What Is the Position Operator if Momentum Is Given by a Specific Operator?

[kemiisto]

Homework Statement

Find the operator for position $$x$$ if the operator for momentum p is taken to be $$\left(\hbar/2m\right)^{1/2}\left(A + B\right)$$, with $$\left[A,B\right] = 1$$ and all other commutators zero.

Homework Equations

Canonical commutation relation
$$\left [ \hat{ x }, \hat{ p } \right ] = \hat{x} \hat{p} - \hat{p} \hat{x} = i \hbar$$

The Attempt at a Solution

Using $$c = \left(\hbar/2m\right)^{1/2}$$
$$\hat{x} \hat{p} f - \hat{p} \hat{x} f = i \hbar$$
$$\hat{x} c \left(\hat{A} + \hat{B}\right) f - c \left(\hat{A} + \hat{B}\right) \hat{x} f = i \hbar$$
$$c \hat{x} \left(\hat{A} + \hat{B}\right) f - c \left(\hat{A} + \hat{B}\right) \hat{x} f = i \hbar$$
$$\hat{x} \hat{A} f + \hat{x} \hat{B} f - \hat{A} \hat{x} f - \hat{B} \hat{x} f = i \hbar / c$$
$$\hat{x} \hat{A} f - c \hat{A} \hat{x} f + \hat{x} \hat{B} f - \hat{B} \hat{x} f = i \hbar / c$$
$$\left [ \hat{x}, \hat{A} \right ] + \left [ \hat{x}, \hat{B} \right ] = i \hbar / c$$
"all other commutators zero"
$$0 + 0 = i \hbar / c$$



Problem 1.2 from http://www.oup.com/uk/orc/bin/9780199274987/" .

 

[arkajad]

If [A,B]=1, then what will be the commutator of A with A+B? Can you go from there?

P.S. Your "f" on the left hand side is unnecessary.

 

[kemiisto]

If [A, B] = 1 then [A, A + B] = 1

[A, A + B] = A(A + B) - (A + B)A = AA + AB - AA - BA = AB - BA = [A, B] = 1

We have p = c(A + B) and [x, p] = iħ.

xp - px = iħ
cx(A + B) - c(A + B)x = iħ
x(A + B) - (A + B)x = iħ/c | multiply on A from the left
Ax(A + B) - A(A + B)x = iħ/c A | A(A + B) = 1 + (A + B)A
Ax(A + B) - (1 + (A + B)A)x = iħ/c A
Ax(A + B) - x - (A + B)Ax = iħ/c A
[Ax, A + B] - x = iħ/c A
x = -iħ/c A

Checking: [x, p] = xp - px = (-iħ/c A)(c(A + B)) - (c(A + B))(-iħ/c A) = -iħ A(A + B) + iħ (A + B) A = -ih

Ouch! Like always accurate within a sign...

Anyway, arkajad, many thanks for guiding me. I didn't have enough sleep last night. Maybe tomorrow I'll find the missing sign...

 

[arkajad]

Get a sleep and you will have it. But, perhaps, first prove for yourself a very useful identity: for any number a, and any operators A,B,C we have that

[A,B+C]=[A,B]+[A,C]
[A+B,C]=[A,C]+[B,C]
[aA,B]=a[A,B]
[A,aB]=a[A,B]

To save some work use [A,B]=-[B,A] as many times as useful. Once you are done - remember these rules, as they will be handy in the future.

 

[paranormal]

I would first clarify the notation used in the problem. The operator for position is commonly denoted as \hat{x}, while the operator for momentum is denoted as \hat{p}. It is also important to note that the commutator of two operators is defined as \left [ \hat{A}, \hat{B} \right ] = \hat{A}\hat{B} - \hat{B}\hat{A}.

Now, to find the operator for position, we can use the canonical commutation relation, \left [ \hat{ x }, \hat{ p } \right ] = i \hbar. Substituting the given operator for momentum, we have:

\left [ \hat{ x }, \left(\hbar/2m\right)^{1/2}\left(A + B\right) \right ] = i \hbar

Expanding the commutator, we get:

\hat{x} \left(\hbar/2m\right)^{1/2}\left(A + B\right) - \left(\hbar/2m\right)^{1/2}\left(A + B\right)\hat{x} = i \hbar

Simplifying, we get:

\hat{x}A + \hat{x}B - A\hat{x} - B\hat{x} = i \hbar

Using the given information that all other commutators are zero, we can simplify further to:

\hat{x}A - A\hat{x} = i \hbar

We can now solve for the operator for position, \hat{x}:

\hat{x} = \frac{i \hbar}{A - A}

However, this leads to an undefined result. This could be because the given operator for momentum, \left(\hbar/2m\right)^{1/2}\left(A + B\right), is not a valid operator for momentum. In quantum mechanics, the operators must be Hermitian and satisfy other important properties. Without more information about the operators A and B, it is not possible to find the operator for position x.