Reading Week 1 - Quantum Cryptography, Computation, Teleportation; Review of Wave Functions and Young's Slit Experiment

Spiller Paper

You can read the whole thing via [[1 Quantum Information Processing Cryptography Computation Teleportation IEEE85 1996 Spiller with reading assignments.pdf]]. A cliff notes version follows:

Introduction

This paper talks about uses of quantum computing, in the sense of:

Cryptography
Computation
Teleportation

We used to use QM to study large systems of particles, but now we use QM on a small group of individual electrons. For example, trapped ion is this method, trapping single atoms (ions) that can be probed with magnetic fields. This allows quantum information technology, where QM dictates how information is stored. It outperforms classical computers in very specific areas, making it useful to study. We'll cover:

Quantum Cryptography: sharing quantum systems allows for a shared random bit string to represent a key that only those communicating on the line can know, and thus helps in encryption.
Quantum Computation: Qubits exist in a superposition between 0 and 1, allowing for multiple inputs states to be computed against in parallel.
Quantum Teleportation: Using classical bits, we can transfer an unknown quantum state through a hostile environment, where quantum computers would have a more fragile existence.

Quantum Cryptography

The Idea

We want to exchange a message between two individuals, without a possible interceptor being able to know what message we sent.

A popular encryption method is via public key cryptosystems, where details of the encryption are agreed upon by both parties first. A common one is RSA (named after Ronals Rivest, Adi Shamir, and Leonard Adleman) which use the difficulty of factorizing large composite integers to make them work. But they are still potentially insecure. It's just inefficient for a classical computer to tackle doing the factorization.

Even if the key is changed every time, akin to a Verman cipher, we don't know if Eve (the interceptor) didn't intercept the key itself.

However, a quantum system is irreversible, so if the key is distributed, then Eve can only determine the key while tampering the system (ie: observing the key forces the system to change state, indicating the presence of the listener).

Alice and Bob share a quantum channel and a public channel. Photons are sent down the quantum channel, and are what Eve wants to try to intercept, but Even can only really listen to what's on the public channel.

Alice sends photons one by one, in states she chooses at random via:

| ⇕ ⟩, | \Leftrightarrow ⟩, . . .

and the diagonal states (/ and \). Bob randomly chooses to measure the rectilinear basis, or the diagonal basis. He records his results and keeps them secret, but let's Alice know the bases he used instead. Alice tells Bob what data to keep, for the vertical and horizontal for the rectilinear basis, or for the diagonal one. They agree that $⇕$ is the 1 state, and $\Leftrightarrow$ is a 0, and similar for the diagonal states. Boom! They've sent their raw quantum transmission (RQT).

Errors and Eavesdropping

The problems with RQT are that:

Errors can occur over time from the environment, flipping the bits on occasion
Eve trying to listen in will cause errors themselves. This corrupts a quarter of the bits she intercepts since on average, half of her bits to Bob are wrong, and half are wrong to Alice.

So we try to solve these issues. They have to build in error correction in the data they send, and also know that if Eve is eavesdropping then she will know some of the correct bits. But they can have systems in place to deal with both of these. The former comes from classical computing, using codes within the message itself to correct it.

Eve's expected information about the key can be reduced to an exponentially small fraction, knowing exactly one bit of it.

Alice and bob can estimate errors in the RQT by public comparison. If the error rate is greater than 25%, then they can suspect that Eve was listening in and thus bin the transmission. But they know that they should do this, which is the big thing here. This is the key distillation procedure.

For error correction, they can do bisective parity checks similar to Hamming Codes (more on them via this awesome video) . These work by taking the number of 0's and 1's in certain "slots" (modulo 2, 3, 4, ...) and determining if there's an even parity (0) or odd parity (1), and appending that to the message. This takes a $2^{k}$ size message and add $k$ parity bits for the code.

After removing the errors, Alice and Bob decrease Eve's knowledge via privacy amplification. From the RQT error rate, Eve's information can be estimated as follows. If Alice and Bob agree on a random subset of bit positions in the corrected RQT, then Eve will only know the parity of that subset if she knows the values of all the bits it contains, which is extremely unlikely. Thus, they use this parity bit as the first bit in a new secret final key. Choosing subsets approximately equal to the corrected RQT minus the number of bits they think Eve knows, they get a number of shared bits, and thus the final perfect secret key. It essentially mixes up what Eve knows with what she doesn't, diluting her knowledge.

As an example, having 4% RQT arrows, Alice and Bob distill 2000 bits down to 754, where Eve will know less than $10^{- 6}$ of a bit. With an error rate of 8%, 2000 bits distill down to 105, with better secrecy.

Quantum Computation

In Turing's model of a Universal Turning Machine (the generalization of a classical computer), a program must possibly be an infinite tape, and a head must read these values and execute based on their values. It must also be able to store values via internal memory. This machine can execute any algorithm, and thus is modellable on a classical computer.

Conventional operations like AND and OR are irreversible; any given output like $0$ cannot determine if the inputs were $01, 10, 00$ in the AND case. But making physical realizations of these processes cost entropy. The entropy change associated with erasing a bit is $Δ S = k \ln (2)$ and at a temperature $T$ it costs $Δ E = T Δ S = k T \ln (2)$ units of energy.

To avoid this, you must have the same number of inputs as outputs to avoid losing information. An example of a quantum gate is the controlled-NOT (ie: XOR) denoted $C_{12}$ . There's two inputs and outputs. Bit one is $ϵ_{1}$ is the control and bit two $ϵ_{2}$ is the target. The operation negates $ϵ_{2}$ if $ϵ_{1} = 1$ and leaves it alone if $ϵ_{1} = 0$ . Namely:

C_{12} | ϵ_{1} ⟩_{1} | ϵ_{2} ⟩_{2} = | ϵ_{1} ⟩_{1} | (ϵ_{1} + ϵ_{2}) mod 2 ⟩_{2}

For example, $C_{12} | 1 ⟩_{1} | 0 ⟩_{2} = | 1 ⟩_{1} | 1 ⟩_{2}$ . The nice thing here is that now, similar to the classical XOR gate, we can reverse a lot of these computations, and thus make a reversible Turing Machine.

Consider calculating a function $f$ that maps a value $x \in [0, 2^{m - 1}]$ to some value $[0, 2^{n - 1}]$ . The computer needs a register of size $m + n$ bits to hold all the data (and definitely need more for processing). $m$ bits are used for the input, and $n$ bits for the output. But in the grander scheme of things, the whole register is updated reversibly to the new set of values, given by operator $F$ :

F | x ⟩_{i} | 0 ⟩_{o} = | x ⟩_{i} | f (x) ⟩_{o}

Here, one of the $2^{m}$ possible inputs $| x ⟩_{i}$ is prepared and $F$ generates its output $f (x)$ , putting this in the output register $| . . . ⟩_{o}$ , which was initially all 0's. If $f$ is one-to-one (namely, if two outputs are equal, then their inputs were the same), then by nature $f$ is reversible so we can determine $x$ purely from the output $f (x)$ . If instead it's not, then we save a copy of $x$ to refer to later on.

An interesting thing is that even though we know that all classical systems that we know are essentially universal turning machines (UTM), it is still an open problem considering if a built quantum computer is equivalent to the Universal Quantum Computer (UQC), where states are not definite and are in fact possibly infinite in variation.

Similar to classical computing, it's believed we build a UQC from a set of quantum gates. We care about reversible gates of similar classical gates since these new gates are the building blocks of quantum machines. XOR can act on a superposition of both the $| 0 ⟩$ and $| 1 ⟩$ states, hence differentiating itself from classical computing. This allows the XOR operation to turn single product states of two qubits into entanglements via, for instance:

C_{12} 2^{- 1 / 2} (| 0 ⟩_{1} + | 1 ⟩_{1}) | 0 ⟩_{2} = 2^{- 1 / 2} (| 0 ⟩_{1} | 0 ⟩_{2} + | 1 ⟩_{1} | 1 ⟩_{2})

The nice thing here is that XOR is now reversible, helping to disentangle states:

C_{12} 2^{- 1 / 2} (| 0 ⟩_{1} | 0 ⟩_{2} \pm | 1 ⟩_{1} | 1 ⟩_{2}) = 2^{- 1 / 2} (| 0 ⟩_{1} \pm 1 ⟩_{1}) | 0 ⟩_{2}

C_{12} 2^{- 1 / 2} (| 0 ⟩_{1} | 1 ⟩_{2} \pm | 1 ⟩_{1} | 0 ⟩_{2}) = 2^{- 1 / 2} (| 0 ⟩_{1} \pm 1 ⟩_{1}) | 1 ⟩_{2}

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But we must do more. We require that the effects of the environment be small enough so that the whole computer evolves in reversible Schrodinger fashion. This is a hard ask! But for the moment, assume their achievable. Consider the $m$ -qubit input part of evaluating $f (x)$ is a superposition of all possible classical inputs:

| ψ ⟩_{i} = 2^{- m / 2} \sum_{x = 0}^{2^{m} - 1} | x ⟩_{i}

where $2^{- m / 2}$ is used for normalization since there's are $2^{m}$ terms in the sum. If all $m$ qubits are prepared in a superposition state $2^{- 1 / 2} (| 0 ⟩ + | 1 ⟩)$ then the single product of all these contains $2^{m}$ states $| x ⟩_{i}$ with equal amplitude $2^{- m / 2}$ :

F | ψ ⟩_{i} | 0 ⟩_{0} = 2^{- m / 2} \sum_{x = 0}^{2^{m} - 1} (| x ⟩_{i} | f (x) ⟩_{o})

giving an entanglement of the input $x$ and it's corresponding output $f (x)$ .

Probable Uses

Note that computing $f$ here is not ideal for a quantum computer! You can only measure one of the $m$ bits, and especially if they're all important, then $f$ only yields one of these at random (with a probability of $1 / 2^{m}$ ). And once you measure, the wavefunction collapses, destroying the other possible measurements in the process!

So what are they good for? In essence, a classical computer, if it had to check all $2^{m}$ possible combinations of input, wouldn't be good for say checking a property of $f$ . But for a quantum system, we could run it on a quantum computer, and by having the inputs be superimposed, then we could have one output bit on the system run indicate if all possible states satisfy $f$ , then just read that one bit. A way of thinking this is that all $2^{m}$ calculations run in parallel via our superimposed states.

Quantum Teleportation

QT relies on Alice and Bob sharing and storing EPR pairs of quantum systems.

Teleportation

A customer gives Alice a system of a single qubit in an arbitrary quantum state. Bob can reproduce the same system some distance away, requiring only classical data to do so. Note that Alice cannot send some measurement on the system to Bob, since the state has to be known to be one of a given basis set. This is because measuring the observed state, referring to some basis, will give needed $λ$ 's and thus we can identify the state as a superposition of these eigenstates.

In layman's Alice is unable to infer the state of the system she's given. Measuring the state internally destroys the state! But we can use EPR pairs to get around this problem.

Suppose Alice has qubit one of a pair and Bob has qubit two. The unknown state of the customer's qubit three is represented as $a | 0 ⟩_{3} + b | 1 ⟩_{3}$ where $a, b \in C$ . Thus, the full state of the three particles is a single product of qubit three with the EPR pair:

| ψ ⟩_{123} = 2^{- 1 / 2} (| 1 ⟩_{1} | 1 ⟩_{2} + | 0 ⟩_{1} | 0 ⟩_{2}) (a | 0 ⟩_{3} + b | 1 ⟩_{3})

Alice measures qubits one and three using the XOR process described in Reading Week 1 - Quantum Cryptography, Computation, Teleportation; Review of Wave Functions and Young's Slit Experiment#^51ee9a through Reading Week 1 - Quantum Cryptography, Computation, Teleportation; Review of Wave Functions and Young's Slit Experiment#^5b388a.

1.5.3: Young's Double-Slit Experiment

Consider a sandblaster gun can randomly fire sand particles at an assembly with two slits. We can close or open slits 1 and 2.

Running the experiment with a closed slit 2, the particles pass through Sl. 1 and pile up in a wave-like pattern, which we represent as $ψ (x)$ , the amplitude at a point $x$ on the screen. Here the probability of a particle hitting the $x$ point is:

P_{1} (x) = ψ^{2}

Pasted image 20240406194833.png

If we close Sl. 1 and open Sl. 2, we get $P_{2} (x) = ψ^{2}$ .

If we open both slits and rerun, the particles can either enter Sl. 1 or 2. The probability $P_{12}$ of a sand particle hitting the screen at $x$ is the sum of the probability of the particle passing through Sl. 1 and separately through Sl. 2:

P_{12} (x) = P_{1} (x) + P_{2} (x) = ψ^{2}

Pasted image 20240406195102.png

Now if we replace the sandblaster with an electron gun, we expect a similar behavior, expecting the following:

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But we get:

Pasted image 20240406195116.png

If electrons are a wave, then we expect wave interference, where we'll get constructive interference at the light fringes, and destructive interference at the dark ones. Even if we do the experiment and release one electron at a time, to make sure that electron collisions don't happen on the other side of the slit, we get the wave-like pattern.

This implies that the electron must somehow pass through both slits and interfere with itself on the other side, which can only happen if the electron acts as a wave.

Pasted image 20240406195428.png

So if $ψ_{1} (x, t)$ is the probability amplitude of the electron passing through Sl. 1 and similar for $ψ_{2} (x, t)$ , then our total probability amplitude is:

ψ (x, t) = ψ_{1} (x, t) + ψ_{2} (x, t)

where $x$ is the position on screen, the electron is found at time $t$ .

1.6: Wavefunctions and Probability Amplitudes

We (hopefully) know that electric fields $\vec{E}$ and magnetic fields $\vec{B}$ are perpendicular to each other. But we can talk about the energy density of the wave. The energy density $u$ of an EM wave is given by:

u = ε_{0} E^{2} = \frac{1}{μ_{0}} B^{2}

where $ε_{0}$ is the permittivity of free space (the electric constant) and $μ_{0}$ is the permeability of free space (the magnetic constant).

Namely, the thing is that higher energy density correlates to a higher probability of finding a photon. Hence, the probability of finding the photon is proportional to the square of the fields, as given by the equations.

In that light, we define wavefunctions/probability amplitude $ψ (x, t)$ . Notice that, like in the double slit experiment, we got a probability distribution like:

Pasted image 20240406195940.png

But if we square that we get:

Pasted image 20240406195955.png

Namely, the height at any point on the square graph is proportional to the probability of finding the electron at that point.

Hence, if we want to know the probability of finding the particle in range $x_{1}$ to $x_{2}$ we can just do an integral:

P_{12} = \int_{x_{1}}^{x_{2}} | ψ (x, t) |^{2} d x

and the probability of finding the particle anywhere should be 1:

P (x, t) = \int_{- \infty}^{\infty} | ψ (x, t) |^{2} d x = 1

In general, to make the $ψ$ work with the Schrodinger equation, we say that:

ψ (x, t) = A \cos (k x - ω t) + i A \sin (k x - ω t)

where $k = \frac{2 π}{λ}$ and $λ = \frac{h}{p}$ and $h$ is Plank's constant, $p$ is the particle's momentum, $ω t$ is the phase factor of the wave, and $ω = 2 π ν$ where $ν$ is the frequency of the wave (oh and $A$ is the amplitude of the wave).

We know that $e^{i θ} = \cos (θ) + i \sin (θ)$ so then we can say that:

ψ (x, t) = A e^{i (k x - ω t)}

But notice that $e^{- i ω t}$ is just a constant for a given time $t$ , so then we usually multiply it into the arbitrary constant $A$ , and thus be ignored:

ψ (x, t) = A e^{i k x}

where $A \in C$ .

The above equation can be rewritten as the following momentum eigenfunction via the Schrodinger Equation:

ψ (x, t) = A e^{\frac{i p_{x} x}{ℏ}}

The probability is still:

P (x, t) = | ψ (x, t) |^{2}

2.2: Complex Numbers

There's a lot of complex number math that I'm familiar with, but here's an overview.

$\sqrt{- 1} = i$
$a + b i = z$
$a = r \cos (θ)$
$b = r \sin (θ)$
$z = r (\cos (θ) + i \sin (θ)) = r e^{i θ}$
(De Moivre's Theorem) $z^{n} = (r (\cos (θ) + i \sin (θ)))^{n} = r^{n} (\cos (n θ) + i \sin (n θ))$
(Complex Conjugate) $z^{*} = a - b i$
(Norm) $| z | = \sqrt{a^{2} + b^{2}}$
$| z | = z z^{*}$
$| z |^{2} = r^{2} = (r e^{i θ}) (r e^{- i θ})$

2.5: Linear Vector Spaces

Here, we learn all about the $| ⟩$ bracket notation. Given by \bracket{} in latex:

| 0 ⟩

A lot of this is review for me, see Chapter 1 - Vector Spaces for more info on what's covered here. In short, we have the following:

A linear vector space $V$ is a collection of vector that meet the following:
- $| X ⟩ + | Y ⟩ = | Y ⟩ + | X ⟩$ (commutativity and closure)
- (associative property)
- There's a null or zero vector where $| 0 ⟩ + | X ⟩ = | X ⟩$ for any $| X ⟩ \in V$
- Each vector $| X ⟩$ has an additive inverse $| - X ⟩$ such that $| X ⟩ + | - X ⟩ = | 0 ⟩$
There's scalar multiplication with distributive properties:
- $a (| X ⟩ + | Y ⟩) = a | X ⟩ + a | Y ⟩$
- $a (b | X ⟩) = a b | X ⟩$
- There's a multiplicative identity 1 where $1 | X ⟩ = | X ⟩$
- $0 | X ⟩ = | 0 ⟩$

The scalars belong to some field $F$ which essentially just denotes the reals $R$ or the complex numbers $C$ .

A set of these vectors is linearly independent if the following sum only is satisfied by all $a_{i} = 0$ :

\sum_{i = 1}^{n} a_{i} | v_{i} ⟩ = | 0 ⟩

Otherwise the vectors are linearly dependent and if we know say $a_{1} \neq 0$ then we know that:

| v_{1} ⟩ = \frac{- a_{2} | v_{2} ⟩ - \dots - a_{n} | v_{n} ⟩}{a_{1}}

A vector space of $n$ linearly-independent vectors is said to have dimension $n$ . It's written $V^{n}$ . A vector $| V ⟩$ in $n$ -dimensional vector space can be written as a linear combination of $n$ linearly independent vectors ${| v_{1} ⟩, . . ., | v_{n} ⟩}$ . This set of $n$ linearly independent vectors is called a basis:

| V ⟩ = \sum_{i = 1}^{n} a_{i} | v_{i} ⟩

Each $a_{i}$ is the component of the state vector $| V ⟩$ on the basis $| v_{i} ⟩$ . For example, the standard basis for $R^{3}$ are the vectors:

\hat{i} = (1, 0, 0), \hat{j} = (0, 1, 0), \hat{k} = (0, 0, 1)

We can add two of these vectors where:

| V_{1} ⟩ = \sum_{i = 1}^{n} a_{i} | v_{i} ⟩, | V_{2} ⟩ = \sum_{i = 1}^{n} b_{i} | v_{i} ⟩ \Rightarrow | V_{1} ⟩ + | V_{2} ⟩ = \sum_{i = 1}^{n} (a_{i} + b_{i}) | v_{i} ⟩

and scaled by some $α \in F$ :

α | V ⟩ = \sum_{i = 1}^{n} (α a_{i}) | v_{i} ⟩

Inner Product

The inner product takes in two vectors $| X ⟩, | Y ⟩$ and outputs a scalar, denoted $⟨ X | Y ⟩$ . Here:

$⟨ X | Y ⟩ = {⟨ Y | X ⟩}^{*}$
$⟨ X | X ⟩ \geq 0$ and is only equal iff $| X ⟩ = | 0 ⟩$
$⟨ α X | Y ⟩ = α ⟨ X | Y ⟩$
$⟨ X + Y | Z ⟩ = ⟨ X | Z ⟩ + ⟨ Y | Z ⟩$ (additivity in the first slot)
Additivity in the second slot
Additivity in both slots

Two vectors are orthogonal if $⟨ X | Y ⟩ = 0$ .

The norm of a vector $| V |$ is given by:

\sqrt{⟨ V | V ⟩} = | V |

If the norm is 1, then $| V ⟩$ is normalized.

If a set of basis vectors are normalized and are orthogonal to one another, they are an orthonormal basis.

Assume $| V_{a} ⟩ = \sum_{i = 1}^{n} a_{i} | v_{i} ⟩$ and similarly $| V_{b} ⟩ = \sum_{i = 1}^{n} b_{i} | v_{i} ⟩$ . The inner product of the two vectors is:

⟨ V_{a} | V_{b} ⟩ = \sum_{i} \sum_{j} a_{i}^{*} b_{j} ⟨ v_{i} | v_{j} ⟩

Where since all the $v_{i}, v_{j}$ 's are orthonormal to each other, then only:

⟨ v_{i} | v_{j} ⟩ = {\begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases} = δ_{i j}

This function $δ_{i j}$ is the Kronecker delta function. Applying this, we get that:

⟨ V_{a} | V_{b} ⟩ = \sum_{i = 1}^{n} a_{i}^{*} b_{i}

We know that $| V_{a} ⟩, | V_{b} ⟩$ are given by their components:

| V_{a} ⟩ = (a_{1}, . . ., a_{n}), | V_{b} ⟩ = (b_{1}, . . ., b_{n})

So then:

⟨ V_{a} | V_{b} ⟩ = [\begin{matrix} a_{1}^{*} & \dots & a_{n}^{*} \end{matrix}] [\begin{matrix} b_{1} \\ ⋮ \\ b_{n} \end{matrix}]

If we instead have some continuous functions $a_{i} = ψ_{a} (x), b_{i} = ψ_{b} (x)$ then:

⟨ V_{a} | V_{b} ⟩ = \int ψ_{a}^{*} (x) ψ_{b} (x) d x

2.6: Using Dirac's Bra-ket Notation

We've seen writing an abstract vector $| V ⟩$ as a column vector in basis $| v_{i} ⟩$ . We write $| V ⟩$ as:

| V ⟩ = [\begin{matrix} a_{1} \\ \dots \\ a_{n} \end{matrix}]

The conjugate transpose vector is $⟨ V |$ on the basis $⟨ v_{i} |$ writing it as:

⟨ V | = [\begin{matrix} a_{1}^{*} & \dots & a_{n}^{*} \end{matrix}]

So for every ket vector $| V ⟩$ there's a corresponding bra vector $⟨ V |$ and vice versa. Together, these vectors form dual vector spaces.

Here we are assuming that:

| v_{i} ⟩ = [\begin{matrix} 0 \\ ⋮ \\ 0 \\ 1 \\ 0 \\ ⋮ \\ 0 \end{matrix}]

at the corresponding $i$ -th position. As such, then:

| V ⟩ = a_{1} [\begin{matrix} 1 \\ 0 \\ ⋮ \\ 0 \end{matrix}] + a_{2} [\begin{matrix} + 0 \\ 1 \\ ⋮ \\ 0 \end{matrix}] + \dots + a_{n} [\begin{matrix} 0 \\ 0 \\ ⋮ \\ 1 \end{matrix}]

Missed Lecture Slides: Hamiltonian

The evolution of a wave function with time is governed by the time-dependent Schrodinger equation:

i ℏ \frac{\partial}{\partial t} Ψ (x, t) = H Ψ (x, t)

Here $H$ is the Hamiltonian and is the energy operator. Namely:

H = \underset{Kinetic energy Operator}{\underset{⏟}{- \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}}} + \underset{potential energy operator}{\underset{⏟}{U (x)}}

In contrast, we have the time-independent Schrodinger equation in 3D:

\nabla^{2} ψ (\vec{r}) = \frac{2 m}{ℏ^{2}} (U (\vec{r}) - E) ψ (\vec{r})

with $\vec{r} = (x, y, z)$ position and $\nabla$ is the Laplace Operator:

\nabla^{2} = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}}

In spherical coordinates it becomes:

![[Physics CPE 345 Quantum Computing Lecture slides Quantum superposition 240408.pdf#page=17]]

The general solution for the wave function in hydrogen is:

Ψ_{n l m} (\vec{r}, t) = R_{n l} (r) Y_{l m} (θ, ϕ) e^{- i \frac{E_{n}}{ℏ} t}

Where $n, l, m$ are the quantum numbers, $R_{n l}, Y_{l m}$ are functions in a LUT from these quantum numbers, and $E_{n}$ is the energy of state $n$ .

As an example, in the ground state of hydrogen we get:

Ψ_{100} (r, t) = \frac{1}{\sqrt{π a_{B}}} e^{- \frac{r}{a_{B}} - i \frac{E_{1}}{ℏ} t}

where $a_{B}$ is the Bohr radius.