Clarification on how bra ket works here

[Kashmir]

"##
\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid E\rangle=E\langle x \mid E\rangle##
is often referred to as the time-independent Schrödinger equation in position space. This equation also results from projecting the energy eigenvalue equation
##\hat{H}|E\rangle=E|E\rangle##
into position space:

##\langle x|\hat{H}| E\rangle=E\langle x \mid E\rangle
##"

How is
##\langle x|\hat{H}| E\rangle
=\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid E\rangle##?

Please help me.

 

[PeterDonis]

"##
\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid E\rangle=E\langle x \mid E\rangle##
is often referred to as the time-independent Schrödinger equation in position space. This equation also results from projecting the energy eigenvalue equation
##\hat{H}|E\rangle=E|E\rangle##
into position space:

##\langle x|\hat{H}| E\rangle=E\langle x \mid E\rangle
##"

Where are you getting this quote from?

 

[Kashmir]

Where are you getting this quote from?

The eigenvalue equation of H has been operated by the bra <x| both sides. The RHS is understandable but the LHS, I'm not getting. Please help me

 

[PeterDonis]

The eigenvalue equation of H has been operated by the bra <x| both sides.

You're not answering my question. My question is, where are you getting the things you posted in the OP from? You put quote marks around them, which implies that you read them somewhere. Where?

 

[PeterDonis]

The eigenvalue equation of H has been operated by the bra <x| both sides.

That gives ##\bra{x} H \ket{E} = E \braket{x | E}##.

 

[Kashmir]

You're not answering my question. My question is, where are you getting the things you posted in the OP from? You put quote marks around them, which implies that you read them somewhere. Where?

From Townsend pg 214.

 

[Kashmir]

That gives ##\bra{x} H \ket{E} = E \bra{X} \ket{E}##.

Yes, but how
##\langle x|\hat{H}| E\rangle

=\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid E\rangle##?

 

[PeterDonis]

From Townsend pg 214.

Do you mean this?

https://www.amazon.com/dp/1891389785/?tag=pfamazon01-20

 

[Kashmir]

Do you mean this?

https://www.amazon.com/dp/1891389785/?tag=pfamazon01-20

Yes . I'm actually studying from a less advanced book 'McIntyre ' but found few passages from Townsend very helpful, I wanted to understand this part but I couldn't do it. My book McIntyre doesn't give me full explanation of this equation so I tried to study it from Townsend.

 

[PeterDonis]

Yes, but how
##\langle x|\hat{H}| E\rangle

=\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid E\rangle##?

I don't have Townsend so I don't know how much explanation he gives for what is being done here. I think the key point is that the equation ##\bra{x} H \ket{E} = E \braket{x | E}## is only valid for eigenstates of ##H##. For any such state, ##H \ket{E} = E \ket{E}##, and since ##E## is just a number we can pull it outside the bracket when we contract with ##\bra{x}##. Then he's just rewriting ##H## in differential operator form.

 

[Kashmir]

I don't have Townsend so I don't know how much explanation he gives for what is being done here. I think the key point is that the equation ##\bra{x} H \ket{E} = E \braket{x | E}## is only valid for eigenstates of ##H##. For any such state, ##H \ket{E} = E \ket{E}##, and since ##E## is just a number we can pull it outside the bracket when we contract with ##\bra{x}##. Then he's just rewriting ##H## in differential operator form.

Yes the equation is for the eigenstates. But we can't differentiate a ket |En> with respect to x ? I think we've to represent the ket as a wavefunction to do so.

 

[PeterDonis]

we can't differentiate a ket |En> with respect to x ? I think we've to represent the ket as a wavefunction to do so.

The bracket ##\braket{x | E}## is the wave function ##\psi(x)## corresponding to the eigenstate ##\ket{E}## of ##H##.

 

[George Jones]

With respect to what Peter said in post #12: equation (6.82) on page 214 of Townsend is
$$\left< x | \psi \left(t\right) \right> = \psi \left(x,t\right).$$

 

[Kashmir]

With respect to what Peter said in post #12: equation (6.82) on page 214 of Townsend is
$$\left< x | \psi \left(t\right) \right> = \psi \left(x,t\right).$$

Thank you, I understand that part. What's troubling me is :

##\begin{aligned}\langle x|\hat{H}| \psi(t)\rangle &=\left\langle x\left|\left[\frac{\hat{p}_{x}^{2}}{2 m}+V(\hat{x})\right]\right| \psi(t)\right\rangle \\ &=\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid \psi(t)\rangle \end{aligned}## in this equation the ##H## operator was in the middle, how did it go to the left in the last expression?

 

[PeterDonis]

how did it go to the left in the last expression?

I already answered that. Go back and read this again:

For any such state, ##H \ket{E} = E \ket{E}##, and since ##E## is just a number we can pull it outside the bracket when we contract with ##\bra{x}##. Then he's just rewriting ##H## in differential operator form.

You are now using ##\ket{\psi(t)}## instead of ##\ket{E}##, but the reasoning is the same.

 

[PeroK]

Thank you, I understand that part. What's troubling me is :

##\begin{aligned}\langle x|\hat{H}| \psi(t)\rangle &=\left\langle x\left|\left[\frac{\hat{p}_{x}^{2}}{2 m}+V(\hat{x})\right]\right| \psi(t)\right\rangle \\ &=\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid \psi(t)\rangle \end{aligned}## in this equation the ##H## operator was in the middle, how did it go to the left in the last expression?

Hint: express ##V(x)## and ##V(\hat x)## as Taylor series!

 

[PeroK]

PS for the other term, you need a proof (e.g. Sakurai, page 54), that:
$$\langle x|\hat p |\psi \rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\psi\rangle$$

 

[PeroK]

Exercise: explain with complete precision and clarity, what the following actually means(!):
$$\langle x|\hat p |\psi \rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\psi\rangle$$Until you can do that, you cannot make further progress!

 

[Kashmir]

Exercise: explain with complete precision and clarity, what the following actually means(!):
$$\langle x|\hat p |\psi \rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\psi\rangle$$Until you can do that, you cannot make further progress!

working on it.

 

[samalkhaiat]

"##
\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid E\rangle=E\langle x \mid E\rangle##
is often referred to as the time-independent Schrödinger equation in position space. This equation also results from projecting the energy eigenvalue equation
##\hat{H}|E\rangle=E|E\rangle##
into position space:

##\langle x|\hat{H}| E\rangle=E\langle x \mid E\rangle
##"

How is
##\langle x|\hat{H}| E\rangle
=\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right]\langle x \mid E\rangle##?

Please help me.

Start from [tex]H|E\rangle = E|E\rangle , \ \ \ \ H = \frac{P^{2}}{2m} + V(Q) ,[/tex] with [tex]P|p\rangle = p|p\rangle , \ \ Q|x\rangle = x|x\rangle .[/tex] Multiply by [itex]\langle x|[/itex] and insert the identity operator [itex]\int dy \ |y\rangle \langle y|[/itex] : [tex]E \langle x|E\rangle = \int dy \ \langle x|H|y\rangle \langle y|E\rangle .[/tex] Rewrite this in terms of the energy eigen-function [itex]\Psi_{E}(x) \equiv \langle x|E\rangle[/itex]: [tex]E \Psi_{E}(x) = \int dy \ \langle x|\left( \frac{P^{2}}{2m} + V(Q) \right)|y\rangle \ \Psi_{E}(y) .[/tex] Now, using [itex]V(Q) \ |y\rangle = V(y) \ |y\rangle[/itex] and [itex]\langle x|y\rangle = \delta (x - y)[/itex], we get [tex]\left( E - V(x) \right) \Psi_{E}(x) = \int dy \ \langle x |\frac{P^{2}}{2m}|y\rangle \ \Psi_{E}(y) .[/tex] Now, if I don’t know (or no one told me) how [itex]P^{2}[/itex] acts on [itex]|y\rangle[/itex], then I should insert the identity operator [itex]\int dp \ |p\rangle \langle p|[/itex], because I know that [itex]P^{2}|p\rangle = p^{2}|p\rangle[/itex]. So, [tex]\left( E - V(x) \right) \Psi_{E}(x) = \int_{\mathbb{R}^{2}} dy \ dp \ \langle x |\frac{P^{2}}{2m}|p\rangle \langle p|y \rangle \ \Psi_{E}(y) ,[/tex] or (by getting rid of the operator [itex]P^{2}[/itex]) [tex]\left( E - V(x) \right) \Psi_{E}(x) = \int_{\mathbb{R}^{2}} dy \ dp \ \frac{p^{2}}{2m} \langle x |p\rangle \langle p|y\rangle \ \Psi_{E}(y) . \ \ \ \ (1)[/tex] Now, suppose I don’t know the expressions for the complex numbers [itex]\langle x |p\rangle[/itex] and [itex]\langle p|y\rangle[/itex]. Again, that is not a problem: I can write the identity [tex]\langle x| \psi \rangle = \int dp \ \langle x|p\rangle \langle p|\psi \rangle ,[/tex] and convert it to a equation between wave-functions: [tex]\psi (x) = \int dp \ \langle x|p\rangle \ \varphi (p) .[/tex] Ah ha, so [itex]\langle x|p\rangle[/itex] must be the “matrix” which transforms a momentum-space wavefunction [itex]\varphi (p)[/itex] into the coordinate-space [itex]\psi (x)[/itex]. But, from the good old Fourier expansion, we know that [tex]\psi (x) = \frac{1}{\sqrt{2\pi \hbar}} \int dp \ e^{\frac{i}{\hbar}px} \ \varphi (p) .[/tex] So, we conclude that [tex]\langle x |p\rangle = \frac{1}{\sqrt{2\pi \hbar}} \ e^{\frac{i}{\hbar}px}.[/tex] Now, Eq(1) becomes [tex]\left( E - V(x) \right) \Psi_{E}(x) = \frac{1}{2 \pi \hbar} \int_{\mathbb{R}^{2}} dy \ dp \ \frac{p^{2}}{2m} \ e^{\frac{i}{\hbar}(x - y)p} \ \Psi_{E}(y) . \ \ \ (2)[/tex] Next, we need to produce the differential operator [itex]\frac{\partial^{2}}{\partial y^{2}}[/itex] inside the integral. Well, using elementary school differentiations, we have the identity [tex]- \hbar^{2} \frac{\partial^{2}}{\partial y^{2}} e^{- \frac{i}{\hbar}yp} = p^{2} \ e^{- \frac{i}{\hbar}yp} .[/tex] Substituting this in (2) we get [tex]\left( E - V(x) \right) \Psi_{E}(x) = \frac{- \hbar^{2}}{2m} \left( \frac{1}{2 \pi \hbar}\right) \int_{\mathbb{R}^{2}} dy \ dp \ \frac{\partial^{2}}{\partial y^{2}} \left( e^{\frac{i}{\hbar}(x - y)p} \right) \ \Psi_{E}(y) .[/tex] Now, integrate (twice) by parts to obtain [tex]\left( E - V(x) \right) \Psi_{E}(x) = \frac{- \hbar^{2}}{2m} \int_{\mathbb{R}} dy \left( \frac{1}{2 \pi \hbar} \int_{\mathbb{R}} dp \ e^{\frac{i}{\hbar}(x - y)p} \right) \ \Psi^{\prime \prime}_{E}(y) .[/tex] Finally, since [tex]\frac{1}{2 \pi \hbar} \int_{\mathbb{R}} dp \ e^{\frac{i}{\hbar}(x - y)p} = \delta (x - y) ,[/tex] you get [tex]\frac{- \hbar^{2}}{2m} \Psi^{\prime \prime}_{E} (x) + V(x)\Psi_{E}(x) = E \Psi_{E}(x) .[/tex] And to learn more about this stuff read my posts in

https://www.physicsforums.com/threa...how-do-we-find-the-hamiltonian-matrix.395680/

Source https://www.physicsforums.com/threa...how-do-we-find-the-hamiltonian-matrix.395680/

 

[vanhees71]

You can derive everything from the Heisenberg algebra,
$$[\hat{x},\hat{p}]=\mathrm{i}\hbar.$$
It's a bit hard to motivate, if you don't have a good understanding of classical canonical mechanics in terms of Poisson brackets, which define a (Lie) derivative and thus admit the understanding of the conserved quantities as generators of symmetry groups (Noether's theorem in Hamiltonian formulation, i.e., there's a one-to-one correspondence between conserved quantities and (one-parameter) Lie symmetries). The math, however is not too difficult:

The only idea you need from the above mentioned point of view of classical mechanics is that momentum is the generator of spatial translations, which is a symmetry of the Galilei-Newton spacetime model. This leads to the corresponding Poisson bracket and then to the above Heisenberg algebra for the position and momentum operators in QT. For simplicity I only consider motion in one dimension, the extension to 3D is straight forward.

The idea with the meaning of the momentum operator of translations leads to the idea to investigate the operator-valued function
$$\hat{X}(\xi) = \exp(\mathrm{i} \hat{p} \xi/\hbar) \hat{x} \exp(-\mathrm{i} \hat{p} \xi/\hbar).$$
You'll see, why this is a good idea, in a moment.

The key point is that you can explicitly evaluate this function. Just take the derivative wrt. ##\xi##:
$$\mathrm{d}_{\xi} \hat{X}(\xi) = \frac{1}{\hbar} \exp(\mathrm{i} \hat{p} \xi/\hbar) (\mathrm{i} \hat{p} \hat{x} - \mathrm{i} \hat{x} \hat{p} \exp(-\mathrm{i} \hat{p} \xi/\hbar) = \frac{1}{\mathrm{i} \hbar} \exp(\mathrm{i} \hat{p} \xi/\hbar) [\hat{x},\hat{p}] \exp(-\mathrm{i} \hat{p} \xi/\hbar) = \hat{1}.$$
Further you have ##\hat{X}(0)=\hat{x}## and thus by integration
$$\hat{X}(\xi)=\hat{x}+\xi \hat{1}.$$
Now consider the (generalized) position eigenstates ##|x \rangle##. Then define
$$|x,\xi \rangle=\exp(-\mathrm{i} \xi \hat{p}/\hbar) |x \rangle,$$
and we have
$$\hat{x} |x,\xi \rangle=\exp(-\mathrm{i} \xi \hat{p}/\hbar) \hat{X}(\xi) |x \rangle = (x+\xi) \exp(-\mathrm{i} \xi \hat{p}/\hbar) |x \rangle=(x+\xi) |x +\xi \rangle,$$
i.e., ##|x,\xi \rangle## is a position eigenstate with eigenvalue ##x,\xi##.

Thus we can define position eigenstates by
$$|x \rangle=\exp(-\mathrm{i} \hat{p} x/\hbar) |x=0 \rangle.$$
Now it's easy to evaluate the scalar product of the so defined position eigenstates with the momentum eigenstates ##|p \rangle##:
$$u_x(p)=\langle p|x \rangle = \langle p|\exp(-\mathrm{i} \hat{p} x/\hbar)|x=0 \rangle = \langle \exp(+\mathrm{i} \hat{p} x \hbar p|x=0 \rangle = \langle \exp(\mathrm{i} p x/\hbar) p|x=0 \rangle = \exp(-\mathrm{i} p x/\hbar) \langle p|x=0 \rangle=\exp(-\mathrm{i} p x) u_0(p).$$
To get ##u_0(p)## we note that we like to have
$$\langle p|p' \rangle=\delta(p-p'). \qquad (*)$$
Assuming the completeness of the position eigenstates constructed above then leads to
$$\langle p |p' \rangle =\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|p' \rangle = \int_{\mathbb{R}} \mathrm{d} x u_0(p) u_0^*(p') \exp[\mathrm{i} (p'-p) x/\hbar] = |u_0(p)|^2 2 \pi \hbar \delta(p-p').$$
With (*) it follows
$$u_0(p)=\frac{1}{\sqrt{2 \pi \hbar}}$$
modulo an irrelevant phase factor. This finally implies that
$$\langle p|x \rangle=\frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} x p/\hbar)$$
and thus
$$\langle x|p \rangle=u_p(x) =\langle p|x \rangle^* = \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} x p/\hbar).$$
This then implies the momentum operator in the position representation as follows:
$$\hat{p} \psi(x) := \langle x |\hat{p} \psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\hat{p} \psi \rangle = \int_{\mathbb{R}} \mathrm{d} p p u_p(x) \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p (-\mathrm{i} \hbar) \mathrm{d}_x u_p(x) \langle p|\psi \rangle = -\mathrm{i} \hbar \mathrm{d}_x \int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle = -\mathrm{i} \hbar \mathrm{d}_x \psi(x).$$
The momentum operator in position representation thus is
$$\hat{p} \psi(x)=-\mathrm{i} \hbar \psi(x).$$
In the same way you can derive the position operator in the momentum representation:
$$\hat{x} \tilde{\psi}(p) = +\mathrm{i} \hbar \mathrm{d}_p \tilde{\psi}(p)$$
and the transformation between position and momentum representation,
$$\tilde{\psi}(p) = \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} p x/\hbar) \psi(x)$$
and
$$\psi(x) = \langle x|\psi \rangle= \int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar) \tilde{\psi}(p).$$

 

[Kashmir]

Start from [tex]H|E\rangle = E|E\rangle , \ \ \ \ H = \frac{P^{2}}{2m} + V(Q) ,[/tex] with [tex]P|p\rangle = p|p\rangle , \ \ Q|x\rangle = x|x\rangle .[/tex] Multiply by [itex]\langle x|[/itex] and insert the identity operator [itex]\int dy \ |y\rangle \langle y|[/itex] : [tex]E \langle x|E\rangle = \int dy \ \langle x|H|y\rangle \langle y|E\rangle .[/tex] Rewrite this in terms of the energy eigen-function [itex]\Psi_{E}(x) \equiv \langle x|E\rangle[/itex]: [tex]E \Psi_{E}(x) = \int dy \ \langle x|\left( \frac{P^{2}}{2m} + V(Q) \right)|y\rangle \ \Psi_{E}(y) .[/tex] Now, using [itex]V(Q) \ |y\rangle = V(y) \ |y\rangle[/itex] and [itex]\langle x|y\rangle = \delta (x - y)[/itex], we get [tex]\left( E - V(x) \right) \Psi_{E}(x) = \int dy \ \langle x |\frac{P^{2}}{2m}|y\rangle \ \Psi_{E}(y) .[/tex] Now, if I don’t know (or no one told me) how [itex]P^{2}[/itex] acts on [itex]|y\rangle[/itex], then I should insert the identity operator [itex]\int dp \ |p\rangle \langle p|[/itex], because I know that [itex]P^{2}|p\rangle = p^{2}|p\rangle[/itex]. So, [tex]\left( E - V(x) \right) \Psi_{E}(x) = \int_{\mathbb{R}^{2}} dy \ dp \ \langle x |\frac{P^{2}}{2m}|p\rangle \langle p|y \rangle \ \Psi_{E}(y) ,[/tex] or (by getting rid of the operator [itex]P^{2}[/itex]) [tex]\left( E - V(x) \right) \Psi_{E}(x) = \int_{\mathbb{R}^{2}} dy \ dp \ \frac{p^{2}}{2m} \langle x |p\rangle \langle p|y\rangle \ \Psi_{E}(y) . \ \ \ \ (1)[/tex] Now, suppose I don’t know the expressions for the complex numbers [itex]\langle x |p\rangle[/itex] and [itex]\langle p|y\rangle[/itex]. Again, that is not a problem: I can write the identity [tex]\langle x| \psi \rangle = \int dp \ \langle x|p\rangle \langle p|\psi \rangle ,[/tex] and convert it to a equation between wave-functions: [tex]\psi (x) = \int dp \ \langle x|p\rangle \ \varphi (p) .[/tex] Ah ha, so [itex]\langle x|p\rangle[/itex] must be the “matrix” which transforms a momentum-space wavefunction [itex]\varphi (p)[/itex] into the coordinate-space [itex]\psi (x)[/itex]. But, from the good old Fourier expansion, we know that [tex]\psi (x) = \frac{1}{\sqrt{2\pi \hbar}} \int dp \ e^{\frac{i}{\hbar}px} \ \varphi (p) .[/tex] So, we conclude that [tex]\langle x |p\rangle = \frac{1}{\sqrt{2\pi \hbar}} \ e^{\frac{i}{\hbar}px}.[/tex] Now, Eq(1) becomes [tex]\left( E - V(x) \right) \Psi_{E}(x) = \frac{1}{2 \pi \hbar} \int_{\mathbb{R}^{2}} dy \ dp \ \frac{p^{2}}{2m} \ e^{\frac{i}{\hbar}(x - y)p} \ \Psi_{E}(y) . \ \ \ (2)[/tex] Next, we need to produce the differential operator [itex]\frac{\partial^{2}}{\partial y^{2}}[/itex] inside the integral. Well, using elementary school differentiations, we have the identity [tex]- \hbar^{2} \frac{\partial^{2}}{\partial y^{2}} e^{- \frac{i}{\hbar}yp} = p^{2} \ e^{- \frac{i}{\hbar}yp} .[/tex] Substituting this in (2) we get [tex]\left( E - V(x) \right) \Psi_{E}(x) = \frac{- \hbar^{2}}{2m} \left( \frac{1}{2 \pi \hbar}\right) \int_{\mathbb{R}^{2}} dy \ dp \ \frac{\partial^{2}}{\partial y^{2}} \left( e^{\frac{i}{\hbar}(x - y)} \right) \ \Psi_{E}(y) .[/tex] Now, integrate (twice) by parts to obtain [tex]\left( E - V(x) \right) \Psi_{E}(x) = \frac{- \hbar^{2}}{2m} \int_{\mathbb{R}} dy \left( \frac{1}{2 \pi \hbar} \int_{\mathbb{R}} dp \ e^{\frac{i}{\hbar}(x - y)p} \right) \ \Psi^{\prime \prime}_{E}(y) .[/tex] Finally, since [tex]\frac{1}{2 \pi \hbar} \int_{\mathbb{R}} dp \ e^{\frac{i}{\hbar}(x - y)p} = \delta (x - y) ,[/tex] you get [tex]\frac{- \hbar^{2}}{2m} \Psi^{\prime \prime}_{E} (x) + V(x)\Psi_{E}(x) = E \Psi_{E}(x) .[/tex] And to learn more about this stuff read my posts in

https://www.physicsforums.com/threa...how-do-we-find-the-hamiltonian-matrix.395680/

Source https://www.physicsforums.com/threa...how-do-we-find-the-hamiltonian-matrix.395680/

This was extremely helpful, Thank you so much.

Also in the equation below your 2nd equation you've forget to write ##p## in the exponent with ##x-y##

 

[Kashmir]

You can derive everything from the Heisenberg algebra,
$$[\hat{x},\hat{p}]=\mathrm{i}\hbar.$$
It's a bit hard to motivate, if you don't have a good understanding of classical canonical mechanics in terms of Poisson brackets, which define a (Lie) derivative and thus admit the understanding of the conserved quantities as generators of symmetry groups (Noether's theorem in Hamiltonian formulation, i.e., there's a one-to-one correspondence between conserved quantities and (one-parameter) Lie symmetries). The math, however is not too difficult:

The only idea you need from the above mentioned point of view of classical mechanics is that momentum is the generator of spatial translations, which is a symmetry of the Galilei-Newton spacetime model. This leads to the corresponding Poisson bracket and then to the above Heisenberg algebra for the position and momentum operators in QT. For simplicity I only consider motion in one dimension, the extension to 3D is straight forward.

The idea with the meaning of the momentum operator of translations leads to the idea to investigate the operator-valued function
$$\hat{X}(\xi) = \exp(\mathrm{i} \hat{p} \xi/\hbar) \hat{x} \exp(-\mathrm{i} \hat{p} \xi/\hbar).$$
You'll see, why this is a good idea, in a moment.

The key point is that you can explicitly evaluate this function. Just take the derivative wrt. ##\xi##:
$$\mathrm{d}_{\xi} \hat{X}(\xi) = \frac{1}{\hbar} \exp(\mathrm{i} \hat{p} \xi/\hbar) (\mathrm{i} \hat{p} \hat{x} - \mathrm{i} \hat{x} \hat{p} \exp(-\mathrm{i} \hat{p} \xi/\hbar) = \frac{1}{\mathrm{i} \hbar} \exp(\mathrm{i} \hat{p} \xi/\hbar) [\hat{x},\hat{p}] \exp(-\mathrm{i} \hat{p} \xi/\hbar) = \hat{1}.$$
Further you have ##\hat{X}(0)=\hat{x}## and thus by integration
$$\hat{X}(\xi)=\hat{x}+\xi \hat{1}.$$
Now consider the (generalized) position eigenstates ##|x \rangle##. Then define
$$|x,\xi \rangle=\exp(-\mathrm{i} \xi \hat{p}/\hbar) |x \rangle,$$
and we have
$$\hat{x} |x,\xi \rangle=\exp(-\mathrm{i} \xi \hat{p}/\hbar) \hat{X}(\xi) |x \rangle = (x+\xi) \exp(-\mathrm{i} \xi \hat{p}/\hbar) |x \rangle=(x+\xi) |x +\xi \rangle,$$
i.e., ##|x,\xi \rangle## is a position eigenstate with eigenvalue ##x,\xi##.

Thus we can define position eigenstates by
$$|x \rangle=\exp(-\mathrm{i} \hat{p} x/\hbar) |x=0 \rangle.$$
Now it's easy to evaluate the scalar product of the so defined position eigenstates with the momentum eigenstates ##|p \rangle##:
$$u_x(p)=\langle p|x \rangle = \langle p|\exp(-\mathrm{i} \hat{p} x/\hbar)|x=0 \rangle = \langle \exp(+\mathrm{i} \hat{p} x \hbar p|x=0 \rangle = \langle \exp(\mathrm{i} p x/\hbar) p|x=0 \rangle = \exp(-\mathrm{i} p x/\hbar) \langle p|x=0 \rangle=\exp(-\mathrm{i} p x) u_0(p).$$
To get ##u_0(p)## we note that we like to have
$$\langle p|p' \rangle=\delta(p-p'). \qquad (*)$$
Assuming the completeness of the position eigenstates constructed above then leads to
$$\langle p |p' \rangle =\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|p' \rangle = \int_{\mathbb{R}} \mathrm{d} x u_0(p) u_0^*(p') \exp[\mathrm{i} (p'-p) x/\hbar] = |u_0(p)|^2 2 \pi \hbar \delta(p-p').$$
With (*) it follows
$$u_0(p)=\frac{1}{\sqrt{2 \pi \hbar}}$$
modulo an irrelevant phase factor. This finally implies that
$$\langle p|x \rangle=\frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} x p/\hbar)$$
and thus
$$\langle x|p \rangle=u_p(x) =\langle p|x \rangle^* = \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} x p/\hbar).$$
This then implies the momentum operator in the position representation as follows:
$$\hat{p} \psi(x) := \langle x |\hat{p} \psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\hat{p} \psi \rangle = \int_{\mathbb{R}} \mathrm{d} p p u_p(x) \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p (-\mathrm{i} \hbar) \mathrm{d}_x u_p(x) \langle p|\psi \rangle = -\mathrm{i} \hbar \mathrm{d}_x \int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle = -\mathrm{i} \hbar \mathrm{d}_x \psi(x).$$
The momentum operator in position representation thus is
$$\hat{p} \psi(x)=-\mathrm{i} \hbar \psi(x).$$
In the same way you can derive the position operator in the momentum representation:
$$\hat{x} \tilde{\psi}(p) = +\mathrm{i} \hbar \mathrm{d}_p \tilde{\psi}(p)$$
and the transformation between position and momentum representation,
$$\tilde{\psi}(p) = \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} p x/\hbar) \psi(x)$$
and
$$\psi(x) = \langle x|\psi \rangle= \int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar) \tilde{\psi}(p).$$

I'm thorough this. Its so good to see from different perspectives :)

 

[Kashmir]

I've studied another way( From Townsends book) to get the required differential equation but it first assumes that we've a translation operator such that ##T(a)|x\rangle \equiv|x+a\rangle## and for infinitesimal translation we've ##\hat{T}(d x)=\mathbb{1}-\frac{i}{\hbar} \hat{p}_{x} d x##

whose action on a position ket is given by
##
\hat{T}(d x)|x\rangle=|x+d x\rangle
##

Can anyone help me understand how for infinitesimal translation we've ##\hat{T}(d x)=\mathbb{1}-\frac{i}{\hbar} \hat{p}_{x} d x##?

 

[vanhees71]

That's pretty much the same approach I've given above. It's maybe more didactical, because you don't need to know the trick with the unitary transformation. I think, it's indeed more elegant.

The idea again is that ##\hat{p}_x \equiv \hat{p}## is the generator of translations, i.e., you define ##\hat{p}## to exactly do an infinitesimal translation, i.e.,
$$\hat{T}(\mathrm{d} x) |x \rangle =(\mathbb{1} - \mathrm{i} \hat{p} /\hbar \mathrm{d} x)|x \rangle \stackrel{\text{def}}{=} |x + \mathrm{d} x \rangle.$$
From this you get
$$\frac{\mathrm{d}}{\mathrm{d} x} |x \rangle = \lim_{\mathrm{d} x \rightarrow 0} \frac{1}{\mathrm{d} x} (|x+\mathrm{d} x \rangle -|x \rangle) = -\mathrm{i} \hat{p}/\hbar |x \rangle.$$
By integration you then formally get
$$|x \rangle = \exp(-\mathrm{i} \hat{p} x/\hbar) |x=0 \rangle.$$

 

[Kashmir]

That's pretty much the same approach I've given above. It's maybe more didactical, because you don't need to know the trick with the unitary transformation. I think, it's indeed more elegant.

The idea again is that ##\hat{p}_x \equiv \hat{p}## is the generator of translations, i.e., you define ##\hat{p}## to exactly do an infinitesimal translation, i.e.,
$$\hat{T}(\mathrm{d} x) |x \rangle =(\mathbb{1} - \mathrm{i} \hat{p} /\hbar \mathrm{d} x)|x \rangle \stackrel{\text{def}}{=} |x + \mathrm{d} x \rangle.$$
From this you get
$$\frac{\mathrm{d}}{\mathrm{d} x} |x \rangle = \lim_{\mathrm{d} x \rightarrow 0} \frac{1}{\mathrm{d} x} (|x+\mathrm{d} x \rangle -|x \rangle) = -\mathrm{i} \hat{p}/\hbar |x \rangle.$$
By integration you then formally get
$$|x \rangle = \exp(-\mathrm{i} \hat{p} x/\hbar) |x=0 \rangle.$$

I've a few doubts here, I'd be grateful if you clarify.

What I understood is that we have a translation operator ##T(a)|x\rangle \equiv|x+a\rangle## and we've defined another operator ##p## such that ##1-\frac{i}{\hbar} \rho d x## is the infinitesimal translation operator. Is that correct?

Also I don't understand the rhs of ##|x \rangle = \exp(-\mathrm{i} \hat{p} x/\hbar) |x=0 \rangle.## What does## |x=0>## inside the ket mean

 

[vanhees71]

##|x=0 \rangle## is the (generalized) position eigenstate with eigenvalue ##0##.

 

[samalkhaiat]

you've forget to write ##p## in the exponent with ##x-y##

Yes, it looks like that I did, thanks, I will correct it. It is good to know that you did read my answer.

 

[samalkhaiat]

Can anyone help me understand how for infinitesimal translation we've ##\hat{T}(d x)=\mathbb{1}-\frac{i}{\hbar} \hat{p}_{x} d x##?

1) Probability conservation: If the state [itex]|\psi \rangle[/itex] is normalized to unity, the translated state [itex]|\psi^{\prime}\rangle = T(a)|\psi\rangle[/itex] must also be normalized to unity. So, [tex]\langle \psi |\psi \rangle = \langle \psi |T^{\dagger}(a)T(a)|\psi \rangle = 1 .[/tex] Thus, [itex]T(a)[/itex] is a unitary operator, [tex]T^{\dagger}(a) = T^{-1}(a) \ \ \ \ \ \ \ \ \ \ (1)[/tex]

2) The set [itex]\{ T(a): \ a \in \mathbb{R} \}[/itex] forms an infinite-dimensional unitary representation of the Abelian (Lie) group of translation. This means that the [itex]T(a)[/itex] satisfies the following group axioms: [tex]T(a)T(b) = T(a + b) , \ \ \ \ \ \ \ \ \ (2)[/tex][tex]T^{-1}(a) = T(-a) , \ \ \ \ \ \ \ \ \ \ \ (3)[/tex][tex]T(0) = 1 , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)[/tex]

3) Since [itex]T(0)|\psi \rangle = |\psi \rangle[/itex], then for an infinitesimal translation [itex]x \to x + \epsilon , \ \ |\epsilon | \ll 1[/itex] the operator [itex]T(\epsilon)[/itex] must (by continuity) equal to [itex]1[/itex] plus term linear in the infinitesimal parameter [itex]\epsilon[/itex]. So, we may write [tex]T(\epsilon) = 1 - i \epsilon \ G . \ \ \ \ \ \ \ (5)[/tex] Now it is a simple exercise to show that [itex]T(\epsilon)[/itex] will satisfies the equations (1)-(4) if we take the operator [itex]G[/itex] to be Hermitian: [itex]G^{\dagger} = G[/itex].

4) Notice that [tex]T(\epsilon)|x \rangle = |x + \epsilon \rangle ,[/tex] implies that [tex]\langle x |T(\epsilon) = \langle x - \epsilon|.[/tex] Therefore [tex]\psi^{\prime}(x) = \left(T_{\epsilon}\psi\right)(x) \equiv \langle x |T(\epsilon)|\psi \rangle = \psi (x - \epsilon) .[/tex] Now Taylor expansion of [itex]\psi (x - \epsilon)[/itex] gives us [tex]\langle x|T(\epsilon)|\psi \rangle = \psi (x) - i \epsilon \ (- i \partial \psi)(x) + \mathcal{O}(\epsilon^{2}).[/tex] Substituting Eq(5) in the LHS, we find (to first order in [itex]\epsilon[/itex]) [tex]\psi (x) - i \epsilon \ (G\psi)(x) = \psi (x) - i \epsilon \ (- i \partial \psi )(x) .[/tex]Thus, in the coordinate representation, the Hermitian operator [itex]G[/itex] is given by [tex]G \psi (x) = - i \partial_{x}\psi (x) .[/tex]

5) The relation between the generator [itex]G[/itex] and the momentum operator [itex]P[/itex]: We would like to obtain the commutation relation [itex]\big[Q , P \big] = i\hbar[/itex]. So, consider [itex]Q|x \rangle = x \ |x \rangle[/itex] and operate with [itex]T(\epsilon)[/itex] from the left: [tex]T(\epsilon)Q|x \rangle = x T(\epsilon)|x \rangle = x |x + \epsilon \rangle .[/tex] Next, consider [itex]T(\epsilon)|x \rangle = |x + \epsilon \rangle[/itex] and operate with position operator [itex]Q[/itex]: [tex]QT(\epsilon)|x \rangle = Q| x + \epsilon \rangle = (x + \epsilon) \ |x + \epsilon \rangle .[/tex] Subtracting the results, we get [tex]\big[ Q , T(\epsilon) \big] \ | x \rangle = \epsilon \ |x + \epsilon \rangle .[/tex] Then, from Eq(5), we obtain [tex]- i \epsilon \ \big[ Q , G \big] | x \rangle = \epsilon \ |x + \epsilon \rangle \approx \epsilon \ | x \rangle .[/tex] Now, [itex]| x \rangle[/itex] can be any position eigen-ket, and we know that the position eigen-kets form a complete set. Thus, we can write the operator identity [tex][Q , G] = i ,[/tex] and make the identification [tex]G = \frac{1}{\hbar} \ P .[/tex] So, the infinitesimal transformation operator becomes [tex]T(\epsilon) = 1 - \frac{i}{\hbar} \epsilon \ P .[/tex] From this infinitesimal form, a finite translation [itex]T(a)[/itex] can be obtained as [tex]T(a) = \mbox{lim}_{n \to \infty} \left( 1 - \frac{i}{\hbar}\left(\frac{a}{n}\right) P \right)^{n} = e^{- \frac{i}{\hbar}aP} .[/tex]

 

[Kashmir]

Yes, it looks like that I did, thanks, I will correct it. It is good to know that you did read my answer.

i read it and did Explicitly all the steps you'd written down. It was immensely helpful. :)