what $⟨\psi |\psi ⟩$ needs to be. The pairing $⟨\psi |\psi ⟩$ must equal the sum of squared magnitudes of the amplitudes,

$$⟨\psi |\psi ⟩=\sum _i|c_i{|}^2,$$

which has to be a real, non-negative number, because it is used directly as (or normalized into) a probability.

Step 2, why a plain transpose fails. If $⟨\psi |$ were only the plain transpose of $|\psi ⟩$ (transposed, but not conjugated), then

$$⟨\psi |\psi ⟩\text{(plain transpose version)}=\sum _i{c}_i^2,$$

which is generally a complex number whenever the amplitudes $c_i$ are complex, and it cannot be interpreted as a probability at all.

Step 3, why conjugation fixes this. Conjugating each amplitude when forming the bra is what forces

$$⟨\psi |\psi ⟩=\sum _i\overline{c_i}{c}_i=\sum _i|c_i{|}^2\ge 0,$$

a real, non-negative number for every state $|\psi ⟩$, with no exceptions. That is why the bra must be the conjugate transpose, not the plain transpose.

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Q4. Why does the ordering of qubits in a tensor-product string such as $|x_1{x}_2{x}_3{x}_4⟩$ matter for applying a gate to a specific wire?

Step 1, the labeling convention. Each position in the string is tied to a specific physical wire by convention: position 1 is wire 1, position 2 is wire 2, and so on. A single-qubit gate “applied to wire $k$” acts only on the factor sitting at position $k$ in the tensor product, and leaves every other position untouched (acts as identity there).

Step 2, what would break without a fixed convention. If the position-to-wire labeling were not fixed, there would be no clear way to say which physical qubit a gate refers to. This matters for every gate that names specific wires: $H$ applied “to wire 2” needs to know clearly which factor is wire 2; a controlled-NOT’s control and target wires need to be distinguishable; and a swap gate explicitly exchanges two named positions (e.g. wires 2 and 3), which only makes sense if positions map to wires consistently.

Conclusion. The fixed position-to-wire convention is what makes “apply this gate to this wire” a well-defined instruction at all, for both single-qubit and multi-qubit gates.

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Q5. Why is normalization (norm 1) required for a physically valid qubit string, and what would an unnormalized state imply?

Step 1, where probabilities come from. Measurement probabilities are computed as $|c_i{|}^2$ for each basis outcome $i$. For this to form a valid probability distribution, the probabilities across all outcomes must sum to exactly 1:

$$\sum _i|c_i{|}^2=⟨\psi |\psi ⟩=1.$$

Step 2, what happens if this fails. An unnormalized vector (where $⟨\psi |\psi ⟩\ne 1$) would give a total probability different from 1, either more than certainty or less than certainty, neither of which is a valid probability distribution over measurement outcomes.

Step 3, the practical fix. In practice, an unnormalized vector is rescaled by dividing by its norm before being treated as a physical state:

$$|{\psi }_{\text{normalized}}⟩=\frac{|\psi ⟩}{\sqrt{⟨\psi |\psi ⟩}}.$$

Math Q5 below works through this rescaling on a concrete example.

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Q6. Why is a multi-qubit basis state such as $|01⟩$ written as the tensor product $|0⟩\otimes |1⟩$ rather than the sum $|0⟩+|1⟩$?

Step 1, what $|0⟩\otimes |1⟩$ represents. The tensor product represents two separate qubits jointly, each in one definite value: the first qubit is 0, the second is 1. This is exactly what a basis state of the joint 4-dimensional two-qubit space must look like, one definite combined configuration.

Step 2, what $|0⟩+|1⟩$ represents instead. This notation describes a single qubit in an (unnormalized) superposition of being 0 or 1, a completely different physical situation: one qubit with two possible outcomes, not two qubits each with one definite outcome.

Step 3, why the two cannot be interchanged. These are states of systems of different sizes, one qubit versus two qubits, so the sum notation cannot stand in for the two-qubit basis label. Superposition ($+$) and joint composition ($\otimes$) are separate operations: $+$ combines different possible amplitudes of the same system, while $\otimes$ combines separate systems into one larger joint system.

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Math questions (Q&A)

Q1. Write $|01⟩$ and $|10⟩$ as tensor products and as 4-dimensional column vectors in the ordered basis $\{|00⟩,|01⟩,|10⟩,|11⟩\}$.

$|01⟩$:

$$|01⟩=|0⟩\otimes |1⟩=\left(\begin{array}{c}1\\ 0\end{array}\right)\otimes \left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}1\cdot 0\\ 1\cdot 1\\ 0\cdot 0\\ 0\cdot 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right).$$

$|10⟩$:

$$|10⟩=|1⟩\otimes |0⟩=\left(\begin{array}{c}0\\ 1\end{array}\right)\otimes \left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}0\cdot 1\\ 0\cdot 0\\ 1\cdot 1\\ 1\cdot 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right).$$

Reading the pattern. $|01⟩$ is the second standard basis vector (index 1, counting from 0) and $|10⟩$ is the third (index 2). This matches the binary value of the two-bit label, read as a base-2 index: $01_2=1$, $10_2=2$.

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Q2. A 3-qubit register is in state $|101⟩$. Write the corresponding 8-dimensional standard basis column vector.

Step 1, convert the label to an index. Interpret $101$ as a binary number, most significant bit first:

$$1\cdot 4+0\cdot 2+1\cdot 1=5(\text{0-indexed}).$$

Step 2, build the vector. An 8-dimensional basis vector has a single $1$ at the position given by that index, and $0$ everywhere else. Positions $0$ through $7$ correspond, in order, to $|000⟩$ through $|111⟩$, so position $5$ is $|101⟩$:

$$|101⟩=(0,0,0,0,0,1,0,0).$$

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Q3. Compute $⟨\varphi |\psi ⟩$ for $|\varphi ⟩=|01⟩$ and $|\psi ⟩=\frac{|00⟩+|11⟩}{\sqrt{2}}$.

Step 1, write both states as column vectors.

$$\varphi =(0,1,0,0),\psi =\left(\frac{1}{\sqrt{2}},0,0,\frac{1}{\sqrt{2}}\right).$$

Step 2, apply the conjugated dot product. Since all entries here are real, conjugation does not change them, and

$$⟨\varphi |\psi ⟩=\overline{\varphi }\cdot \psi =0\cdot \frac{1}{\sqrt{2}}+1\cdot 0+0\cdot 0+0\cdot \frac{1}{\sqrt{2}}=0.$$

Interpretation. The two states are orthogonal. $|01⟩$ has zero overlap with the Bell state $\frac{|00⟩+|11⟩}{\sqrt{2}}$, because the Bell state has no $|01⟩$ component at all, only $|00⟩$ and $|11⟩$ appear in it.

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Q4. Given the 4-qubit string $|x_1{x}_2{x}_3{x}_4⟩=|0110⟩$, state which wire each digit corresponds to, then give the resulting string after a swap gate is applied to wires 2 and 3.

Step 1, map positions to wires.

$$\text{position 1}\to \text{wire 1}\to \text{value }0,\text{position 2}\to \text{wire 2}\to \text{value }1,$$ $$\text{position 3}\to \text{wire 3}\to \text{value }1,\text{position 4}\to \text{wire 4}\to \text{value }0.$$

Step 2, apply swap(2,3). Swap exchanges the values held on wires 2 and 3. Here both wires already hold the same value ($1$ and $1$), so exchanging them changes nothing:

$$|0110⟩\stackrel{\text{swap}(2,3)}{\to }|0110⟩.$$

A worked contrast (to show swap does something in the general case). If instead the starting string had been $|0100⟩$ (wire 2 value $1$, wire 3 value $0$, different values from each other), applying swap(2,3) would exchange them and give $|0010⟩$, the second and third digits trade places. The example in this problem happened to have matching values, which is why nothing visibly changed.

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Q5. Normalize the unnormalized vector $v=3|0⟩+4i|1⟩$.

Step 1, compute the norm.

$$\Vert v\Vert =\sqrt{|3{|}^2+|4i{|}^2}=\sqrt{9+16}=\sqrt{25}=5.$$

Step 2, divide every amplitude by the norm.

$$v_{\text{normalized}}=\frac{v}{5}=\frac{3}{5}|0⟩+\frac{4i}{5}|1⟩.$$

Step 3, check the result is normalized.

$${\left|\frac{3}{5}\right|}^2+{\left|\frac{4i}{5}\right|}^2=\frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1.$$

The rescaled vector has norm exactly $1$, confirming it is now a valid, physically meaningful qubit string.

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Q6. Expand $|\psi ⟩=(a|0⟩+b|1⟩)\otimes (c|0⟩+d|1⟩)$ into the four-term computational-basis expansion, and give the coefficient of each basis ket.

Step 1, distribute the tensor product over both sums, using that $\otimes$ distributes over addition on each side independently:

$$|\psi ⟩=ac|00⟩+ad|01⟩+bc|10⟩+bd|11⟩.$$

Step 2, read off the coefficients.

$$\text{coeff. of }|00⟩=ac,\text{coeff. of }|01⟩=ad,\text{coeff. of }|10⟩=bc,\text{coeff. of }|11⟩=bd.$$

Step 3, normalization check. Suppose each factor is separately normalized, i.e. $|a{|}^2+|b{|}^2=1$ and $|c{|}^2+|d{|}^2=1$. Then

$$|ac{|}^2+|ad{|}^2+|bc{|}^2+|bd{|}^2=(|a{|}^2+|b{|}^2)(|c{|}^2+|d{|}^2)=1\cdot 1=1.$$

So a tensor product of two normalized single-qubit states is automatically a normalized two-qubit state, no extra rescaling step is ever needed when combining independently-normalized registers.

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Q7. Simplify $⟨\mathrm{cis}(\theta )\psi \mid \mathrm{cis}(\theta )\psi ⟩$ step by step, using Dirac-notation rules for conjugation and linearity.

Step 1, write out what the ket means. The ket $\mathrm{cis}(\theta )|\psi ⟩$ is shorthand for the vector $e^{i\theta }|\psi ⟩$.

Step 2, form the corresponding bra. The bra is the conjugate transpose, so pulling the scalar $e^{i\theta }$ out of a ket and into a bra conjugates it:

$$⟨e^{i\theta }\psi |=\overline{e^{i\theta }}⟨\psi |=e^{-i\theta }⟨\psi |,$$

because conjugation of a scalar factor flips the sign of its phase exponent.

Step 3, combine bra and ket.

$$⟨e^{i\theta }\psi \mid e^{i\theta }\psi ⟩=\left(e^{-i\theta }⟨\psi |\right)\left(e^{i\theta }|\psi ⟩\right)=e^{-i\theta }{e}^{i\theta }⟨\psi |\psi ⟩.$$

Step 4, simplify the phase factors.

$$e^{-i\theta }{e}^{i\theta }=e^0=1,$$

so

$$⟨e^{i\theta }\psi \mid e^{i\theta }\psi ⟩=⟨\psi |\psi ⟩.$$

Conclusion. The two phase factors cancel exactly, leaving the original norm untouched. This matches, symbolically this time rather than by direct numerical computation, the global-phase invariance already verified by direct computation in Question 2, Math Q5.