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In Townsend's A Modern Approach to Quantum Mechanics he states:

"Although it is not possible to obtain a single value for the measurement of the position of [a] particle, nonetheless kets such as $|x\rangle$ in which the particle has a single position are very useful. We may think of the physical states that occur in nature as a superposition of these position eigenstates. We are then presented with the following incorrect assumption:

$$|\psi\rangle=\sum_i|x_i\rangle\langle x_i|\psi\rangle$$

but rather that

$$\int_{-\infty}^\infty|x \rangle\langle x|\space dx = 1 $$

I'm new to quantum mechanics and for some reason I can't understand exactly what the integrand means, and I rationalize it to myself in this way: any position vector can be expressed as a linear combination of the position eigenvectors. But what are these position eigenvectors? Are they all of the possible positions an object could take, since any possible position must be an "eigenposition"?

Furthermore, why are we using the projection operator at all? Is it because the position of whatever we're measuring is exactly equal to one and only one eigenposition, and all other eigenpositions are orthogonal? Therefore taking the inner product between all possible positions and the position of the particle must, at some point, be one, since the particle has to exist somewhere? However, in light of the Heinsenberg uncertainty principle, an object doesn't have a definite position, so how can we even discuss eigenpositions at all?

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  • $\begingroup$ For anyone who will answer: this question has nothing to do with quantum mechanics. It's just linear algebra and notation. $\endgroup$ – DanielSank Apr 25 '17 at 1:35
  • $\begingroup$ The validity of the main equation in question relies explicitly on the uncertainty principle, and I ask a question at the end related to it. Aside from this, yes, it is a linear algebra question $\endgroup$ – john morrison Apr 25 '17 at 1:38
  • $\begingroup$ This is an issue with resolution of identity, as per en.wikipedia.org/wiki/… . The uncertainty principle doesn't enter into this. $\endgroup$ – ZeroTheHero Apr 25 '17 at 2:40
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I think it's best when first studying quantum mechanics to make sense of the expressions in a purely formal way. It takes sophisticated mathematics to express rigorously, and it doesn't really give you much physics back.

The operator "1" does what's on the tin. There is no projection involved: It is a "do-nothing" operator which takes a wavefunction and gives you back that same wavefunction. One very convenient resolution of the operator "1" is $\int_{\mathbb{R}}|x \rangle\langle x|\ dx$, which is to be understood in a purely formal sense, and is completely analogous to how the identity matrix can be expressed as $[1,0]^T[1,0]+[0,1]^T[0,1]$, just taken to its logical extreme: instead of the diagonal of the identity matrix being labeled by $(1,2)$, it is now labeled by $\mathbb{R}$.

I don't think it's good to interpret this resolution of the identity as any physical statement about the world. Once you've accepted that you have a wavefunction $|\psi\rangle$ at all, you can make whatever linear operators or mathematical formalism you want.

If you made an observation (so as to ask a question about physics and not linear algebra), you could find the expectation value of the operator $|y\rangle\langle y|$. This would give you something physical: the probability density of finding a particle at position $y$. However, if you find the expectation value of the operator in question, $\int_{\mathbb{R}}|x \rangle\langle x|\ dx$, you get $\langle\psi|\psi\rangle=1$. This gives you no information at all: an observation of the $1$ operator does nothing.

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  • $\begingroup$ It's a good way to think about this: as a purely formal object. $\endgroup$ – ZeroTheHero Apr 25 '17 at 2:18
  • $\begingroup$ When we expressed spin states as linear combinations of eigenstates of the spin operator, that made sense. Why can't we do it for something which has a continuous basis such as position? $\endgroup$ – john morrison Apr 25 '17 at 3:19
  • $\begingroup$ @johnmorrison We do, but that is a totally unrelated question to your post as far as I see it! The operator $\hat{x}=\int_{\mathbb{R}}x|x \rangle\langle x|\ dx$ sends the wavefunction $\psi(x)$ to the function $x\psi(x)$. The eigenstates of this operator are $|x\rangle$ with eigenvalues $x$. Alternatively and equivalently, the eigenfunctions are $\psi(x)=\delta(x-\lambda)$ with eigenvalue $\lambda$. The wavefunction can be expressed as a linear combination of these eigenfunctions: $|\psi\rangle=\int_{\mathbb{R}}\psi(x)|x \rangle\ dx$. Though mathematical rigor is admittedly out the window. $\endgroup$ – user12029 Apr 25 '17 at 3:38
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In QM, we treat functions as vectors - (in the sense of abstract vector spaces, functions are vectors belonging to a Hilbert space). In 'standard' linear algebra, any vector $\mathbf{v}$ can be expanded in terms of an arbitrary orthonormal basis $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ by writing $\mathbf{v} = c_1 \mathbf{e}_1 + c_2 \mathbf{e}_2 + c_3 \mathbf{e}_3$

In standard linear algebra, we find the coefficients by taking the dot product of both sides: $\mathbf{e}_1 \bullet \mathbf{v} = \mathbf{e}_1 \bullet (c_1 \mathbf{e}_1 + c_2 \mathbf{e}_2 + c_3 \mathbf{e}_3) = c_1$. Dotting with $\mathbf{e}_1$ 'pulls out' the coefficient thanks to orthonormality of the basis vectors.

With functions, we (usually) define the inner product (a generalised 'dot') by some form of integral. For example, we might define $\langle f, g \rangle = \frac{1}{2L}\int_{-L}^L f(x) g^*(x) dx$

For example, in the theory of Fourier series and Fourier transforms, we use the complex exponentials as the 'basis vectors' to expand an arbitrary function $\psi(x)$ as $$\psi(x)=\sum_{k=-\infty}^\infty c_k e^{2\pi ikx}$$ To actually determine the $c_k$'s, we take the inner product with $e^{2\pi i k x}$ $$\langle \psi, e^{2\pi ikx}\rangle = \frac{1}{2L}\int_{-L}^L \psi(x)e^{-2πikx}dx = \sum_{k=-\infty}^\infty \langle c_k e^{2\pi ikx}, e^{2\pi ikx}\rangle = c_k$$

QM applies the same idea to position wave functions - we express it as a weighted sum of orthogonal functions, where the $c_k$ coefficients now reflect the overlap integral of the wavefunction with the position eigenfunctions, or informally, "the amount that the wavefunction looks like this eigenfunction" - the position eigenfunctions are no more physically significant than their linear combinations, they just happen to be mathematically convenient.

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It is the generalisation of the discrete spectrum case:

$$ \sum_i |\alpha_i\rangle \langle\alpha_i| = 1 \longleftrightarrow \int_{-\infty}^\infty |x\rangle\langle x| dx = 1 $$

What both these relations are telling you is that the $|x\rangle$s or the $|\alpha_i\rangle$s form a complete basis by themselves and that any (wave)function can be expressed in either of these basis. If the basis are not complete then they are somewhat useless for expressing a wave function because you cannot account for all the possibilities. Linear Algebra allows you to describe vectors as more than just arrows. Anything that follows the laws of vector spaces is a vector. And functions happen to follow all those laws and we get much (not all) of the power of linear algebra for free.

The discrete case is what you have already encountered in the spin $\frac{1}{2}$ states of an electron. When a wavefunction is expanded in the basis of spin $\frac{1}{2}$ states then it is said to be expressed in the spinor space (space is a fancy word for all possible values for a vector).

Now, as you can never have a precise value for the position you ask what does that basis even mean? The best classical analogue that I can now offer to you is the Maxwell Boltzmann distribution of speeds for an ideal gas.

$$ f(v) = \sqrt{\Big(\frac{m}{2 \pi kT}\Big)^3} 4\pi v^2 e^{\frac{-mv^2}{2kT}} $$

Here, if I ask you what is the probability that you will find a gas molecule travelling at precisely 2.5 m/s you will say 0. That is because you allow infinite number of values and choose only one precise value from them.

In very much the same way then, if we express our wavefunction in the position basis, we will end up with some sort of distribution. That distribution will be our particle's position probability distribution. Formally, we shall call it wavefunction expressed in the position space. And even when you are unable to obtain precise values of the position itself you will be able to calculate the probability of locating the particle in a range.

Let's again visit the discrete case briefly. To determine the probability of obtaining a particular eigenvalue you sum across all projections that can give you the eigenvalues you are interested in. Formally you will have a state $|\psi\rangle = \sum_i c_i |a_i\rangle$ and let's say states $j \in J $ all have the desired eigenvalues you are interested in then the total probability of obtaining a desired eigenvalue in the set $J$ will be found by applying the projections for all those eigenstates that yield those eigenvalues and summing up the coeffecients. Formally, the state "collapses" post measurement (projection) to:

$$ \sum_j |a_j\rangle \langle a_j| \psi \rangle = \sum_j |a_j\rangle \langle a_j| \sum_i c_i |a_i\rangle = \sum_j |a_j\rangle \sum_i c_i \delta_{ij} = \sum_j c_j |a_j\rangle $$

And the probability of being in this state (as calculated before measurement) is therefore, $\sum_j |c_j|^2$.

Similarly, in the position space the probability of finding the particle between between $x- \frac{dx}{2}$ and $x+ \frac{dx}{2}$ is $|x\rangle \langle x | \psi \rangle dx$. Now, let's say you want to find out the probability of this particle being in a range $x_0 + \Delta$ and $x_0 - \Delta$ then you should sum this projection formula over that range (sum is integral when we deal with infinitesimal bucket. Formally, this becomes:

$$\int_{x_0-\Delta}^{x_0+\Delta} |x \rangle \langle x| \psi \rangle dx$$

Edit: If you have the time, you may find reading the first three chapters of Hoffman and Kunze very rewarding.

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protected by Qmechanic Apr 25 '17 at 11:13

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