What are energy values and how do they relate to eigenvectors in quantum states?

[Gbox]

Summary: Finding state at t=0, energy values and more

So this is my first question in quantum mechanics (please understand).

1. So we have a system, and to describe the state of the system we have to measure, A is an hermitian matrix, that each physical measurable quantity has.
To find the state after the measurement we first need to find the eigenvectors of A which are for λ1 = 0.5 v1=1/sqrt(2)(1 , 1) and for λ2 = 1.5 v1=1/sqrt(2)(-1 , 1)

Now we have to normalize the vector, but because the eigenvalues are non degenerate we know that

So

As it the superposition of the two states.

2. what are energy values? are they the eigenvalues?

 

[eys_physics]

Summary: Finding state at t=0, energy values and more

Now we have to normalize the vector, but because the eigenvalues are non degenerate we know that

This is not correct. You have that $$|v_1> = \frac{1}{\sqrt{2}} |x> + \frac{1}{\sqrt{2}}|y>$$
and
$$|v_2> = -\frac{1}{\sqrt{2}} |x> + \frac{1}{\sqrt{2}}|y>$$ . What is then $|x>$ and #|y># in terms of $|v_1>$ and $|v_2>$?

Summary: Finding state at t=0, energy values and more

what are energy values? are they the eigenvalues

Yes, if $A$ is the Hamiltonian? But, it is not obvious from the information you have given?

 

[Gbox]

Why

$\left| v_1 \right>= \frac{1}{\sqrt{2}} \left| x \right>+\frac{1}{\sqrt{2}}\left| y \right>$

And

$\left| v_2 \right>= -\frac{1}{\sqrt{2}}\left| x \right>+\frac{1}{\sqrt{2}} \left| y \right>$
?

We took the dot product of the eigenvalues of $A$? Did we use the fact that $\left| v_1 \right>$ and $\left| v_2 \right>$ are orthonormal?

 

[eys_physics]

You have two states $|x>$ and $|y>$, which forms an orthonormal basis. Your matrix $A$ is written in this basis. You are solving for the eigenvectors, $|v_1>$ and $|v_2>$, in this basis.
That is,
$$|v_1> = a |x> + b|y>$$
$$|v_2> = -a |x> + b|y>,$$
with $a^2+b^2=1$.
The notation $|v_1> = \frac{1}{\sqrt{2}}(1,1)$, thus means $a=b=\frac{1}{\sqrt{2}}$

 

[Gbox]

Thanks,
I do not understand the physics the led to the two equations.
maybe I will go a step back $\left| x \right>$ and $\left| y \right>$ are the state of the system after a measure is done?
The eigenvectors of $A$ are the state of the system before the mesure?
So now we are measuring, knowing that there are 2 possible states (Which are orthonormal basis) $\left| x \right>$ and $\left| y \right>$ those are the states we will get after a measure.
So we are asked to find the state of the system before the measurement using the matrix $A$?

We know that the for $\lambda_1=0.5$ we have $\frac{1}{\sqrt{2}}(1, 1)$ and for $\lambda_2=1.5$ we have $\frac{1}{\sqrt{2}}(-1 ,1)$

So what we have basically done next is to say that the state in $t=0$ (which are the eigenvectors of $A$) are a linear combination of $\left| x \right>$ and $\left| y \right>$ or:

$\left| v_1 \right> =a \left| x \right>+b \left| y \right>$

$\left| v_2 \right> =b \left| x \right>+c \left| y \right>$

which is:

$\frac{1}{\sqrt{2}}(1, 1)=a \left| x \right>+b \left| y \right>$

$\frac{1}{\sqrt{2}}(-1, 1) =c \left| x \right>+d \left| y \right>$

Now due to $\left| x \right>$ and $\left| y \right>$ being orthonormal we can conclude that:

$a=-c=\frac{1}{\sqrt{2}}$ and $b=d=\frac{1}{\sqrt{2}}$

Thanks for your patience I am new to those problems

 

[PeroK]

Thanks,
I do not understand the physics the led to the two equations.
maybe I will go a step back $\left| x \right>$ and $\left| y \right>$ are the state of the system after a measure is done?
The eigenvectors of $A$ are the state of the system before the mesure?
So now we are measuring, knowing that there are 2 possible states (Which are orthonormal basis) $\left| x \right>$ and $\left| y \right>$ those are the states we will get after a measure.
So we are asked to find the state of the system before the measurement using the matrix $A$?

Not quite. You are given first some general information about the system: that it is a system that can be described as a linear combination of only two states. What that means in terms of linear algebra is that it is two-dimensional. What it means in terms of the physics ... let's wait and see.

In $|x \rangle, |y \rangle>$ you are given an orthonormal basis for the system. That is all. You are not told what $|x \rangle, |y \rangle$ are physically.

Next, you are given a Hermitian operator $A$, which presumably represents some important physical observable of your system. But, you are not told which.

As $A$ is Hermitian, its eigenvectors form an orthonormal basis. Part 1) asks you to find these and their relationship to $|x \rangle, |y \rangle$.

That's all general stuff that is more linear algebra that physics.

Now, you are also told that the system at time $t=0$ is at state $|x \rangle$. You're not told how we know this. It's possible that we know this for one of two reasons:

a) We have just measured the system using an observable for which $|x \rangle$ is an eigenstate (and got the approriate eigenvalue). That's one way to know the state of a system.

b) Perhaps we have prepared a large number of systems in the same (identical) way and measured them and they are always found in state $|x \rangle$. We concluded, therefore, that the preparation process always puts the system in this state.

Either way, as far as the problem is concerned, we simply know the state of the system at time $t=0$.

That's the background to the problem. I hope that helps.

To the question:

We can see from parts 2-4 that $A$ is indeed the Hamiltonian. I'm not sure why they didn't say this explicitly. In any case, you are nearly there for part 1). Can you finish that first?

Start from here:

$\left| v_1 \right>= \frac{1}{\sqrt{2}} \left| x \right>+\frac{1}{\sqrt{2}}\left| y \right>$

And

$\left| v_2 \right>= -\frac{1}{\sqrt{2}}\left| x \right>+\frac{1}{\sqrt{2}} \left| y \right>$

Hint: this is now basic linear algebra to solve for $|x \rangle$.

 

[PeroK]

Summary: Finding state at t=0, energy values and more

View attachment 245007

PS

This problem is quite loosely worded. What it should say is that the operator $A$ is given by that matrix in the $|x \rangle, |y \rangle$ basis.

 

[Gbox]

Thanks!

So mathematically, are we doing a basis change? as how else can we arrive to

$\left| v_1 \right>= \frac{1}{\sqrt{2}} \left| x \right>+\frac{1}{\sqrt{2}}\left| y \right>$

And

$\left| v_2 \right>= -\frac{1}{\sqrt{2}}\left| x \right>+\frac{1}{\sqrt{2}} \left| y \right>$

Using the hint I have arrive to:

$\left| x \right>= \frac{1}{\sqrt{2}}(\left| v_1 \right>+\left| v_2 \right>)$

and

$\left| y \right>= \frac{1}{\sqrt{2}}(\left| v_1 \right>-\left| v_2 \right>)$

 

[eys_physics]

Yes, you either write $|x>$ and $|y>$ in terms of $|v_1>$ and $|v_2|$, or the other way around. They are connected by a linear transformation (mapping). Similarly, the matrix $A$ can be expressed in the $\{v_1, v_2\}$ basis. In that basis $A$ is diagonal.

 

[PeroK]

Thanks!

So mathematically, are we doing a basis change? as how else can we arrive to

$\left| v_1 \right>= \frac{1}{\sqrt{2}} \left| x \right>+\frac{1}{\sqrt{2}}\left| y \right>$

And

$\left| v_2 \right>= -\frac{1}{\sqrt{2}}\left| x \right>+\frac{1}{\sqrt{2}} \left| y \right>$

Using the hint I have arrive to:

$\left| x \right>= \frac{1}{\sqrt{2}}(\left| v_1 \right>+\left| v_2 \right>)$

That's odd. When I add $v_1 + v_2$ I get $\sqrt{2} y$.

 

[eys_physics]

Yes, the formula for $|x>$ is wrong. The plus sign should be a minus sign. That is, $|x> = \frac{1}{\sqrt{2}}(|v_1> - |v_2>)$

 

[Gbox]

Yes, you either write $|x>$ and $|y>$ in terms of $|v_1>$ and $|v_2|$, or the other way around. They are connected by a linear transformation (mapping). Similarly, the matrix $A$ can be expressed in the $\{v_1, v_2\}$ basis. In that basis $A$ is diagonal.

I am maybe confused with the dirac notations, but how did you get to this expression?

 

[eys_physics]

I am maybe confused with the dirac notations, but how did you get to this expression?

Sorry, it was a mistake in the first sentence. It should read "|v_1>" and "|v_2>".

 

[Gbox]

I mean here:

You have two states $|x>$ and $|y>$, which forms an orthonormal basis. Your matrix $A$ is written in this basis. You are solving for the eigenvectors, $|v_1>$ and $|v_2>$, in this basis.
That is,
$$|v_1> = a |x> + b|y>$$
$$|v_2> = -a |x> + b|y>,$$
with $a^2+b^2=1$.
The notation $|v_1> = \frac{1}{\sqrt{2}}(1,1)$, thus means $a=b=\frac{1}{\sqrt{2}}$

Why you have $a, b, -a, b$ and not $a,b,c,d$?

 

[eys_physics]

Because, for the eigenvectors are (or can be constructed) orthonormal , i.e.
$$<v_1|v_2>=<v_2|v_1> =0, $$
and
$$<v_1|v_1>=<v_2|v_2> =1 $$.

 

[PeroK]

I am maybe confused with the dirac notations, but how did you get to this expression?

Let's take a step back. All we really wanted were the eigenvalues and eigenvectors of a 2x2 matrix. Dirac notation isn't relevant.

You've got those now. As an aside, you chose $v_2 = (-1/\sqrt{2}, 1/\sqrt{2})$. Let's stick with this.

However, note that normalisation is up to an arbitrary, complex constant of unit modulus. In other words, your normalised eigenvectors are not totally unique. In particular, you could equally well have chosen $v_2 = (1/\sqrt{2}, -1/\sqrt{2})$.

I only mention this because it's possible that if you see a solution to this problem it may well have done things this way round. I would tend to make the x-coordinate positive for no good reason other than it comes first.

I suggest you stick with what you have and move on!

 

[PeroK]

Sorry, I can not see why, using the orthonormal property I got to

$$a^2 |x>^2+b^2 |y>^2=1\Rightarrow a^2 +b^2 =1$$

$$c^2 |x>^2+d^2 |y>^2=1\Rightarrow c^2 +b^2 =1$$

$$ac |x>^2+bd |y>^2=0\Rightarrow ac+bd=0$$

You appear to be looking for any orthonormal pair of vectors here. $a, b, c, d$ are determined (up to a common factor) by the eigenvector property.

To be honest, in the case where $a=b$, you should be able to write down $a = b = \frac{1}{\sqrt{2}}$. Unless QM is the first time you've encountered unit vectors?

 

[Gbox]

You appear to be looking for any orthonormal pair of vectors here. $a, b, c, d$ are determined (up to a common factor) by the eigenvector property.

To be honest, in the case where $a=b$, you should be able to write down $a = b = \frac{1}{\sqrt{2}}$. Unless QM is the first time you've encountered unit vectors?

Not it is just that my background is in math (done linear algebra 1+2) so I am checking each step form math (and now physics) point of view