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So from what I understand $\langle w | v \rangle=\vec w^* \cdot \vec v$. Ok. I'm fine with that notation. But then I've seen $\langle x | y \rangle=\delta(x-y)$ and $\langle x | p \rangle=e^{-ixp/\hbar}$. I can see that these are the eigenfunctions of position and momentum respectively, but I don't see how they're derived. Could one of you explain this or point me to a resource which explains it? Thanks.

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    $\begingroup$ $\langle w | v \rangle=\vec w^* \cdot \vec v$ only holds when there are finitely many mutually disjoint states (i.e. when you're in a finite dimensional Hilbert space). More generally, we have (computationally speaking) $$\langle f \mid g \rangle = \int f^*(x)g(x)\,dx$$ $\endgroup$ – Omnomnomnom Sep 15 '14 at 22:40
  • $\begingroup$ In any case, you should begin by trying to figure out what the vectors $|x\rangle,|y\rangle,$ and $|p\rangle$ actually look like. $\endgroup$ – Omnomnomnom Sep 15 '14 at 22:43
  • $\begingroup$ Should ask this in the physics forum. $\endgroup$ – Argyll Sep 15 '14 at 22:59
  • $\begingroup$ If I am to explain this in an answer, can I begin by assuming that you already understand that $$ \hat x | \phi \rangle = x \cdot \phi(x,t)\\ \hat p | \phi \rangle = -i \hbar \frac{\partial \phi(x,t)}{\partial x}? $$ $\endgroup$ – Omnomnomnom Sep 15 '14 at 22:59
  • $\begingroup$ @Omnomnomnom Yes I do understand that. And sorry posting it here: I had both this site up and physics.SE and posted to the wrong one -- then I went to class. If you like I can delete this and repost to the physics site. $\endgroup$ – got it--thanks Sep 16 '14 at 0:04
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The thing to keep in mind is that the objects we primarily work with in quantum mechanics are state-vectors, AKA kets. These are vectors that are manipulated by operators, and compared with an inner product.

In the context of (one-dimensional) position and momentum, our kets can be represented as $$ |\psi \rangle = \psi : \Bbb R \to \Bbb C $$ That is, $\psi$ is a complex-valued function of $x$ (position). We can compare two kets with the following inner product: $$ \langle \phi|\psi \rangle = \int_{\Bbb R} \phi^*(x)\,\psi(x)\,dx $$ where $\phi^*(x)$ denotes the complex conjugate of $\phi(x)$.

Our position and momentum operators can be defined by

$$ \hat x | \phi \rangle = x \cdot \phi(x) \\ \hat p | \phi \rangle = -i \hbar \frac{d \phi}{dx}(x) $$ so that the expected values of each would be given by $$ \langle \hat x\rangle = \int_{\Bbb R} \phi^*(x)\,[x \cdot \phi(x)]\,dx = \int_{\Bbb R} x |\phi(x)|^2\,dx \\ \langle \hat p\rangle = - i\hbar\int_{\Bbb R} \phi^*(x)\,\frac{d \phi}{dx}(x)\,dx = $$ and that's all fine and well.

With that in mind, we want $|x\rangle = \phi_x(x)$ and $|p\rangle = \phi_p(x)$ to be eigenfunctions of the above operators. So, $\phi_{x,\lambda}(x)$ (the eigenfunction of $\hat x$ associated with $\lambda$) should be some function such that $\hat x(\phi_{x,\lambda}) = \lambda |x \rangle$. That is, $$ x \cdot \phi_x(x) = \lambda \phi_{x,\lambda}(x) $$ Why does $\phi_{x,\lambda}(x) = \delta(x-\lambda)$ work here?

For the next, $\phi_{p,\lambda}(x)$ should be some function such that $$ -i \hbar \frac{d}{dx}\phi_{p,\lambda}(x) = \lambda \phi_{p,\lambda}(x) $$ Why does $\phi_{p,\lambda}(x) = e^{-ix\lambda/\hbar}$ work here?

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  • $\begingroup$ Why is $\hat p | \phi \rangle = -i \hbar \frac{d \phi}{dx}(x)$ true? $\endgroup$ – Argyll Sep 16 '14 at 0:35
  • $\begingroup$ @Argyll that is a definition $\endgroup$ – Omnomnomnom Sep 16 '14 at 0:37
  • $\begingroup$ It can be considered as a definition for the purpose of learning. But it is not an a priori definition of momentum. $\endgroup$ – Argyll Sep 16 '14 at 0:40
  • $\begingroup$ @Argyll is there some other, canonical "a priori definition" of momentum used to derive this operator? I've only ever seen heuristic justifications like this, and I would be genuinely interested if you could link me to some sort of derivation. My own curiosity aside, since OP accepts the notion that $\hat p$ and $\hat x$ should be as they are generally given, I decided to begin from there. $\endgroup$ – Omnomnomnom Sep 16 '14 at 0:44
  • $\begingroup$ I think it is brilliant to start from the assumption! I somehow didn't think that way. As for deriving the form of the canonical relation between position and momentum, I am afraid I can't find any good resources. Sakurai's book referenced at the bottom of my post is the closest I can find. $\endgroup$ – Argyll Sep 16 '14 at 0:49
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Please post this in physics next time. But anyway..


First of all, $\langle x | y \rangle=\delta(x-y)$ and $\langle x | p \rangle=e^{-ixp/\hbar}$ are not "eigenfunctions" of any thing.

$\langle x | y \rangle=\delta(x-y)$ is by definition. It is a part of our attempt at characterizing systems in physical space with linear algebra. For our purpose, we can consider $\delta$ as an attempt to characterize instantaneous impulses with the expected properties $$ \begin{gather} 1. &\int_A\delta(x-a)=\begin{cases}1\quad&\text{if}\quad a\in A \\0\quad&\text{if}\quad a\not\in A\end{cases} \\2.&\int_A f(x)\delta(x-a)=\begin{cases}f(a)\quad&\text{if}\quad a\in A \\0\quad&\text{if}\quad a\not\in A\end{cases} \end{gather} $$ Intuitively, consider $A$ as a time interval $[b,c]$ in 1D and $a\in(b,c)$, and we traverse from one end of the interval to another. At first, the integral is zero, as we pass time $a$ the delta function switch on $f$ momentarily and intensely and then switch off right away so that the integral becomes $f(a)$ and keeps at $f(a)$ afterwards.

Now there is no function that possess the above properties. So the proper definition of the delta function had to wait until measure theory and Lebesgue integration came around. The definition section of the Wikipedia page has more details. Note that the animation at the beginning is incorrect mathematically (within the scope of my knowledge anyway).

$\langle x | p \rangle=e^{-ixp/\hbar}$ is a bit more involved. Let me give you some explanations here and conclude with a textbook reference. The explanations are tailored for students in mathematics, however.

Fundamentally, momentum is the cause of translation. Now that we are using vector spaces, we need to make clear the meaning of translation. Consider 1D for now, in quantum mechanics, the location of an particle is assumed to be fully described by an abstract vector space with a basis parameterized by $\mathbb{R}$, or $V=span\{v_x\}_{x\in\mathbb{R}}$. A particular location is represented as a particular vector $v$ from such a vector space $V$. Translation by $L$ then means to map $v_x$ to $v_{x-L}$.

For various reasons, we further assume that only countably many elements from $\{v_x\}_x$ are needed. So we only keep a sequence from the previously set, say $\{v_{x_i}\}^\infty_{i=1}$. Moreover. we require that $V$ is complete and normed.

Now consider the one parameter family of translation maps defined by $$ T[L](v_x)=v_{x-L} $$ with parameter $L\in\mathbb{R}$.

About $T[L]$.

  1. We expect $T[L]$ to be norm-preserving because norm is associated with probability of observations and physically translation preserves that. As a result, $T[L]$ is unitary $\forall L$.
  2. We expect that the result of the translation depends solely on how much you translate, and not on the initial position. As such, we expect the result of two success translations is the same as one translation of the equal amount. ie. $T[L_1+L_2](v_x)=T[L_2]\circ T[L_1](v_x) \quad\forall L_1, L_2\in\mathbb{R}\quad\forall x$.
  3. Importantly we expect $T$ to be linear. I do not know why

Note that expectation 2 means the inverse of $T[L]$ is $T[-L]$ and $T[0]=\mathbb{I}$.

1, 2, 3 together make the set $\{T[L]\}_L$ an one parameter group of unitary operators and by Stone's theorem bearing the same name, we arrive at the canonical conjugation relations between position and momentum operators. Details omitted here. The accepted answer to this post offers some insight.

Another thing is what position operator is. As a consequence of one of the fundamental postulates in quantum mechanics, any observation on position is represented by an operator on $V$ whose eigenvectors are $v_x$ with eigenvalues $x$. So the position operator $X$ is $$ X=\sum_{i=1}^\infty v_{x_i}\cdot v_{x_i}^* $$

Subsequently, the momentum kets are defined as the eigenvectors of $T$. From that, the projection relation between position and momentum kets can be obtained.

See Modern Quantum Mechanics by J. J. Sakurai (2e) chapter 1 for a different perspective.

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