Vector Spaces: From Arrows to Kets

I'm working through Leonard Susskind and Art Friedman's Quantum Mechanics: The Theoretical Minimum^[1]. Chapter 1 is about spin and simple experiments, but before the physics can go anywhere it pauses for a mathematical interlude on complex numbers and vector spaces, because the state of a quantum system is a vector in a vector space. Susskind's definition is short and abstract. This post is me making it concrete.

The objective is simple:

1. Understand what a vector space is, starting from an ordinary arrow in three dimensions.
2. See how the same idea extends to as many dimensions as you like, even when you can't draw them.
3. See a vector made of three complex numbers, the kind quantum mechanics actually uses.

Every picture below can be dragged, nudged or rebuilt. The notation matters, but the aim is that each symbol has a picture behind it.

A vector is an arrow, and a list of numbers

Start with one dimension. A vector on a line is a single number, $a_1$: how far to walk, and in which direction (the sign). If $\hat{e}_1$ is a step of length one along the line, the vector is

$$\vec{A} = a_1 \hat{e}_1.$$

Now add a second direction, at right angles to the first. To describe a vector in the plane, you need two numbers: walk $a_1$ along the first direction, then $a_2$ along the second.

$$\vec{A} = a_1 \hat{e}_1 + a_2 \hat{e}_2.$$

Add a third direction, at right angles to both, and you need a third number:

$$\vec{A} = a_1 \hat{e}_1 + a_2 \hat{e}_2 + a_3 \hat{e}_3 = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix},$$

where

$$\hat{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\; \hat{e}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\; \hat{e}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.$$

That's the whole pattern: each new dimension is one more independent direction, and one more number in the list. The column on the right is just the list of how far to walk in each direction. The unit steps $\hat{e}_1, \hat{e}_2, \hat{e}_3$ are called basis vectors, and the numbers $a_i$ are the vector's components in that basis.

Build one up below. Start in 1D, then switch to 2D and 3D and watch the picture gain a direction each time. The coloured arrows are the components, laid head-to-tail; the plain arrow is the vector they add up to.

The length of the vector comes from Pythagoras, applied twice: once across the floor ($a_1$ and $a_2$), and once up the wall ($a_3$):

$$|\vec{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.$$

What makes it a space

A single arrow isn't the interesting part. What makes a collection of vectors a vector space is what you can do with them. There are exactly two operations.

Addition, component by component. Geometrically, put the arrows head-to-tail:

$$\vec{A} + \vec{B} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix}.$$

Multiplication by a number (a scalar), again component by component. Geometrically, stretch the arrow, shrink it, or flip it round if the number is negative:

$$\lambda \vec{A} = \begin{pmatrix} \lambda a_1 \\ \lambda a_2 \\ \lambda a_3 \end{pmatrix}.$$

A vector space is any collection of objects where these two operations always give you back something in the collection, and behave the way you'd hope. Written out, for any vectors $\vec{A}, \vec{B}, \vec{C}$ in the space and any scalars $\lambda, \mu$:

• Closure: $\vec{A} + \vec{B}$ and $\lambda\vec{A}$ are in the space too.
• The order of adding doesn't matter: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$.
• Nor does the grouping: $(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})$.
• There's a zero vector, $\vec{0}$, with $\vec{A} + \vec{0} = \vec{A}$.
• Every step can be undone: each $\vec{A}$ has a $-\vec{A}$ with $\vec{A} + (-\vec{A}) = \vec{0}$.
• Scaling distributes: $\lambda(\vec{A} + \vec{B}) = \lambda\vec{A} + \lambda\vec{B}$ and $(\lambda + \mu)\vec{A} = \lambda\vec{A} + \mu\vec{A}$.
• Scaling is consistent: $\lambda(\mu\vec{A}) = (\lambda\mu)\vec{A}$, and $1\vec{A} = \vec{A}$.

None of this mentions arrows or columns of numbers. That's the point Susskind is making: anything that obeys these rules is a vector space, and everything we prove using only these rules applies to all of them. Arrows are just the example we can draw.

Put the two operations together and you get a linear combination, $\alpha\vec{A} + \beta\vec{B}$. Try it: scale $\vec{A}$ and $\vec{B}$ with the sliders, and watch the sum move. Then switch on the span.

Some things to try:

• Set $\beta$ to 0 and sweep $\alpha$ from $-2$ to $2$. The tip of $\alpha\vec{A}$ traces out a line through the origin.
• Show the span. Every combination $\alpha\vec{A} + \beta\vec{B}$ lies on one flat plane through the origin. However you set the sliders, two vectors can't get you off it.
• Type B = 4, 1, 0 (twice $\vec{A}$). Now $\vec{B}$ adds no new direction, and the span collapses to a line.
• Turn the picture until you're looking along the plane edge-on, and it becomes a line. That's the third dimension the plane is missing.

This is what "three-dimensional" really means. Two vectors pointing in different directions reach a plane. To reach every point in space you need a third that isn't in that plane. The dimension of a vector space is the number of independent directions you need to reach every vector in it, and any set of that many independent vectors (such as $\hat{e}_1, \hat{e}_2, \hat{e}_3$) is a basis.

Past three dimensions

We can't draw four mutually perpendicular axes, but notice that nothing in the algebra above cared about the number three. A vector in $n$ dimensions is a list of $n$ real numbers, and the set of all of them is called $\mathbb{R}^n$:

$$\vec{A} = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} = \sum_{i=1}^{n} a_i \hat{e}_i.$$

Addition and scaling are still component by component, every axiom still holds, and the length formula just grows more terms:

$$|\vec{A}| = \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} = \sqrt{\sum_{i=1}^{n} a_i^2}.$$

The trick for building intuition is to stop picturing the arrow and picture the list. Draw each component as a bar. For $n \le 3$ the bars and the arrow are two pictures of the same thing; past that, the arrow runs out but the bars carry on happily. Add components below and watch the arrow go first from a line to a plane to space, then disappear, while the bars and the length don't care.

High-dimensional vectors are everywhere once you look. A yield curve quoted at ten maturities is a single point in $\mathbb{R}^{10}$; a greyscale photo with a million pixels is a point in $\mathbb{R}^{1{,}000{,}000}$. You can still add them, scale them, and measure how far apart they are.

One more tool before we leave real vectors. The inner product (or dot product) of two vectors multiplies matching components and adds them up:

$$\vec{A} \cdot \vec{B} = \sum_{i=1}^{n} a_i b_i.$$

It measures how much two vectors point the same way: it's zero when they're at right angles (orthogonal), and the inner product of a vector with itself is its length squared, $\vec{A} \cdot \vec{A} = |\vec{A}|^2$. The basis vectors are each of length one and orthogonal to each other, which makes them an orthonormal basis, and the components of any vector are just its inner products with them: $a_i = \hat{e}_i \cdot \vec{A}$.

Complex numbers, briefly

A complex number $z = x + iy$, with $i^2 = -1$, is a pair of real numbers, so it's natural to draw it as a point (or an arrow) in a plane: the real part $x$ across, the imaginary part $y$ up. The same point can be described by its distance from the origin, $r$, and its angle, $\theta$, called its phase:

$$z = x + iy = r(\cos\theta + i\sin\theta) = r e^{i\theta}.$$

These are two ways of describing the same point, not two different numbers. The Cartesian form, $x + iy$, says how far across and how far up. The polar form, $r e^{i\theta}$, says how far out and in which direction. Each can be turned into the other with a little trigonometry:

$$x = r\cos\theta, \quad y = r\sin\theta \qquad\Longleftrightarrow\qquad r = \sqrt{x^2 + y^2}, \quad \tan\theta = \frac{y}{x}.$$

(For $\theta$, take the angle that puts the point in the right quarter of the plane: $\tan\theta$ alone can't tell $1 + i$ from $-1 - i$.) Drag $z$ in either picture below, or move either pair of sliders, and watch the other description follow.

Each form is better at something. Adding complex numbers is easiest in Cartesian form: add the $x$s and add the $y$s. Multiplying is easiest in polar form, as we're about to see. Which is why, from here on, you'll see both.

Two facts do almost all the work in what follows:

• The complex conjugate $z^* = x - iy = r e^{-i\theta}$ flips the sign of the imaginary part: a reflection in the real axis. Multiplying a number by its conjugate gives a real, non-negative number, the length squared:

$$z^* z = (x - iy)(x + iy) = x^2 + y^2 = r^2 = |z|^2.$$

• Multiplying complex numbers multiplies their lengths and adds their angles: $r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 \, e^{i(\theta_1 + \theta_2)}$. So multiplying by $i = e^{i\pi/2}$ is a quarter turn, multiplying by $-1 = e^{i\pi}$ is a half turn, and multiplying by $e^{i\varphi}$ turns by $\varphi$ without changing the length.

A complex 3-vector

Now the step Susskind takes. Keep everything about a vector space exactly as it was, but let the components, and the scalars you multiply by, be complex numbers. Following Susskind (and Dirac), we write the vector as a ket, $|A\rangle$:

$$|A\rangle = \begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{pmatrix} = \alpha_1 |1\rangle + \alpha_2 |2\rangle + \alpha_3 |3\rangle, \qquad \alpha_i \in \mathbb{C}.$$

Here $|1\rangle, |2\rangle, |3\rangle$ play the role $\hat{e}_1, \hat{e}_2, \hat{e}_3$ did before: a column with a 1 in one place and 0 elsewhere. Adding kets is still component by component, scaling $|A\rangle$ by a complex number $z$ multiplies every component by $z$, and every axiom from earlier holds word for word. This space is called $\mathbb{C}^3$.

So why can't we just draw it as an arrow? Count the real numbers. Each component $\alpha_i = x_i + i y_i$ is two real numbers, so $|A\rangle$ holds six. As far as the number of dials to turn goes, $\mathbb{C}^3$ is as big as $\mathbb{R}^6$, and we've just seen that six dimensions won't fit in one picture.

But the bar-chart trick still works, upgraded. Instead of a bar for each component, give each component its own complex plane, and draw it there as an arrow with a length and a phase. Three components, three little planes. Drag the arrows around, and try the buttons, which multiply the whole vector by a complex number.

Things to try:

• Press × i. All three arrows turn a quarter turn together. Each component's phase changes, but the angles between them (shown under the picture) don't, and neither does any length.
• Press × −1. A half turn, all together. In a real vector space, multiplying by $-1$ flips an arrow; here you can see it's really a rotation through $180°$ that happens to land on the opposite side.
• Press × 2, then × ½. Scaling by a real number stretches every component equally, leaving all the phases alone.
• Press Make real. Every arrow drops onto its real axis. What's left is three real numbers: an ordinary vector in $\mathbb{R}^3$, like the one at the top of this post. The real vectors sit inside the complex ones.
• Show the bra. More on this next.

Bras, and the length of a complex vector

How long is $|A\rangle$? The obvious guess is to reuse the real formula, $\sqrt{\alpha_1^2 + \alpha_2^2 + \alpha_3^2}$. But try it on the one-component vector $(i)$: we get $i^2 = -1$, a negative length squared, which is nonsense.

The fix is the conjugate. Every ket $|A\rangle$ has a partner, the bra $\langle A|$, which is a row whose entries are the conjugates of the ket's components:

$$\langle A| = \begin{pmatrix} \alpha_1^* & \alpha_2^* & \alpha_3^* \end{pmatrix}.$$

Switch on the bra in the widget above and you'll see each $\alpha_i^*$ as the reflection of $\alpha_i$ in its real axis. The inner product of a bra with a ket (a "bra-ket", or bracket) multiplies matching entries and adds them up:

$$\langle B|A\rangle = \beta_1^* \alpha_1 + \beta_2^* \alpha_2 + \beta_3^* \alpha_3 = \sum_{i} \beta_i^* \alpha_i.$$

With a vector and itself, each term is $\alpha_i^* \alpha_i = |\alpha_i|^2$, which is real and never negative:

$$\langle A|A\rangle = |\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 \;\ge\; 0.$$

That's the length squared of $|A\rangle$, and it's the number shown under the widget. Each component contributes its own length squared, the strip at the bottom shows how the total is shared out between them, and the Normalise button divides $|A\rangle$ by $\sqrt{\langle A|A\rangle}$ so that the total is exactly 1.

A small worked example. Take

$$|A\rangle = \begin{pmatrix} 1 \\ i \\ 0 \end{pmatrix}, \qquad |B\rangle = \begin{pmatrix} i \\ 1 \\ 1 \end{pmatrix}.$$

Then $\langle A|A\rangle = 1 \cdot 1 + (-i)(i) + 0 = 1 + 1 = 2$, and

$$\langle B|A\rangle = (i)^* \cdot 1 + (1)^* \cdot i + (1)^* \cdot 0 = -i + i + 0 = 0.$$

Neither vector looks perpendicular to the other at a glance, but their inner product is zero, so they are orthogonal. Without the conjugate, we'd have got $i + i = 2i$ instead. One consequence of the conjugate is that the order now matters, a little: swapping bra and ket conjugates the answer, $\langle A|B\rangle = \langle B|A\rangle^*$.

Everything we built for real vectors carries over. The basis kets are orthonormal, $\langle i|j\rangle = 1$ if $i = j$ and $0$ otherwise, and the components of any ket are its inner products with them: $\alpha_i = \langle i|A\rangle$.

Two real arrows inside one complex vector

The dials group the six real numbers in $|A\rangle$ one component at a time. They can be grouped the other way too. Write each component as its real and imaginary parts, $\alpha_k = x_k + i y_k$, and lay them out in a table:

	Real part	Imaginary part
$\alpha_1$	$x_1$	$y_1$
$\alpha_2$	$x_2$	$y_2$
$\alpha_3$	$x_3$	$y_3$

Each row is one dial. Each column is a list of three real numbers, which is an ordinary real 3-vector:

$$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \qquad \mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}, \qquad |A\rangle = \mathbf{x} + i\,\mathbf{y}.$$

Both $\mathbf{x}$ and $\mathbf{y}$ are real: the $i$ sits outside $\mathbf{y}$, not inside it. So both can be drawn as arrows in one 3D picture, just like the vectors at the start of this post, with the three axes now the directions $|1\rangle, |2\rangle, |3\rangle$. Reading across dial $k$ gives the $k$-th component of $\mathbf{x}$; reading up it gives the $k$-th component of $\mathbf{y}$. For example, $(1, i, 0)$ is $\mathbf{x} = (1, 0, 0)$ plus $i$ times $\mathbf{y} = (0, 1, 0)$: two arrows at right angles.

The length splits neatly between the two arrows. Each $|\alpha_k|^2 = x_k^2 + y_k^2$, so

$$\langle A|A\rangle = \sum_k |\alpha_k|^2 = \sum_k x_k^2 + \sum_k y_k^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2.$$

A phase mixes the two arrows

Now multiply $|A\rangle$ by a phase, $e^{i\varphi} = \cos\varphi + i\sin\varphi$, and gather up the real and imaginary parts:

$$\begin{aligned} e^{i\varphi}|A\rangle &= (\cos\varphi + i\sin\varphi)(\mathbf{x} + i\,\mathbf{y}) \\ &= (\mathbf{x}\cos\varphi - \mathbf{y}\sin\varphi) + i\,(\mathbf{x}\sin\varphi + \mathbf{y}\cos\varphi). \end{aligned}$$

So after the phase, the real and imaginary arrows are

$$\mathbf{x}' = \mathbf{x}\cos\varphi - \mathbf{y}\sin\varphi, \qquad \mathbf{y}' = \mathbf{x}\sin\varphi + \mathbf{y}\cos\varphi.$$

Each new arrow is a blend of the two old ones. As $\varphi$ turns, they swing round together, always staying in the plane the original two share. At $\varphi = 90°$, multiplying by $i$, the blend is complete: $\mathbf{x}' = -\mathbf{y}$ and $\mathbf{y}' = \mathbf{x}$. The real arrow has become the imaginary one, and the imaginary arrow has become (minus) the real one.

In general, neither arrow keeps its length. Square them, using $|\mathbf{u} + \mathbf{v}|^2 = |\mathbf{u}|^2 + 2\,\mathbf{u}\cdot\mathbf{v} + |\mathbf{v}|^2$:

$$\begin{aligned} |\mathbf{x}'|^2 &= |\mathbf{x}|^2\cos^2\varphi - 2\,(\mathbf{x}\cdot\mathbf{y})\sin\varphi\cos\varphi + |\mathbf{y}|^2\sin^2\varphi, \\ |\mathbf{y}'|^2 &= |\mathbf{x}|^2\sin^2\varphi + 2\,(\mathbf{x}\cdot\mathbf{y})\sin\varphi\cos\varphi + |\mathbf{y}|^2\cos^2\varphi. \end{aligned}$$

But add them, and the middle terms cancel, while $\cos^2\varphi + \sin^2\varphi = 1$ collects the rest:

$$|\mathbf{x}'|^2 + |\mathbf{y}'|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2 = \langle A|A\rangle.$$

Length flows back and forth between the real arrow and the imaginary arrow, but the total never changes. That's the length of $|A\rangle$, and a phase can't touch it.

See it happen below. The dials at the top are the components of $e^{i\varphi}|A\rangle$; the 3D picture shows the real arrow $\mathbf{x}$ (solid) and the imaginary arrow $\mathbf{y}$ (dashed) built from them; and the bar at the bottom splits $\langle A|A\rangle$ into $|\mathbf{x}|^2$ and $|\mathbf{y}|^2$.

Things to try:

• Press Play, or drag the phase slider. All three dials turn together. In 3D, $\mathbf{x}$ and $\mathbf{y}$ swing round, growing and shrinking. In the bar, the dividing line between $|\mathbf{x}|^2$ and $|\mathbf{y}|^2$ slides back and forth, but the total, marked in pink, stays exactly where it is.
• Break down into components shows $\mathbf{x}$ (or $\mathbf{y}$) laid head-to-tail along $|1\rangle, |2\rangle, |3\rangle$, as in the first widget. Each step is one dial's reading: across for $\mathbf{x}$, up for $\mathbf{y}$.
• Press × i and check the algebra: the solid arrow moves to where the dashed one was, and the dashed one moves to minus where the solid one was.
• Start from $(1, 1, 0)/\sqrt{2}$, a real vector. At $\varphi = 0$, $\mathbf{y}$ is zero and all the length is in $\mathbf{x}$. Turn to $90°$ and it's all in $\mathbf{y}$. Both arrows stay on one line, and the ellipse is squashed flat.
• Start from $(1, i, 0)/\sqrt{2}$. Now $\mathbf{x}$ and $\mathbf{y}$ are the same length and at right angles, so $\mathbf{x}\cdot\mathbf{y} = 0$, the middle terms above vanish, and the split stays at exactly half and half. The arrows just rotate, and the ellipse is a circle.
• Watch the ellipse. As the phase turns, the tip of $\mathbf{x}$ runs round it, and so does the tip of $\mathbf{y}$, a quarter turn behind. The ellipse itself doesn't move, whatever the phase: it's a picture of $|A\rangle$ that ignores the overall phase. Its size is set by $\langle A|A\rangle$, while its shape (flat, oval or round) comes from the relative phases between the components.
• Drag a dial. That changes $|A\rangle$ itself, not just its phase, so now the total does change.

The inner product can be written in the same terms. Multiplying out $\sum_k (x_{B,k} - i y_{B,k})(x_{A,k} + i y_{A,k})$ gives

$$\langle B|A\rangle = (\mathbf{x}_B\cdot\mathbf{x}_A + \mathbf{y}_B\cdot\mathbf{y}_A) + i\,(\mathbf{x}_B\cdot\mathbf{y}_A - \mathbf{y}_B\cdot\mathbf{x}_A),$$

a complex inner product built from four ordinary dot products. For the worked example above, $|A\rangle = (1, i, 0)$ has $\mathbf{x}_A = (1, 0, 0)$ and $\mathbf{y}_A = (0, 1, 0)$, and $|B\rangle = (i, 1, 1)$ has $\mathbf{x}_B = (0, 1, 1)$ and $\mathbf{y}_B = (1, 0, 0)$. The real part is $0 + 0 = 0$, the imaginary part is $1 - 1 = 0$, and $\langle B|A\rangle = 0$, as before.

Why physics bothers

Susskind uses all of this straight away. The spin of an electron, the system at the heart of his first chapters, is described by a vector in a two-dimensional complex vector space, $\mathbb{C}^2$, with a basis of two kets he calls $|u\rangle$ (up) and $|d\rangle$ (down). A general spin state is a linear combination

$$|A\rangle = \alpha_u |u\rangle + \alpha_d |d\rangle,$$

which is the widget above with one fewer plane. The lengths $|\alpha_u|^2$ and $|\alpha_d|^2$ turn out to be probabilities, which is why normalising (making them add up to 1) matters. And the phases, the part with no counterpart in real vectors, turn out to be where interference, and much of what makes quantum mechanics strange, lives. Notice what the × i button showed: turning every component by the same angle changes none of the lengths and none of the angles between components. That's a hint of something Susskind comes back to: multiplying a state by an overall phase $e^{i\varphi}$ changes nothing you can measure.

Real and complex, side by side

	Real vectors, $\mathbb{R}^n$	Complex vectors, $\mathbb{C}^n$
Components	$n$ real numbers	$n$ complex numbers ($2n$ real numbers)
Scalars	real	complex
Picture	one arrow (if $n \le 3$), or a bar per component	an arrow in its own complex plane per component
Basis	$\hat{e}_1, \dots, \hat{e}_n$	$\vert 1\rangle, \dots, \vert n\rangle$
As real arrows	one arrow	two: $\vert A\rangle = \mathbf{x} + i\,\mathbf{y}$
Inner product	$\vec{A} \cdot \vec{B} = \sum_i a_i b_i$	$\langle B \vert A\rangle = \sum_i \beta_i^* \alpha_i$
Length squared	$\sum_i a_i^2$	$\langle A \vert A\rangle = \sum_i \lvert\alpha_i\rvert^2 = \lvert\mathbf{x}\rvert^2 + \lvert\mathbf{y}\rvert^2$
Swapping order	$\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$	$\langle A \vert B\rangle = \langle B \vert A\rangle^*$

Exercises

A few to check your understanding. Answers are hidden; try them first.

1. Is the set of all real 3-vectors whose components add up to 1 a vector space? What about those whose components add up to 0?

Answer

The first is not. $(1, 0, 0)$ and $(0, 1, 0)$ are both in it, but their sum $(1, 1, 0)$ has components adding up to 2, so it isn't closed under addition (and it doesn't contain $\vec{0}$ either). The second is: if two vectors' components each add up to 0, so do those of their sum and of any multiple. Geometrically, it's a plane through the origin, just like the span in the second widget.

2. How many real numbers does it take to specify a vector in $\mathbb{C}^4$? In $\mathbb{C}^2$, the space of electron spin states?

Answer

Eight, and four: two real numbers (a real and an imaginary part, or a length and a phase) for each complex component.

3. Find $\langle A|A\rangle$ for $|A\rangle = (1 + i,\; 2,\; -i)^T$, then normalise $|A\rangle$.

Answer

$|1 + i|^2 = 1^2 + 1^2 = 2$, $|2|^2 = 4$ and $|-i|^2 = 1$, so $\langle A|A\rangle = 7$. Dividing by $\sqrt{7}$ gives the normalised ket $\tfrac{1}{\sqrt{7}}(1 + i,\; 2,\; -i)^T$.

4. Show that multiplying $|A\rangle$ by $e^{i\varphi}$ leaves $\langle A|A\rangle$ unchanged, as the × i button suggested.

Answer

Every component becomes $e^{i\varphi}\alpha_i$, and its conjugate becomes $e^{-i\varphi}\alpha_i^*$. So each term of the inner product becomes $e^{-i\varphi}\alpha_i^* \, e^{i\varphi}\alpha_i = \alpha_i^* \alpha_i = |\alpha_i|^2$, exactly as before.

5. Using the widget, make $|A\rangle$ real (press Make real), then press × i. Is the result still a real vector? Is it still in $\mathbb{C}^3$?

Answer

No, and yes. Every component is now purely imaginary, so it's no longer in $\mathbb{R}^3$: multiplying a real vector by a complex scalar can take you out of the real vectors. But it's still a perfectly good vector in $\mathbb{C}^3$. That's why the choice of scalars is part of the definition of a vector space: $\mathbb{R}^3$ is closed under multiplication by real numbers, not complex ones.

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1. Susskind, Leonard, and Art Friedman. Quantum Mechanics: The Theoretical Minimum. Basic Books, 2014. ↩︎