0
$\begingroup$

Currently I'm reading Wolfgang Scherer's Mathematics of Quantum Computing, An Introduction.

I am really enjoying this book as it treats the subject from a mathematical point of view, which for me is helpful since it lets me understand how complicated physical questions link back to the mathematics I'm used to. For some context, I am a studying pure mathematics and going into fourth year, so have a background on real analysis, group theory, ring theory, metric spaces, topology and linear algebra to name a few.

Right now I'm at the point where he introduces the postulates used in Quantum Mechanics, and having compared to online materials, I find it hard to understand the link between his definitions, and what I've found online. Here is how he phrases the first postulate:

Postulate 1, (Observables and Pure states): An observable, that is, a physically measurable quantity of a quantum system is represented by a self-adjoint operator on a Hilbert space $\mathbb{H}$. If the preparation of a statistical ensemble is such that for any observable represented by its self-adjoint operator A the mean value of the observable can be calculated with the help of a vector $\vert \psi\rangle \in \mathbb{H}$ satisfying $\vert\vert\psi\vert\vert=1$ $$\langle A\rangle_{\psi}:=\langle\psi\vert A\psi\rangle$$ then the preparation is said to dscribe a pure state represented by the vector $\vert \psi\rangle \in \mathbb{H}$. One calls $\vert \psi\rangle$ the state vector or simply the state, and $\langle A\rangle_{\psi}$ is alled the (quantum mechanical) expectation value of the observable A in the pure state $\vert\psi\rangle$. The space $\mathbb{H}$ is said to be the HILBERT space of the quantum system.

Compare this to a definition for the first postulate found here, for example:

Postulate 1: The state of a quantum mechanical system is completely specified by the function $\psi(\mathbf{r},t)$ that depends on the coordinates of the particle, $\mathbf{r}$ and the time $t$. This function is called the wavefunction or state function and has the property that $\psi^{*}(\mathbf{r}, t)\psi(\mathbf{r},t)d\tau$ is the probability that the particle lies in the volume element $d\tau$ located at $\mathbf{r}$ and time $t$.

I am not including the full postulate, but they go on to describe the normalisation condition as an integral over $\mathbb{R}$ of $\psi^{*}(\mathbf{r}, t)\psi(\mathbf{r},t)$, which evaluates to $1$.

I have to clarify that, in his book, Scherer only uses finite-dimensional Hilbert spaces, whereas in classical QM, I am to understand that infinite-dimensional Hilbert spaces are employed instead.

My questions are: How do these two definitions relate? What differences are there between QC and QM? Should I learn QM before venturing into QC?

Thanks in advance.

$\endgroup$
4
  • $\begingroup$ The formulation of "Postulate 1" in the linked pdf is misleading, to say the least. At any given time, the square of the absolute value of the wave function $\psi$ is a probability density. It is not a probability density for integration over space and time. $\endgroup$
    – MaoWao
    May 31, 2022 at 11:52
  • $\begingroup$ @MaoWao As someone who doesn't have a background in QM I would appreciate it if you would point me towards a correct formulation of "Postulate 1". $\endgroup$
    – ClassyPies
    May 31, 2022 at 11:56
  • $\begingroup$ @ClassyPies The Wikipedia formulation looks good to me $\endgroup$ May 31, 2022 at 11:59
  • $\begingroup$ @BenGrossmann Wikipedia's formulation looks similar to what Scherer presents. However compare that to those found here, here, and here $\endgroup$
    – ClassyPies
    May 31, 2022 at 12:07

1 Answer 1

2
$\begingroup$

It is not strictly necessary to learn quantum mechanics (QM) as a prerequisite to quantum computing (QC), since QC questions can be framed purely in terms of linear algebra; typically about unitary and self-adjoint operators over finite dimensional Hilbert spaces.

In their full generality, the postulates of QM describe systems whose state space could be an arbitrary Hilbert space. However, in practice one tends to consider either finite dimensional Hilbert spaces (essentially $\Bbb C^n$) or the Hilbert space $L^2(\Bbb R^n)$ of square-integrable complex-valued functions over $\Bbb R^n$. In this second case, the state-vector $\psi = \psi(\mathbf r,t)$ (describing the state of the system at a given time $t$) is a square-integrable "wave function" over $\Bbb R^n$, and the relevant inner product is given by $$ \langle \phi | \psi \rangle = \int_{\Bbb R^n} \phi^*(\mathbf r,t)\psi(\mathbf r,t)\,d\mathbf r. $$ If one focuses exclusively on this second case, then it is fine to say that the state-vector is always (within the context of interest) a wave-function of this kind.

$\endgroup$
2
  • $\begingroup$ Thank you! Now it makes sense. If I understand correctly, the general definition is modified to suit the specific system, where in one case you deal with qubits over a finite-dimensional Hilbert space, or wavefunctions in $L^{2}(\mathbb{R}^{n})$ (in regards to the examples discussed). $\endgroup$
    – ClassyPies
    Jun 1, 2022 at 9:28
  • $\begingroup$ @ClassyPies That's right $\endgroup$ Jun 1, 2022 at 12:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .