Quantum Computation Probabilities

Homework Statement

Let a one-qubit system be in the state $\left|ψ\right\rangle$ = $\frac{\sqrt{15}\left|0\right\rangle + i\left|1\right\rangle}{4}$. If we perform a measurement to see whether the qubit is in the state $\left|x_{+}\right\rangle$ = $\frac{\left|0\right\rangle + \left|1\right\rangle}{\sqrt{2}}$ or in the orthogonal state $\left|x_{-}\right\rangle$ = $\frac{\left|0\right\rangle - \left|1\right\rangle}{\sqrt{2}}$, what is the probability of each of these two outcomes?

The Attempt at a Solution

I know the given state can be written as $\left|ψ\right\rangle$ = $\frac{\sqrt{15}}{4}$$\left|0\right\rangle$ + $\frac{i}{4}$$\left|1\right\rangle$

So therefore α = $\sqrt{15}$/4, and β = i/4. And $\left|α^{2}\right|$ + $\left|β^{2}\right|$ = 1 (right?). But those are only for states |0> or |1>, right?

So I basically have no idea how to do this.

Can anybody help me or put me in the right direction?

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Hi jumi,

To determine the probability of finding a system described by $| \psi \rangle$ in the state $|x_{\alpha} \rangle$, we use $P(\alpha) = | \langle x_{\alpha}| \psi \rangle |^2$. I hope that helps!

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Hi jumi,

To determine the probability of finding a system described by $| \psi \rangle$ in the state $|x_{\alpha} \rangle$, we use $P(\alpha) = | \langle x_{\alpha}| \psi \rangle |^2$. I hope that helps!
Thanks for the reply. So in this instance, I would take the dot product of the two states and square the absolute value?

Thanks for the reply. So in this instance, I would take the dot product of the two states and square the absolute value?
Right. But just to make sure we're on the same page: you're taking the inner product of the two states with $\psi$ - not with each other. So you're making two calculations.

Right, $\left|x_{+}\right\rangle$$\cdot$$\left|ψ\right\rangle$ and $\left|x_{-}\right\rangle$$\cdot$$\left|ψ\right\rangle$, correct?

Yep. You've got it.

gabbagabbahey
Homework Helper
Gold Member
Right, $\left|x_{+}\right\rangle$$\cdot$$\left|ψ\right\rangle$ and $\left|x_{-}\right\rangle$$\cdot$$\left|ψ\right\rangle$, correct?
In bra-ket notation, what you've written doesn't make sense (isn't defined). You want to compute the inner products between the kets $\langle x_{\pm} |$ and the bra $|\psi \rangle$; $\langle x_{\pm} |\psi \rangle$ and their adjoints; (complex conjugates) $\langle \psi | x_{\pm} \rangle = \langle x_{\pm} |\psi \rangle^{ \dagger }$ in order to find the probabilities

$$P_{\pm} =\langle x_{\pm} |\psi \rangle \langle \psi | x_{\pm} \rangle = ||\langle x_{\pm} |\psi \rangle||^2$$

Good catch gabbagabbahey. For some reason, I entirely glossed over that.

In bra-ket notation, what you've written doesn't make sense (isn't defined). You want to compute the inner products between the kets $\langle x_{\pm} |$ and the bra $|\psi \rangle$; $\langle x_{\pm} |\psi \rangle$ and their adjoints; (complex conjugates) $\langle \psi | x_{\pm} \rangle = \langle x_{\pm} |\psi \rangle^{ \dagger }$ in order to find the probabilities

$$P_{\pm} =\langle x_{\pm} |\psi \rangle \langle \psi | x_{\pm} \rangle = ||\langle x_{\pm} |\psi \rangle||^2$$
Ok, you've lost me now...

I know the 'braket' notation thing was wrong; I just wrote it that way for my ease of understanding. However, other than "complex conjugates", I don't really understand your reply, sorry...

Which part confuses you in particular?

Remember that $P_{\alpha} = | \langle x_{\alpha} | \psi \rangle |^2 = \langle x_{\alpha} | \psi \rangle^* \langle x_{\alpha} | \psi \rangle = \langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle$. Does that help?

Which part confuses you in particular?

Remember that $P_{\alpha} = | \langle x_{\alpha} | \psi \rangle |^2 = \langle x_{\alpha} | \psi \rangle^* \langle x_{\alpha} | \psi \rangle = \langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle$. Does that help?
Sort of. How exactly does the operation $\langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle$ work?

(P.S. Maybe I should have mentioned in the beginning (and I'm not sure if that's relevant here), but I've never had a full Linear Algebra course.)

$\langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle$ says mathematically: You're going to multiply two scalars (obtained from inner products) in the following way. To obtain the scalar $\langle \psi | x_{\alpha} \rangle$, you're going to take the bra $\langle \psi |$ and act it on the ket $| x_{\alpha} \rangle$. Remember that bras are the adjoints of their respective kets -- that is, we take the complex conjugate of the ket $| \psi \rangle$ (a column vector) and then transpose it so that it becomes the bra $\langle \psi |$ (a row vector). We're also going to do the same operation for $\langle x_{\alpha} | \psi \rangle$, except that $\langle x_{\alpha} |$ is now the bra and $| \psi \rangle$ is now the ket for this inner product.

Once we obtain these two inner products, we multiply them together.

However, we can dramatically simplify this process. We notice that $\langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle = \langle x_{\alpha} | \psi \rangle^* \langle x_{\alpha} | \psi \rangle$ (in words: the product of the inner products is actually the product of one of the inner products and its complex conjugate). So we really only need to calculate one inner product -- $\langle x_{\alpha} | \psi \rangle$ or $\langle \psi | x_{\alpha} \rangle$ (take your pick) -- and then multiply that by its complex conjugate (which is actually the same process I described in length above, but requires much less thought), thereby obtaining $| \langle x_{\alpha} | \psi \rangle |^2 = P(\alpha)$.

I suspect that, while you may not be familiar with bra-ket notation, your natural inclinations for calculating the inner product were essentially correct. You just need to get the notation straight, since it's important. Bra-ket notation hides a lot of deeper mathematics that you'll eventually have to learn (and that I still don't fully understand either!).

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$\langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle$ says mathematically: You're going to multiply two scalars (obtained from inner products) in the following way. To obtain the scalar $\langle \psi | x_{\alpha} \rangle$, you're going to take the bra $\langle \psi |$ and act it on the ket $| x_{\alpha} \rangle$. Remember that bras are the adjoints of their respective kets -- that is, we take the complex conjugate of the ket $| \psi \rangle$ (a column vector) and then transpose it so that it becomes the bra $\langle \psi |$ (a row vector). We're also going to do the same operation for $\langle x_{\alpha} | \psi \rangle$, except that $\langle x_{\alpha} |$ is now the bra and $| \psi \rangle$ is now the ket for this inner product.

Once we obtain these two inner products, we multiply them together.

However, we can dramatically simplify this process. We notice that $\langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle = \langle x_{\alpha} | \psi \rangle^* \langle x_{\alpha} | \psi \rangle$ (in words: the product of the inner products is actually the product of one of the inner products and its complex conjugate). So we really only need to calculate one inner product -- $\langle x_{\alpha} | \psi \rangle$ or $\langle \psi | x_{\alpha} \rangle$ (take your pick) -- and then multiply that by its complex conjugate (which is actually the same process I described in length above, but requires much less thought), thereby obtaining $| \langle x_{\alpha} | \psi \rangle |^2 = P(\alpha)$.

I suspect that, while you may not be familiar with bra-ket notation, your natural inclinations for calculating the inner product were essentially correct. You just need to get the notation straight, since it's important. Bra-ket notation hides a lot of deeper mathematics that you'll eventually have to learn (and that I still don't fully understand either!).
Awesome reply, thanks. I don't have time to really sit down and mess around with the math tonight, but I'll take a look at it tomorrow and see what I can come up with. I'll make another reply if I'm still having trouble.

Otherwise, I really appreciate the help. Thanks.

$\langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle$ says mathematically: You're going to multiply two scalars (obtained from inner products) in the following way. To obtain the scalar $\langle \psi | x_{\alpha} \rangle$, you're going to take the bra $\langle \psi |$ and act it on the ket $| x_{\alpha} \rangle$. Remember that bras are the adjoints of their respective kets -- that is, we take the complex conjugate of the ket $| \psi \rangle$ (a column vector) and then transpose it so that it becomes the bra $\langle \psi |$ (a row vector). We're also going to do the same operation for $\langle x_{\alpha} | \psi \rangle$, except that $\langle x_{\alpha} |$ is now the bra and $| \psi \rangle$ is now the ket for this inner product.

Once we obtain these two inner products, we multiply them together.

However, we can dramatically simplify this process. We notice that $\langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle = \langle x_{\alpha} | \psi \rangle^* \langle x_{\alpha} | \psi \rangle$ (in words: the product of the inner products is actually the product of one of the inner products and its complex conjugate). So we really only need to calculate one inner product -- $\langle x_{\alpha} | \psi \rangle$ or $\langle \psi | x_{\alpha} \rangle$ (take your pick) -- and then multiply that by its complex conjugate (which is actually the same process I described in length above, but requires much less thought), thereby obtaining $| \langle x_{\alpha} | \psi \rangle |^2 = P(\alpha)$.

I suspect that, while you may not be familiar with bra-ket notation, your natural inclinations for calculating the inner product were essentially correct. You just need to get the notation straight, since it's important. Bra-ket notation hides a lot of deeper mathematics that you'll eventually have to learn (and that I still don't fully understand either!).
Ok, I had a chance to review the math a bit.

So knowing, $P(\alpha) = | \langle x_{\alpha} | \psi \rangle |^2 = \langle x_{\alpha} | \psi \rangle ^* \langle x_{\alpha} | \psi \rangle = \langle \psi | x_{\alpha} \rangle \langle x_{\alpha} | \psi \rangle$, I can simply multiply the row vector given by $\left\langle x_{\alpha}\right|$ times the column vector given by $\left| \psi \right\rangle$ and square the absolute value? Or do I need to take the conjugate transpose of both since I'm doing an inner product?

I can simply multiply the row vector given by $\left\langle x_{\alpha}\right|$ times the column vector given by $\left| \psi \right\rangle$ and square the absolute value? Or do I need to take the conjugate transpose of both since I'm doing an inner product?
For the underlined portion: This is correct. Remember to take the complex conjugate of the ket once you transpose it into the bra.

For your second statement: Remember, you're effectively multiplying two inner products together when you're calculating the probability, but this is the same process as what you described in your first statement. But when you say "take the conjugate transpose of both since I'm doing an inner product," what do you mean by "take the conjugate transpose of both"? Both of what?

For the underlined portion: This is correct. Remember to take the complex conjugate of the ket once you transpose it into the bra.

For your second statement: Remember, you're effectively multiplying two inner products together when you're calculating the probability, but this is the same process as what you described in your first statement. But when you say "take the conjugate transpose of both since I'm doing an inner product," what do you mean by "take the conjugate transpose of both"? Both of what?
Oh, alright, I get it now. Disregard that second statement. I knew the complex conjugate showed up in some capacity, but you answered it in your first sentence.

So all in all (sorry for continually pestering you...), I would take $\left|x_{+}\right\rangle = \frac{\left|0\right\rangle + \left|1\right\rangle}{\sqrt{2}}$, take it's adjoint so it's a row vector, and multiply that by the column vector given by $\left|ψ\right\rangle = \frac{\sqrt{15}\left|0\right\rangle + i\left|1\right\rangle}{4}$?

So I would have:
$\left\langle x_{+} \right| = ( \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}})$ and $\left| \psi \right\rangle = (\stackrel{\frac{\sqrt{15}}{4}}{\frac{i}{4}} )$ (don't know how to write a column vector with this math editor, sorry).

Therefore, $\left\langle x_{+} | \psi \right\rangle = (\frac{1}{\sqrt{2}})(\frac{\sqrt{15}}{4}) + (\frac{1}{\sqrt{2}})(\frac{i}{4}) = \frac{\sqrt{15}}{4 \sqrt{2}} + \frac{i}{4 \sqrt{2}}$

Then multiply $\frac{\sqrt{15}}{4 \sqrt{2}} + \frac{i}{4 \sqrt{2}}$ by its complex conjugate and get: 15/32 + 1/32 = 0.5?

How's that look?

Looks correct! Good work.

Thanks so much! I really appreciate all the help.

gabbagabbahey
Homework Helper
Gold Member
$\left\langle x_{+} \right| = ( \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}})$ and $\left| \psi \right\rangle = (\stackrel{\frac{\sqrt{15}}{4}}{\frac{i}{4}} )$ (don't know how to write a column vector with this math editor, sorry).

Therefore, $\left\langle x_{+} | \psi \right\rangle = (\frac{1}{\sqrt{2}})(\frac{\sqrt{15}}{4}) + (\frac{1}{\sqrt{2}})(\frac{i}{4}) = \frac{\sqrt{15}}{4 \sqrt{2}} + \frac{i}{4 \sqrt{2}}$

Then multiply $\frac{\sqrt{15}}{4 \sqrt{2}} + \frac{i}{4 \sqrt{2}}$ by its complex conjugate and get: 15/32 + 1/32 = 0.5?

How's that look?
It's not incorrect, but it is unnecessary to convert stuff to column and row vectors, when you could do this problem entirely in bra-ket notation:

$$\langle x_{\pm} | = | x_{\pm} \rangle^{ \dagger } = \left( \frac{|0\rangle \pm |1\rangle}{\sqrt{2}} \right)^{ \dagger } = \frac{|0\rangle^{ \dagger } \pm |1\rangle^{ \dagger }}{\sqrt{2}} = \frac{\langle 0 | \pm \langle 1 |}{\sqrt{2}}$$

Switching between different types of notation in a problem can introduce some difficult to find errors. Consider, for example, what would happen if the basis states $|0\rangle$ and $|1\rangle$ weren't orthogonal (but still spanned the Hilbert space), and you were given the inner products between them. You would get an incorrect result if you naively represented $\alpha|0\rangle + \beta|1\rangle$ by a column vector $\begin{pmatrix} \alpha \\ \beta \end{pmatrix}$, its adjoint by the conjugate transpose of the column vector, and the inner product of the state with itself as the matrix product of the two.

It's not incorrect, but it is unnecessary to convert stuff to column and row vectors, when you could do this problem entirely in bra-ket notation:

$$\langle x_{\pm} | = | x_{\pm} \rangle^{ \dagger } = \left( \frac{|0\rangle \pm |1\rangle}{\sqrt{2}} \right)^{ \dagger } = \frac{|0\rangle^{ \dagger } \pm |1\rangle^{ \dagger }}{\sqrt{2}} = \frac{\langle 0 | \pm \langle 1 |}{\sqrt{2}}$$

Switching between different types of notation in a problem can introduce some difficult to find errors. Consider, for example, what would happen if the basis states $|0\rangle$ and $|1\rangle$ weren't orthogonal (but still spanned the Hilbert space), and you were given the inner products between them. You would get an incorrect result if you naively represented $\alpha|0\rangle + \beta|1\rangle$ by a column vector $\begin{pmatrix} \alpha \\ \beta \end{pmatrix}$, its adjoint by the conjugate transpose of the column vector, and the inner product of the state with itself as the matrix product of the two.
I don't follow how this expression makes it easier: $$\langle x_{\pm} | = | x_{\pm} \rangle^{ \dagger } = \left( \frac{|0\rangle \pm |1\rangle}{\sqrt{2}} \right)^{ \dagger } = \frac{|0\rangle^{ \dagger } \pm |1\rangle^{ \dagger }}{\sqrt{2}} = \frac{\langle 0 | \pm \langle 1 |}{\sqrt{2}}$$

gabbagabbahey
Homework Helper
Gold Member
I don't follow how this expression makes it easier: $$\langle x_{\pm} | = | x_{\pm} \rangle^{ \dagger } = \left( \frac{|0\rangle \pm |1\rangle}{\sqrt{2}} \right)^{ \dagger } = \frac{|0\rangle^{ \dagger } \pm |1\rangle^{ \dagger }}{\sqrt{2}} = \frac{\langle 0 | \pm \langle 1 |}{\sqrt{2}}$$
Let's say you have something like $\langle v | = v_0 \langle 0 | + v_1 \langle 1 |$ and $| u \rangle = u_0 | 0 \rangle + u_1 | 1 \rangle$, then you can use the distributive and asscoiative properties of bras and kets to easily compute the inner product $\langle v | u \rangle$ as follows:

$$\langle v | u \rangle = \left( v_0 \langle 0 | + v_1 \langle 1 | \right) \left( u_0 | 0 \rangle + u_1 | 1 \rangle \right) = v_0u_0\langle 0 |0 \rangle + v_0u_1 \langle 0 |1 \rangle + v_1u_0\langle 1 |0 \rangle + v_1u_1\langle 1 |1 \rangle$$

Now, if the basis states $|0 \rangle$ & $|1 \rangle$ are orthonormal, then you have $\langle 0 |1 \rangle = \langle 1 |0 \rangle = 0$ & $\langle 0 |0 \rangle = \langle 1 |1 \rangle = 1$, so $\langle v | u \rangle = v_0u_0 + v_1u_1$, which is the same as what you would get by representing $\langle v |$ as the row vector $\left( v_0, v_1 \right)$ and $| u \rangle$ as the column vector $\begin{pmatrix} u_0 \\ u_1 \end{pmatrix}$ and computing the inner product using the usual rules of matrix multiplication.

However, if the basis states aren't orthonormal, say, for example $\langle 0 |1 \rangle = \langle 1 |0 \rangle = \frac{1}{2}$ and $\langle 0 |0 \rangle = \langle 1 |1 \rangle = 3$, then using bra-ket notation will give you the correct result $\langle v | u \rangle = 3v_0u_0 + \frac{1}{2} \left( v_0u_1 + v_1u_0\right) + 3v_1u_1$, but naively switching to row and column vectors as before will give you the now incorrect result $\langle v | u \rangle = v_0u_0 + v_1u_1$.

Moreover, later in your studies you will likely deal with many-particle systems, where the system states are given by tensor products of the individual component states. When that happens, even if the basis states are orthonormal, trying to keep track of which row vectors act on which column vectors will become much more difficult than simply sticking to bra-ket notation.