# Continuous (non-discrete) Quantum States

1. Jun 30, 2014

### kq6up

I am watching James Binney's QM lectures on iTunes University, and also going through his free textbook. He is a tough teacher, but I love how many misconceptions he points out, and some of the points he makes are very subtle and mind blowing when the lightbulb comes on.

I am confused on this point in his lectures:

He states that $| \psi \rangle =\int _{ -\infty }^{ \infty }{ dx\psi (x)| x \rangle }$ is an analogy to the discrete state $| \psi \rangle=\sum _{ i }^{ all }{ a_i | E_i \rangle }$. Binney uses the lower case psi to describe the complete state that is formed out of the basis static states $E_i$.

I am sure he is correct, but I need to some baby steps to make a conceptual bridge between the two equations.

Anyone care to flush this out for me?

Thanks,
Chris

2. Jun 30, 2014

### WannabeNewton

The continuous basis is given by $| x \rangle$ and the projection of $|\psi \rangle$ along the basis $| x \rangle$ is just $\psi(x) \equiv \langle x | \psi \rangle$. The correspondence from the discrete to continuous case is then $| n \rangle \rightarrow |x \rangle$ and $a_n \rightarrow \psi(x)$. Lastly, in $|\psi \rangle = \sum _n a_n |n \rangle$ we are summing over the discrete label $n$ in discrete steps $n \rightarrow n +1$ corresponding to the discrete spectrum $|n \rangle$. For $|x \rangle$ we have a continuous label $x$ in infinitesimal "steps" $dx$ so we simply replace the sum with an integral over $dx$ since, very loosely speaking, a sum over a continuous label is nothing more than an integral. This correspondence gives us $|\psi \rangle = \int dx \psi |x \rangle$.

This replacement of sums with integrals for continuous cases is not special to QM of course. We do it all the time in physics, both classical and quantum. In fact in statistical mechanics one makes this replacement as if it was second nature when calculating partition functions of various quantum systems or when carrying a partition function into the classical limit. However in QM since we are often dealing with operators, one must be much more careful when making such heuristic transitions from discrete to continuous spectra.

3. Jun 30, 2014

### kq6up

I get the part below, but I am struggling with your first line. Because the example that pops into my mind for the discrete static states $|n \rangle$ are the different eigenfunctions for the harmonic oscillator. I am not sure how you would make those states continuous. The functions jump between integer powers of x as you increase n. Is there an example of discrete basis states that can turn into a continuous state on some sort of limit like you suggest?

This part is ok. I understand the analogy between the integral and the summation.

Thanks,
Chris

4. Jun 30, 2014

### WannabeNewton

Consider a particle in a finite box of size $L$. As usual we impose periodic boundary conditions on the solutions of the Schrodinger equation in this box. This results in a discrete spectrum of eigenfunctions $\psi_{\vec{n}}(\vec{x}) = \frac{1}{L^{3/2}}e^{2\pi i (\vec{n}\cdot \vec{x})/ L}$. Note that $\vec{P} \psi_{\vec{n}} = -i\hbar \vec{\nabla}\psi_{\vec{n}} = \frac{2\pi \hbar}{L}\vec{n}\psi_{\vec{n}}$ so these are also eigenfunctions of the momentum operator with eigenvalues $\vec{p} = \frac{2\pi \hbar}{L}\vec{n}$ or equivalently $\vec{k} =\frac{2\pi }{L}\vec{n}$ in terms of the wave-vector $\vec{k}$ of each Fourier mode.

We can therefore rewrite the eigenfunctions explicitly as Fourier modes $\psi_{\vec{k}} = \frac{1}{L^{3/2}}e^{i (\vec{k}\cdot \vec{x})}$; note that the distance between adjacent momentum eigenvalues goes like $|\Delta k_i | \sim \frac{1}{L}$. Now consider the limit $L \rightarrow \infty$. Then in this limit $|\Delta k_i| \ll |k_i|$ i.e. the distance between adjacent momentum eigenvalues becomes vanishingly small compared to the momentum eigenvalues themselves when we let the box size go to infinity. This means $\Delta k_i \sim dk_i$ so $\psi(x) = \sum_{\vec{k}}a_{\vec{k}}\psi_{\vec{k}} \sim \int d^3 k a_{\vec{k}}\psi_{\vec{k}}$ which is in fact a solution to the Schrodinger equation for a free particle. The above limiting procedure is called box normalization.

5. Jun 30, 2014

### kq6up

Thanks, that does help indeed. Is the x operator always treated as continuous? That is -- under the integrand and not the summation?

Chris

6. Jun 30, 2014

### kq6up

This one is killer. I could just accept it and move on, but I am sure there may be some juicy insight into not just taking it to the bank because Dr. Benny said so. That is:

$I|\psi \rangle=\int{dx|x\rangle \langle x | \psi \rangle}$ where I is the identity operator. I get this if I use an orthonormal vector basis (e.g. $I=|\hat{e_n}\rangle \langle \hat{e_m}|$ where $e_n e_m=0 \quad when \quad m\neq n$).

However, understanding this when $\hat{x}$ is continuous is hard for me to conceive.

Chris

7. Jun 30, 2014

### WannabeNewton

$|x \rangle$ forms a complete set of eigenstates so $\int dx |x \rangle \langle x| = \mathbb{1}$. This is a result of the spectral theorem for self-adjoint operators (which the position operator $X$ is on suitable domains for given boundary conditions): http://en.wikipedia.org/wiki/Self-adjoint_operator#Spectral_theorem

And yes, $| x \rangle$ is always a continuous basis.

8. Jul 3, 2014

### samalkhaiat

Use
$$\delta ( x - y ) = \langle x | y \rangle \ ,$$
$$\psi ( x ) = \langle x | I | \psi \rangle = \langle x | \psi \rangle ,$$
and
$$\psi ( y ) = \langle y | \psi \rangle \ ,$$
in
$$\psi ( x ) = \int d y \ \delta ( x - y ) \ \psi ( y ) .$$
you find
$$\langle x | \left( I \right) | \psi \rangle = \langle x | \left( \int d y \ | y \rangle \langle y | \right) | \psi \rangle \ .$$

Sam