# I Some simple questions about <Ψ|P|Ψ>

1. Mar 4, 2017

### mike1000

I would like to exploit his formula to ask some very basic questions about the meaning of this formula.

Here are my questions....

1. Is | Ψg> the wave function for the ground state of the system?

1a. Looked at in terms of simple linear algebra, is it just a column vector?
1b. What does each element in the column vector represent?

2. When you operate on the ground state with the momentum operator, as in P|Ψg>, is the result another column vector?
2a. What do the elements of that column vector represent?

3. In terms of linear algebra, what does the following Dirac notation mean.... |A,B>?

I may have more questions...

2. Mar 4, 2017

### blue_leaf77

1. It depends on how the notation $|\psi_g\rangle$ was defined. It doesn't have to be the ground state, it can be any state.
1a. If you want to work in terms of matrix, a ket becomes column vector and a bra becomes a row vector.
1b. Each element in a column vector representing a ket is the weighting factor of various basis state used to represent that particular ket. Use different set of basis states, you will get an entirely different elements in the column matrix.

2. Yes. By the definition of an operator in linear algebra, it's a function that linearly maps a vector into another vector in the same space.
2a. Same as 1b.

3. Usually that means that that particular state is specified by two quantum numbers A and B.

3. Mar 4, 2017

### Staff: Mentor

1) It depends on the context. Somewhere near where that ket is first introduced you should find something along the lines of "... where | Ψg> is ...." or "Let | Ψg> be ....."
1a) A ket is a vector in a vector space, and any vector can be represented as 1xN matrix (this is why we call these matrices "column vectors") after you've chosen a basis. So it's not quite right to say that it "is" a column vector, but it can be conveniently represented as one. In many situations N will be infinite (we have an infinite-dimensional Hilbert space).
1b) Each element of the column vector is just the coefficient of the corresponding base vector in whatever basis you've chosen.
2) Yes, with the same caution about the matrix not quite exactly being the vector, but rather a convenient way of representing it in a given basis.
2a) In this case the basis will be the momentum eigenstates. If the system is such that the momentum operator has a continuous spectrum, there will be some additional mathematical subtleties because these "eigenstates" are not properly square-integrable. You can often ignore these and still get out alive.
3) Unless the context tells you otherwise, that is a vector that is an eigenstate of two different (but generally commuting) operators, with eigenvalues A for one operator and B for the other: $\hat{A}|A,B\rangle=A|A,B\rangle$ and $\hat{B}|A,B\rangle=B|A,B\rangle$.

Last edited: Mar 5, 2017
4. Mar 5, 2017

### Staff: Mentor

Which thread? It might provide some of the context that the previous posters mention.

5. Mar 5, 2017

6. Mar 6, 2017

### mike1000

Thank you for the replies.

I understand that Ψ is a vector in vector space and can be represented as a column vector. Is it correct to say that the elements of Ψ have units of probability, or, equivalently, it is dimensionless?

I also understand that P|Ψ> has units of momentum.

Does that imply that <Ψ|P|Ψ> has a single value result ( from the dot product) and also has units of momentum? And what does it represent physically?

7. Mar 6, 2017

### Staff: Mentor

Yes, yes, and it is the expectation value of the momentum of a quantum system that has been prepared in the state $|\psi\rangle$. If I prepare many systems in this state and measure the momentum of each one, the average of the results will be the expectation value.

8. Mar 6, 2017

### blue_leaf77

P|Ψ> is a vector, not a physical quantity. Hence it's not quite right to associate a unit with it since it's just a mathematical object.