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I am confused about how one measures the dynamical variables (eg position) of a particle. I thought the wave function $\Psi(x,t)$ was the probability amplitude and $|\Psi(x,t)|^2$ represents the probability of finding a particle at that a given location and time.

But this information only gives us probabilities about position. What if I wanted to measure momentum? Do I get this info from the wave function too?

Qmechanic
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Stan Shunpike
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1 Answers1

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An observable that has a definite value is an eigenvalue of the operator. If $A$ is a hermitian matrix of which $|\psi\rangle$ is an eigenstate, we have $$\tag{1}A|\psi,a\rangle=a|\psi,a\rangle$$ You asked about the wave function, not the state vector, though. We can still get all the information we need from the wave function $\psi(x)=\langle x|\psi\rangle$, but we need to find a differential operator that corresponds to $A$. To do this, Fourier transform to $a$ space from $x$ space and find the operator that fulfills (1).

Suppose however that $|\psi\rangle$ is not an eigenstate. Then there exists no $a$ such that $$A|\psi\rangle=a|\psi\rangle$$ Instead, we have a probability distribution function $$P(a)=|\langle a|\psi\rangle|^2$$ where $|a\rangle$ is an eigenstate. This tells us the probability of our particle having $a$ between $a$ and $a+da$. To get $\psi(a)=\langle a|\psi\rangle$ is the trick then. We insert a full set of position eigenstates: $$\langle a|\psi\rangle=\int \langle a|x\rangle\langle x|\psi\rangle dx=\int \langle a|x\rangle\psi(x)dx$$ What is $\langle a|x\rangle$? We use the fact that $|x\rangle$ is Dirac normalizable: $$\int\frac{da}{2\pi}e^{ia(x-y)}=\delta(x-y)=\langle x|y\rangle=\int \langle x|a\rangle\langle a|y\rangle da$$ By inspection, we see $$\langle a|y\rangle=\frac{e^{-iay}}{\sqrt{2\pi}}$$ Thus $$\psi(a)=\int \frac{dx}{\sqrt{2\pi}}e^{-iax}\psi(x)$$ Which is of course just the Fourier transform.

Ryan Unger
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  • So, given an operator $A$, if an eigenstate exists, we write that eigenstate as $$A|\psi\rangle = a|\psi\rangle$$ by definition. Does an eigenstate always correspond to an observable? By "definite value", what do you mean? – Stan Shunpike Feb 15 '15 at 19:17
  • We preferably write $$A|\psi,a\rangle=a|\psi,a\rangle$$ or something similar. What you have is fine too. The $a$ labels the eigenstate. An observable refers to the operator. If there is no definite value, we get a PDF. For example, a wave packet does not have a definite position value. – Ryan Unger Feb 15 '15 at 19:19
  • What does the $,a$ add? Like what's its purpose? Also, so in my book by Griffiths, he uses $$Q\left(x,-i\hbar \frac{\partial}{\partial x}\right)$$ to denote an observable. Is it common to write the stuff in parentheses? I take it $A$ is our operator but I don't usually see $$A\left(x,-i\hbar \frac{\partial}{\partial x}\right)$$. I usually use hats because it keeps me certain what is an operator and what isn't. – Stan Shunpike Feb 15 '15 at 19:28
  • The $a$ lets us distinguish between a state $|\psi,a\rangle$ and a state $|\psi,a'\rangle$. I haven't read Griffiths' QM book, but I can hazard a guess as to what his notation means. $Q$ is the position operator. In the position basis it acts as $x$ and in the momentum basis it acts as $-i\hbar\partial_x$. – Ryan Unger Feb 15 '15 at 19:30
  • I bet you are right. I haven't seen him use it as a momentum operator. He just never explicitly called it the position operator when he introduced it. – Stan Shunpike Feb 15 '15 at 19:33
  • @StanShunpike: Careful: $Q$ is not a momentum operator, but it can be used in the momentum basis. – Ryan Unger Feb 15 '15 at 19:42
  • You are right! I just learned this. I was so happy! I couldn't figure out where $e^{ipx}$ came from and thought it too trivial to ask. But I found http://physics.stackexchange.com/q/35746/66165 and this guy explain how to shift bases. I still don't really understand what we mean by a position basis, but at least now I can move back and forth between position and momentum bases. What does it mean to have a position basis of position space? – Stan Shunpike Feb 15 '15 at 19:49
  • @StanShunpike: You can think of the various bases as being analogous to coordinate systems in a sense. We can work a problem in Cartesian coordinates. But sometimes it might be more convenient to use polar coordinates. In QM we have a similar situation. We might want to use a position or momentum or even energy basis. Ballentine's book has a chapter for problems in the position basis and another chapter for problems in the momentum basis. – Ryan Unger Feb 15 '15 at 20:31
  • Hey champ, care to chime in here: http://math.stackexchange.com/q/1155285/197705 its a basic question but I'd like a more thorough answer than the one I got – Stan Shunpike Feb 19 '15 at 04:05