If ${ˆ T}_{L}$ $hatT_L$ is the translation operator ,which acts on a function like this ${ˆ T}_{L} f (x) = f (x - L)$ $hatT_Lf(x) = f(x-L)$ and $ˆ x$ $hatx$ which acts on $ˆ x f (x) = x f (x)$ $hatxf(x)=xf(x)$ then prove $[{ˆ T}_{L}, ˆ x] = - L T_{L}$ $[hatT_L ,hatx]= -LT_L$ ? Also how will translation operator act on the fourier transform of $f (x)$ $f(x)$

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If ${ˆ T}_{L}$ $hatT_L$ is the translation operator ,which acts on a function like this ${ˆ T}_{L} f (x) = f (x - L)$ $hatT_Lf(x) = f(x-L)$ and $ˆ x$ $hatx$ which acts on $ˆ x f (x) = x f (x)$ $hatxf(x)=xf(x)$ then prove $[{ˆ T}_{L}, ˆ x] = - L T_{L}$ $[hatT_L ,hatx]= -LT_L$ ? Also how will translation operator act on the fourier transform of $f (x)$ $f(x)$

Q2)IF $[A, B] = C$ $[A,B]=C$ and A,B are linear operators prove that C is a linear operator too
Suppose A and B share a common eigenfunction, $ϕ_{a b}$ $phi_(ab)$ ,show that
$C ϕ_{a b} = 0$ $Cphi_(ab)=0$
More info on ${ˆ T}_{L}$ $hatT_L$ https://en.wikipedia.org/wiki/Translation_operator_(quantum_mechanics)

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Answer 1 · 2018-02-24T04:45:04

I'll answer question 2 first, and question 1 later.

If the commutator of $ˆ A$ $hatA$ and $ˆ B$ $hatB$ , them being linear operators, is some operator $ˆ C$ $hatC$ , then we are to show that $ˆ C$ $hatC$ is linear.

Now, define $ϕ_{A B}$ $phi_(AB)$ as a linear combination of trial functions $ψ_{i}$ $psi_i$ and expansion coefficients $c_{i}$ $c_i$ :

$ϕ_{A B} = \sum i c_{i} ψ_{i}$ $phi_(AB) = sum_i c_i psi_i$

If $ˆ A$ $hatA$ and $ˆ B$ $hatB$ are linear, then the composite operators $ˆ A ˆ B$ $hatAhatB$ and $ˆ B ˆ A$ $hatBhatA$ are also linear (they might not be identical, but they would both be linear).

Apparently, this is "obvious", but I'll briefly prove it.

Without the necessity for commutation between $ˆ A$ $hatA$ and $ˆ B$ $hatB$ , since both operators act from the left (as they do normally), if $ˆ B$ $hatB$ and $ˆ A$ $hatA$ are both linear, then the first operator distributes, and then the second operator also distributes through addition. Thus, the composite operator (which also acts from the left) is linear.

Next, we note that the operation of $ˆ A$ $hatA$ or $ˆ B$ $hatB$ in each sequence gives:

$ˆ A ˆ B (\sum i c_{i} ψ_{i})$ $hatAhatB(sum_i c_i psi_i)$

$= \sum i c_{i} (ˆ A ˆ B ψ_{i})$ $= sum_i c_i (hatAhatBpsi_i)$

$ˆ B ˆ A (\sum i c_{i} ψ_{i})$ $hatBhatA(sum_i c_i psi_i)$

$= \sum i c_{i} (ˆ B ˆ A ψ_{i})$ $= sum_i c_i (hatBhatApsi_i)$

where it is not necessarily true that $ˆ A ˆ B (\sum i c_{i} ψ_{i}) = ˆ B ˆ A (\sum i c_{i} ψ_{i})$ $hatAhatB(sum_i c_i psi_i) = hatBhatA(sum_i c_i psi_i)$ .

If their commutator is $[ˆ A, ˆ B] = ˆ C$ $[hatA,hatB] = hatC$ , then

$ˆ A ˆ B - ˆ B ˆ A = ˆ C$ $hatAhatB - hatBhatA = hatC$

This implies that:

$ˆ C ϕ_{A B} = ˆ C (\sum i c_{i} ψ_{i})$ $color(blue)(hatCphi_(AB)) = hatC(sum_i c_i psi_i)$

$= (ˆ A ˆ B - ˆ B ˆ A) \sum i c_{i} ψ_{i}$ $= (hatAhatB - hatBhatA)sum_i c_i psi_i$

Here we use the linearity of the composite operators:

$= ˆ A ˆ B (\sum i c_{i} ψ_{i}) - ˆ B ˆ A (\sum i c_{i} ψ_{i})$ $= hatAhatB(sum_i c_i psi_i) - hatBhatA(sum_i c_i psi_i)$

$= \sum i c_{i} (ˆ A ˆ B ψ_{i}) - \sum i c_{i} (ˆ B ˆ A ψ_{i})$ $= sum_i c_i (hatAhatBpsi_i) - sum_i c_i (hatBhatApsi_i)$

Writing this out explicitly, it is then clear that:

$= (c_{1} ˆ A ˆ B ψ_{1} + c_{2} ˆ A ˆ B ψ_{2} + . . .) - (c_{1} ˆ B ˆ A ψ_{1} + c_{2} ˆ B ˆ A ψ_{2} + . . .)$ $= (c_1hatAhatBpsi_1 + c_2hatAhatBpsi_2 + . . . ) - (c_1hatBhatApsi_1 + c_2hatBhatApsi_2 + . . . )$

$= c_{1} ˆ A ˆ B ψ_{1} - c_{1} ˆ B ˆ A ψ_{1} + c_{2} ˆ A ˆ B ψ_{2} - c_{2} ˆ B ˆ A ψ_{2} + . . .$ $= c_1hatAhatBpsi_1 - c_1hatBhatApsi_1 + c_2hatAhatBpsi_2 - c_2hatBhatApsi_2 + . . .$

The operators act from the left, and constants always commute, so use the linearity of the composite operators to get:

$= c_{1} (ˆ A ˆ B - ˆ B ˆ A) ψ_{1} + c_{2} (ˆ A ˆ B - ˆ B ˆ A) ψ_{2} + . . .$ $= c_1(hatAhatB - hatBhatA)psi_1 + c_2(hatAhatB - hatBhatA)psi_2 + . . .$

$= \sum i c_{i} (ˆ A ˆ B - ˆ B ˆ A) ψ_{i}$ $= sum_i c_i (hatAhatB - hatBhatA)psi_i$

$= \sum i c_{i} ˆ C ψ_{i}$ $= color(blue)(sum_i c_i hatCpsi_i)$

Therefore, $ˆ C$ $hatC$ is a linear operator. Now, if $ˆ A$ $hatA$ and $ˆ B$ $hatB$ share a common eigenfunction, it follows that both their eigenvalues can be simultaneously observed, i.e. they do NOT obey the Heisenberg Uncertainty Principle.

In other words, if you operate on the system with $ˆ B$ $hatB$ , the state is not altered by the time you operate on the system with $ˆ A$ $hatA$ . If that is true, then it must also be true the other way around.

That can only be if their commutator is zero.

Thus:

$[ˆ A, ˆ B] = 0$ $[hatA, hatB] = 0$

and

$ˆ C ϕ_{A B} = 0$ $color(blue)(hatCphi_(AB) = 0)$ .

Answer 2 · 2018-02-24T07:01:34

$Please see proofs below.$ $"Please see proofs below."$

Explanation:

$"We are asked to show 4 things. With" \ \ \ hat{T_L}, \ hat{x} \ \ \ "as defined in"$
$"the problem:"$

$"1)" \quad [ hat{T_L}, hat{x} ] \ = - L hat{T_L} \quad.$

$"2) How will the translation operator act on the Fourier"$
$\qquad \ "Transform of" \ \ f(x) ?$

$"3) If" \ \ A ,B \ \ "are linear operators, and" \ \ C = [ A, B ], "then:"$

$\qquad "a) " C \ \ "is a linear operator."$

$\qquad "b) If" \ \ A ,B \ \ "share a common eigenfunction," \ \ \phi_{ab}, "then"$
$\qquad \qquad \ \ C \phi_{ab} = 0.$

$"Ok, let's begin:"$

$"1) Let" \ \ g \ \ "be a function. We compute the LHS of (1),"$
$\qquad \ "acting on" \ \ g:$

$\qquad \qquad [ hat{T_L}, \hat{x} ] g \ = \ ( hat{T_L} \hat{x} - \hat{x} \hat{T_L} ) g \ = \ hat{T_L} \hat{x} g - \hat{x} \hat{T_L} g$

$\qquad \qquad \qquad \qquad \qquad \qquad = \ hat{T_L} ( x g(x) ) - hat{x} ( g( x - L ))$

$\qquad \qquad \qquad \qquad \qquad \qquad = \ ( x - L ) g( x - L ) - x g( x - L )$

$\qquad \qquad \qquad \qquad \qquad \qquad = \ color{red}cancel{ x g( x - L ) } - L g( x - L ) -color{red}cancel{ x g( x - L ) }$

$\qquad \qquad \qquad \qquad \qquad \qquad = \ - L g( x - L )$

$\qquad \qquad \qquad \qquad \qquad \qquad = \ - L hat{T_L} g \quad.$

$\qquad \quad "So, we have:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad [ hat{T_L}, \hat{x} ] g \ = \ - L hat{T_L} g \quad.$

$\qquad \quad "Thus, as" \ \ g \ \ "was taken to be an arbitrary function, we have:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ [ hat{T_L}, \hat{x} ] \ = \ - L hat{T_L} \quad.$

$\qquad \quad "This proves (1)."$

$"2) Let's see how" \ \ hat{T_L} \ \ "acts on the Fourier Transform of" \ \ f(x).$
$\qquad \"Let" \ \ hat{F} \ \ "be the Fourier Transform operator. We compute"$
$\qquad hat{T_L} hat{F}, \ "acting on" \ f:$

$\qquad \qquad \qquad \qquad hat{T_L} hat{F} f \ = \ hat{T_L} 1/{ 2 pi } \int_{ x=- infty }^{ infty } \ e^{ - i \omega x } f(x) \ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ hat{T_L} 1/{ 2 pi } underbrace{\int_{ x=- infty }^{ infty } \ e^{ - i \omega x } f(x) \ dx }_{"function of" \ \omega}$

$\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 1/{ 2 pi } \int_{ x=- infty }^{ infty } \ e^{ - i ( \omega - L ) x } f(x) \ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 1/{ 2 pi } \int_{ x=- infty }^{ infty } \ e^{ - i \omega x }e^{ i L x } f(x) \ dx$

$\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ hat{F} ( e^{ i L x } f(x) )$

$\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ hat{F} hat{ e^{ i L x } } f(x); \qquad "where:" \quad hat{ e^{ i L x } } g(x) \ = \ e^{ i L x } g(x).$

$\qquad \quad "So, we have:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad hat{T_L} hat{F} f(x) \ = \ hat{F} hat{ e^{ i L x } } f(x).$

$\qquad \quad "Thus, as" \ \ f \ \ "was taken to be an arbitrary function, we have:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ hat{T_L} hat{F} \ = \ hat{F} hat{ e^{ i L x } }.$

$\qquad \quad "In words:"$

$\qquad \qquad \qquad "The Fourier Transform of a function, shifted right"$
$\qquad \qquad \qquad "by" \ \ L, "is the same as the Fourier Transform of the"$
$\qquad \qquad \qquad "product of" \ \ e^{ i L x } \ \ "and the function".$

$\qquad \quad "This answers (2)."$

$"3) Let" \ A ,B \ \ "be linear operators, and let" \ \ C = [ A, B ].$

$\qquad "a) We want to show" \ \ C \ \ "is a linear operator, too".$

$\qquad \qquad \ "We have:" \ \ C \ = \ A B - B A .$

$\qquad \qquad \ "The linearity of" \ \ C \ \ "follows immediately from the facts"$
$\qquad \qquad \ "that the product (composition) and sum of linear"$
$\qquad \qquad \ "operators is linear. However, let us show this directly."$

$\qquad \qquad \ "Let" \ \ f, g \ \ "be functions;" \ \ \alpha, \beta \ \ "be constants."$
$\qquad \qquad \ "To show the linearity of" \ \ C, "we compute:"$

$\qquad \qquad \qquad \qquad \qquad \qquad C( \alpha f + \beta g ) \ = \ ( A B - B A )( \alpha f + \beta g )$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ A B( \alpha f + \beta g ) - B A( \alpha f + \beta g )$

$\qquad \quad "continuing, using the linearity of" \ B, "then" \ A:$

$\qquad \qquad \qquad \qquad \qquad \qquad \ = \ A ( B ( \alpha f ) + B ( \beta g ) ) - B ( A ( \alpha f ) + A ( \beta g ) )$

$\qquad \qquad \qquad \qquad \qquad \qquad \ = \ A ( \alpha ( B f ) + \beta ( B g ) ) - B ( \alpha ( A f ) + \beta ( A g ) )$

$\qquad \quad "continuing, now using the linearity of" \ A, "then" \ B:$

$\qquad \qquad \qquad \qquad \ \ = \ A ( \alpha ( B f ) ) + A ( \beta ( B g ) ) - B ( \alpha ( A f ) ) - B ( \beta ( A g ) )$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ \alpha ( A B f ) + \beta ( A B g ) - \alpha ( B A f ) - \beta ( B A g )$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ \alpha ( A B f ) - \alpha ( B A f ) + \beta ( A B g ) - \beta ( B A g )$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ \alpha ( A B f - B A f ) + \beta ( A B g - B A g )$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ \alpha ( A B - B A ) f + \beta ( A B - B A ) g$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ \alpha C f + \beta C g.$

$\qquad \quad "So, we have:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad C( \alpha f + \beta g ) \ = \ \alpha C f + \beta C g.$

$\qquad \quad "Thus, as" \ \ f, g \ \ "were taken to be arbitrary functions, and"$
$\qquad \quad \alpha, \beta \ \ "were taken to be arbitrary constants, we have:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad C \ \ "is a linear operator".$

$\qquad \quad "This proves (3a)."$

$\qquad "b) Let" \ \ \phi_{ab} \ \ "be a common eigenfunction of" \ A, B.$

$\qquad \qquad \quad "Thus:" \ \ A \phi_{ab} \ = \ \lambda_A \phi_{ab}, \qquad "for some constant" \ \ \lambda_A.$

$\qquad \qquad \quad "Thus:" \ \ B \phi_{ab} \ = \ \lambda_B \phi_{ab}, \qquad "for some constant" \ \ \lambda_B.$

$\qquad \qquad \quad "We compute" \ \ C \phi_{ab}:$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad C \phi_{ab} \ = \ ( A B - B A ) \phi_{ab}$

$\qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ A B \phi_{ab} - B A \phi_{ab}$

$\qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ A ( \lambda_B \phi_{ab} ) - B ( \lambda_A \phi_{ab} )$

$\qquad \qquad "continuing, using the linearity of" \ A, "then" \ B:$

$\qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \quad \ = \ \lambda_B ( A \phi_{ab} ) - \lambda_A ( B \phi_{ab} )$

$\qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ \lambda_B \lambda_A \phi_{ab} - \lambda_A \lambda_B \phi_{ab}$

$\qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ ( \lambda_B \lambda_A - \lambda_A \lambda_B ) \phi_{ab}$

$\qquad \qquad "continuing, as" \ \ \lambda_A, \lambda_B \ \ "are constants, and so commute:"$

$\qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ ( 0 ) \phi_{ab}$

$\qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ 0.$

$\qquad \quad "Thus, we have:"$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad C \phi_{ab} \ = \ 0.$

$\qquad \quad "This proves (3b)."$

$"Thus, we have shown (1), (2), (3a), (3b)." \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ square "$

Answer 3 · 2018-02-24T07:13:01

Now I'll answer question 1... This is also shown here in question 1b and 1c, via a different approach.

With foreknowledge of quantum step potentials and time-dependent wave functions in general, we already know that there exists a phase shift operator

$hatT_L = e^(ihatp_xL//ℏ)$

such that $hatT_Lf(x) = e^(ihatp_xL//ℏ)f(x) = f(x-L)$ . Again, see my reference, which merely has $L = a$ as the only difference.

Next, we will have to correct your commutator, which is backwards... it should be:

$[hatx, hatT_L] = -LhatT_L$

In the notation of the reference to make it completely obvious,

$[hatx, e^(ihatp_xa//ℏ)] = -ae^(ihatp_xa//ℏ)$

which the reference has described as an exponential operator which has "performed a spatial translation of the position eigenstate."

Explicitly (in our notation), we are proving:

$[hatx, hatT_L] stackrel(?" ")(=) -LhatT_L$

By definition, the operators act on the eigenfunction $f(x_0)$ in some position $x_0$ to give:

$hatx f(x_0)= x_0 f(x_0)$

$hatx (hatT_L f(x_0)) = hatxhatT_L f(x_0) ne hatT_L(hatxf(x_0)$

The coordinate operator $hatx$ returns the eigenvalue observed after applying the shift operator in the manner shown as $hatT_L f(x_0)$ .

Therefore, we are supposed to already know that $hatT_L f(x_0) -= f(x_0 - L)$ is an eigenfunction of $hatx$ , i.e. that

$hatxf(x_0 - L)$

$= hatx(hatT_L f(x_0))$

$-= (x_0 - L)f(x_0 - L)$

$= (x_0 - L) overbrace(hatT_L f(x_0))^(f(x_0 - L)$

since the eigenvalue of this state in position $x_0 - L$ returned by $hatx$ is $x = x_0 - L$ .

If we start from there, we can work back to where the commutator is necessary...

$(x_0 - L)hatT_Lf(x_0)$

$= -LhatT_Lf(x_0) + hatT_Lx_0f(x_0)$

$= -LhatT_Lf(x_0) + hatT_Lhatxf(x_0)$

$= [ . . . ]$

$= hatx(hatT_L f(x_0))$

At this point, we are supposed to connect what happened in between (we cannot simply invoke the commutator before we prove it).

So we work from our known definition backwards to get:

$hatx(hatT_L f(x_0)) = hatxhatT_Lf(x_0)$

$= (hatxhatT_L - hatT_Lhatx + hatT_Lhatx)f(x_0)$

$= ([hatx,hatT_L] + hatT_Lhatx)f(x_0)$

$= [hatx,hatT_L]f(x_0) + hatT_Lhatxf(x_0)$

By inspection, we find that from this definition of the shift operator,

$color(green)(-LhatT_Lf(x_0)) + hatT_Lhatxf(x_0) = color(green)([hatx,hatT_L]f(x_0)) + hatT_Lhatxf(x_0)$

Therefore:

$color(blue)([hatx,e^(ihatp_xL//ℏ)] -= [hatx, hatT_L] = -Le^(ihatp_xL//ℏ) = -LhatT_L)$

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