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For the definition of the momentum operator $$\hat{P } = -i \hbar \nabla$$ in quantum mechanics, as I understand you can derive this by either considering a more general definition of momentum, i.e. 'canonical momentum' which is an operator and then apply this operator to wave functions. This is shown here in this wiki entry. We can alternatlively start with translations and use the face that the momentum operator is a generator of translations as is done here in this wiki entry.

What I am interested in is what is the more fundamental derivation for the position operator: $$\hat{X} = x.$$ To this point I have considered that the motivation for defining position operator is from the definition of the expectation value $$\langle x \rangle = \int dx x |\psi(x)|^2 = \langle \psi|x | \psi \rangle$$ where $| \psi \rangle$ is normalised. Is this the full extent of the motivation or the main point to consider? Is there another motivation for the position operator or is it taken from this motivation and confirmed by experiments to be acceptable?

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  • $\begingroup$ I believe this is more of a definition. Values of observables are given by the eigenvalues of their corresponding operator. Thus, this is what you would expect the position operator to do - give you the position. Equivalently, $p$ of course operates on the $\mid p \rangle$ states to give the momentum, $S_z$ gives the spin projection, etc. $\endgroup$ – Kyle Arean-Raines Aug 8 '16 at 14:29
  • $\begingroup$ You're making confusion between operators and their representations onto (some) bases. The answers below correctly address the issues. $\endgroup$ – gented Aug 8 '16 at 14:43
  • $\begingroup$ @GennaroTedesco: Because additional answers of varying quality might be added later, you might not want post a comment that commends "the answers below" without specifying which answers. $\endgroup$ – WillO Aug 8 '16 at 16:42
  • $\begingroup$ See e.g. Stone - von Neumann theorem on Wikipedia. If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Aug 8 '16 at 18:11
  • $\begingroup$ @WillO Oh yes, sure, that was silly :D. $\endgroup$ – gented Aug 8 '16 at 18:48
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${\hat X} = x$ because we choose to work in a basis of eigenvectors of ${\hat X}$, i.e. wave-functions.

We have a Hilbert space ${\cal H}$ which is a vector space on which we can choose any basis we wish. We most often choose to work in a basis that diagonalizes the position operator ${\hat X}$. Basis states satisfy $$ {\hat X} | x \rangle = x | x \rangle $$ Once such a basis is chosen, any state $|\Psi\rangle$ in ${\cal H}$ can be expanded in it, i.e. we can write $$ | \Psi \rangle = \int dx | x \rangle \Psi(x) $$ The function $\Psi(x)$ is called the wave-function. We can invert this to find $$ \Psi(x) = \langle x | \Psi \rangle $$

Now, what is the meaning of $\big({\hat X} \Psi \big)(x)$? By definition it is the wave-function of the state ${\hat X} | \Psi \rangle$, i.e. it is $\langle x | {\hat X} | \Psi \rangle$. It is then immediately obvious that $$ \langle x | {\hat X} | \Psi \rangle = x \langle x | \Psi \rangle = x \Psi(x) $$ Thus, we find that $\big({\hat X} \Psi \big)(x) = x \Psi(x)$. For this reason, we write for convenience ${\hat X} = x$, but one must remember that this is only true in the coordinate basis.

Momentum operator in coordinate basis: You might then ask how one can derive the expression for the momentum operator ${\hat P}$ in the coordinate basis. This is done as follows.

By definition $\big( {\hat P} \Psi \big)(x) = \langle x | {\hat P} | \Psi \rangle $. Then, $$ \big( {\hat P} \Psi \big)(x) = \int \frac{dp}{2\pi \hbar} \langle x | p \rangle \langle p | {\hat P} | \Psi \rangle = \int \frac{dp}{2\pi \hbar} \langle x | p \rangle \langle p | \Psi \rangle p = \int dx' \frac{dp}{2\pi \hbar} \langle x | p \rangle \langle p | x' \rangle \langle x' |\Psi \rangle p $$ Next, using $\langle x | p \rangle = e^{\frac{i}{\hbar} p x }$, we have $$ \big( {\hat P} \Psi \big)(x) = \int dx' \frac{dp}{2\pi \hbar}e^{\frac{i}{\hbar} p (x-x') } p \Psi(x') $$ We now write $$ e^{\frac{i}{\hbar} p (x-x') } p = - i \hbar \frac{\partial}{\partial x}e^{\frac{i}{\hbar} p (x-x') } $$ Using this, we can explicitly perform the integral over $p$ and we find $$ \big( {\hat P} \Psi \big)(x) = - i \hbar \frac{\partial}{\partial x} \Psi(x) $$ Thus, in coordinate basis, we write $$ {\hat P} = - i \hbar \frac{\partial}{\partial x} $$

PS - Of course, we are free to choose any basis we want. We could for instance work in momentum basis. In this basis, we would have $$ {\hat P} = p~, \qquad {\hat X} = i \hbar \frac{\partial}{\partial p} ~. $$

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  • $\begingroup$ Any chance we could discuss this in chat? The starting point of your definition is $$\hat{X}|x\rangle = x |x \rangle.$$ Is the thinking here that we start with the postulate of QM which states "any observable is an Hermitian operator which has eigenstates which are a basis for the Hilbert space". It doesn't make sense to not start from that point, since in the first two equations of yours you already assumed that $|x \rangle$ is an eigenstate of an operator that you had not yet defined, and in the second equation you assumed the eigenvectors $|x \rangle$ are complete. $\endgroup$ – Alex Aug 8 '16 at 17:07
  • $\begingroup$ What is wrong with deriving the definition of momentum (represented in the position basis) as was done in the links I provided (using the translation operator)? Is the problem just that it is limited to presentation in the position basis? $\endgroup$ – Alex Aug 8 '16 at 17:07
  • $\begingroup$ Yes to your second comment. We have already defined the Hermitian operators ${\hat X}$ and ${\hat P}$ by their commutation relation $[ {\hat X} , {\hat P} ] = i \hbar$. With this definition and the postulate of QM that you have stated, everything follows. I'm available on chat if you wish. $\endgroup$ – Prahar Aug 8 '16 at 17:09
  • $\begingroup$ Okay thanks, if we run out of comments will move it to chat. What do you mean we 'already defined the operators $\hat{X}$ and $\hat{P}$ by their commutation relation', don't you get the commutation relation from the definitions of $\hat{X}$ and $\hat{P}$? $\endgroup$ – Alex Aug 8 '16 at 17:18
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    $\begingroup$ Yes. That is right. $\endgroup$ – Prahar Aug 8 '16 at 23:47
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Note that $\hat{X} = x$ is strictly not correct. The left hand side is the operator acting on the Hilbert states of quantum states. The right-hand side is a real number.

The RHS is the position-representation of the LHS. As the OP recognised correctly this is a tautology to some extent, since it is the defining representation of the operator via the eigenvalue equation

$$\hat{X} |x\rangle = x |x\rangle.$$

So the a priori definition is this one with the notion that the $|x\rangle$-states are localised delta-functions, i.e.

$$\langle x'|x\rangle = \delta^{(3)}\left(x-x'\right).$$

Note that the second equation above follows from the first equation (up to a constant) under the assumption that $\hat{X}$ is hermitian and that its spectrum is continuous.

So far this was only some maths, if you would like to read about this in a more precise manner I recommend Quantum Optics in Phase Space by W. Schleich.

So far this was only maths. The important thing which determines the physics is quantisation, which consists of imposing the relation:

$$\left[\hat{X}, \hat{P} \right] = i\hbar$$

which is also how the first relation in the question (i.e. the momentum operator in position representation) is obtained.

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    $\begingroup$ Does $\langle x' | x \rangle = \delta(x' - x)$ follow from the following reasoning: $\langle x'| \hat{X} | x \rangle= \langle x| \hat{X} | x' \rangle^* = x' \langle x | x' \rangle^* = x'\langle x'|x \rangle = x \langle x'| x \rangle$. This holds for any $x, x'$ hence we must have that $\langle x'| x \rangle = \delta(x-y)$? $\endgroup$ – user100411 Aug 8 '16 at 18:51
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    $\begingroup$ @JohnDoe yep you are right. the last two equated imply that it is non-zero everywhere except when x=x'. and then to get complete states you need it to be a delta function (which is why i said "up to a constant" in the question; the just mentioned fixes that constant) $\endgroup$ – Wolpertinger Aug 8 '16 at 18:56
  • $\begingroup$ So just to confirm, the fact that the operators corresponding to observables are Hermtitian is a postulate rather than something derived? $\endgroup$ – user100411 Aug 8 '16 at 19:09
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    $\begingroup$ @JohnDoe yes, indeed. that is actually one of the axioms of quantum mechanics. it is loosely motivated by the fact that non-hermitian operators can have complex eigenvalues, which would be unphysical for observables. $\endgroup$ – Wolpertinger Aug 8 '16 at 22:06
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In QM you have a general postulate, that requires physical observables (momentum, angular momentum, coordinates) to be represented by Hermitian operators. Actually, in QM you have different bases as well as different notion of time evolution. Those are position and coordinate basis and Schrödinger and Heisenberg representations. Where the difference between the last two is that in the Heisenberg picture operators are time dependent. Your question is for the Schrödinger representation coordinate basis so lets concentrate on it.

In the coordinate representation, the basis is formed by an infinite set of vectors $| \bf{x} >$ for which the following holds $$ \hat{X}| \bf{x} > = \bf{x} | \bf{x} > $$
Every state vector $| \psi >$ can be expressed in this basis as $$ | \psi > = \int d\bf{x} |\bf{x}><\bf{x}|\psi>, $$ where we used the usual expression $ \int d\bf{x}|\bf{x}><\bf{x}|=\hat{I}$. So the position operator was easy, what about the momentum operator? Here you need more calculus. Similarly, for the momentum representation we have $$ \hat{P}| \bf{p} > = \bf{p} | \bf{p} > $$
and $$ | \psi > = \int d\bf{p} |\bf{p}><\bf{p}|\psi>. $$ To relate the two we note: $$ \psi(\bf{x})=<\bf{x}|\psi>=\int d\bf{p} <\bf{x}|\bf{p}><\bf{p}|\psi>=\int d\bf{p} <\bf{x}|\bf{p}>\phi(\bf{p}), $$ which means that you go from one base to the other via a Fourier transformation. Where the Fourier transform is given by $$ \psi(\bf{x})=\frac{1}{(2\pi \hbar)^{\frac{2}{3}}}\int d\bf{p}e^{\frac{i\bf{x}\cdot\bf{p}}{\hbar}}\phi(\bf{p}) $$ In this base, for the momentum operator $\hat{P}$ we have $$ <\bf{x}|\hat{P}|\psi>=<\bf{x}|\hat{P}\int d\bf{p}|\bf{p}><\bf{p}|\psi>= \frac{1}{(2\pi \hbar)^{\frac{2}{3}}}\int d\bf{p}\bf{p}e^{\frac{i\bf{x}\cdot\bf{p}}{\hbar}}\phi(\bf{p}) $$ Since $$ \bf{p}e^{\frac{i\bf{x}\cdot\bf{p}}{\hbar}}=-i\hbar \nabla e^{\frac{i\bf{x}\cdot\bf{p}}{\hbar}}, $$ we have $$ <\bf{x}|\hat{P}|\psi>=-i\hbar \nabla<\bf{x}|\psi>. $$ Or simply $\hat{P}=-i\hbar \nabla$ and for the other operator $\hat{X}| \bf{x} > = \bf{x} | \bf{x} >$ by definition, because we are working in the coordinate basis.

You can easily check that both $\hat{P}$ and $\hat{X}$ are hermitian and satisfy the Heisenberg relation.

P.S. You can have also the Dirac (interaction) picture where both states and operators depend on time.

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protected by Qmechanic Aug 8 '16 at 17:57

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